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Last modified on August 3rd, 2023

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Triangle sum theorem, what is the triangle sum theorem.

The triangle sum theorem states that the sum of the three interior angles in a triangle adds up to 180°. It is also called the angle sum theorem.

Given below is a triangle ABC, having three interior angles ∠a, ∠b, and ∠c. According to the triangle sum theorem, ∠a + ∠b + ∠c = 180°

Prove Triangle Sum Theorem

∠CBA + ∠BAC + ∠ACB = 180 °

A line PQ is drawn parallel to BC passing through the point A

Hence proved that, the sum of the three interior angles in a triangle adds up to 180°.

Solved Examples

Let us solve some problems to understand the theorem better.

As we know, according to the triangle sum theorem, x + 38° + 32° = 180° => x = 180° – (38° + 32°) => x = 110°

Two interior angles of a triangle measure 30 ° and 80 ° . What is the third interior angle of the triangle?

Let the third interior angle be x As we know, according to the triangle sum theorem, x + 30° + 80° = 180° x = 180° – (30° + 80°) x = 70°

As we know, according to the triangle sum theorem, (8x – 1)° + (4x + 8)° + (3x + 8)° = 180° 15x + 15 = 180° 15x = 165 x = 11°

Corollary to the Triangle Sum Theorem

Two corollaries to the triangle sum theorem are:

Corollary 1 : The acute angles of a right triangle are complementary (add up to 90°)

Hypothesis: From the triangle sum theorem, the sum of all three angles equals 180°

Again, from the definition of a right triangle, we have one of its angles to be a right angle, making the remaining angles to be acute whose sum equals (180° – 90°) is 90°

Conclusion: The acute angles of a right triangle are complementary

Corollary 2 : Each angle in an equilateral triangle measures 60°

Again, from the definition of an equilateral triangle, all angles are of equal measure. Adding up all the angles, we get,

⇒ x + x + x = 180°

⇒ 3x = 180°

Conclusion: Each angle in an equilateral triangle measures 60°

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Explanation:

The precise statement of the conjecture is:.

Conjecture ( Triangle Sum ): The sum of the interior angles in any triangle is 180 degrees.

Interactive Sketch Pad Demonstration:

Linked activity:.

Angle Sum Property

The angle sum property of a triangle states that the sum of the angles of a triangle is equal to 180º. A triangle has three sides and three angles, one at each vertex. Whether a triangle is an acute, obtuse, or a right triangle, the sum of its interior angles is always 180º.

The angle sum property of a triangle is one of the most frequently used properties in geometry. This property is mostly used to calculate the unknown angles.

What is the Angle Sum Property?

According to the angle sum property of a triangle , the sum of all three interior angles of a triangle is 180 degrees. A triangle is a closed figure formed by three line segments, consisting of interior as well as exterior angles. The angle sum property is used to find the measure of an unknown interior angle when the values of the other two angles are known . Observe the following figure to understand the property.

Angle Sum Property Formula

The angle sum property formula for any polygon is expressed as, S = (n − 2) × 180°, where 'n' represents the number of sides in the polygon. This property of a polygon states that the sum of the interior angles in a polygon can be found with the help of the number of triangles that can be formed inside it. These triangles are formed by drawing diagonals from a single vertex. However, to make things easier, this can be calculated by a simple formula, which says that if a polygon has 'n' sides, there will be (n - 2) triangles inside it. For example, let us take a decagon that has 10 sides and apply the formula. We get, S = (n − 2) × 180°, S = (10 − 2) × 180° = 10 × 180° = 1800°. Therefore, according to the angle sum property of a decagon , the sum of its interior angles is always 1800°. Similarly, the same formula can be applied to other polygons. The angle sum property is mostly used to find the unknown angles of a polygon.

Proof of the Angle Sum Property

Let's have a look at the proof of the angle sum property of the triangle. The steps for proving the angle sum property of a triangle are listed below:

• Step 1: Draw a line PQ that passes through the vertex A and is parallel to side BC of the triangle ABC.
• Step 2: We know that the sum of the angles on a straight line is equal to 180°. In other words, ∠PAB + ∠BAC + ∠QAC = 180°, which gives, Equation 1: ∠PAB + ∠BAC + ∠QAC = 180°
• Step 3: Now, since line PQ is parallel to BC. ∠PAB = ∠ABC and ∠QAC = ∠ACB. (Interior alternate angles), which gives, Equation 2: ∠PAB = ∠ABC, and Equation 3: ∠QAC = ∠ACB
• Step 4: Substitute ∠PAB and ∠QAC with ∠ABC and ∠ACB respectively, in Equation 1 as shown below.

Equation 1: ∠PAB + ∠BAC + ∠QAC = 180°. Thus we get, ∠ABC + ∠BAC + ∠ACB = 180°

Hence proved, in triangle ABC, ∠ABC + ∠BAC + ∠ACB = 180°. Thus, the sum of the interior angles of a triangle is equal to 180°.

Important Points

The following points should be remembered while solving questions related to the angle sum property.

• The angle sum property formula for any polygon is expressed as, S = ( n − 2) × 180°, where 'n' represents the number of sides in the polygon.
• The angle sum property of a polygon states that the sum of the interior angles in a polygon can be found with the help of the number of triangles that can be formed inside it.
• The sum of the interior angles of a triangle is always 180°.

Impoprtant Topics

• Exterior Angle Theorem

Solved Examples

Example 1: Sam needs to find the measure of the third angle of a triangle ABC in which ∠ABC = 45° and ∠ACB = 55°. Can you help him?

We know that ∠ABC = 45° and ∠ACB = 55°. Using the Angle Sum Property of a triangle, ∠A + ∠B + ∠C = 180, ∠A + 45 + 55° = 180°, ∠A + 100° = 180°, ∠A = 180° -100°, ∠A = 80°. Therefore, the third angle: ∠A = 80°

Example 2: If the angles of a triangle are in the ratio 3:4:5, determine the value of the three angles.

Let the angles be 3x, 4x and 5x. According to the Angle Sum Property of a triangle, 3x + 4x + 5x = 180°, 12x = 180, x = 15. Thus, the three angles will be: 3x = 3 × 15 = 45°, 4x = 4 × 15 = 60°, 5x = 5 × 15 = 75°. Therefore, the three angles are: 45°, 60°, 75°.

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Practice Questions

Faqs on angle sum property, what is the angle sum property of a polygon.

The angle sum property of a polygon states that the sum of all the angles in a polygon can be found with the help of the number of triangles that can be formed in it. These triangles are formed by drawing diagonals from a single vertex. However, this can be calculated by a simple formula, which says that if a polygon has 'n' sides, there will be (n - 2) triangles inside it. The sum of the interior angles of a polygon can be calculated with the formula: S = (n − 2) × 180°, where 'n' represents the number of sides in the polygon. For example, if we take a quadrilateral and apply the formula using n = 4, we get: S = (n − 2) × 180°, S = (4 − 2) × 180° = 2 × 180° = 360°. Therefore, according to the angle sum property of a quadrilateral, the sum of its interior angles is always 360°. Similarly, the same formula can be applied to other polygons . The angle sum property is mostly used to find the unknown angles of a polygon.

What is the Angle Sum Property of a Triangle?

The angle sum property of a triangle says that the sum of its interior angles is equal to 180°. Whether a triangle is an acute, obtuse, or a right triangle, the sum of the angles will always be 180°. This can be represented as follows: In a triangle ABC, ∠A + ∠B + ∠C = 180°.

What is the Angle Sum Property of a Hexagon?

According to the angle sum property of a hexagon , the sum of all the interior angles of a hexagon is 720°. In order to find the sum of the interior angles of a hexagon, we multiply the number of triangles in it by 180°. This is expressed by the formula: S = (n − 2) × 180°, where 'n' represents the number of sides in the polygon. In this case, 'n' = 6. Therefore, the sum of the interior angles of a hexagon = S = (n − 2) × 180° = (6 − 2) × 180° = 4 × 180° = 720°.

What is the Angle Sum Property of a Quadrilateral?

According to the angle sum property of a quadrilateral , the sum of all its four interior angles is 360°. This can be calculated by the formula, S = (n − 2) × 180°, where 'n' represents the number of sides in the polygon. In this case, 'n' = 4. Therefore, the sum of the interior angles of a quadrilateral = S = (4 − 2) × 180° = (4 − 2) × 180° = 2 × 180° = 360°.

What is the Exterior Angle Sum Property of a Triangle?

The exterior angle theorem says that the measure of each exterior angle of a triangle is equal to the sum of the opposite and non-adjacent interior angles.

What is the Formula of Angle Sum Property?

The formula for the angle sum property is, S = ( n − 2) × 180°, where 'n' represents the number of sides in the polygon. For example, if we want to find the sum of the interior angles of an octagon , in this case, 'n' = 8. Therefore, we will substitute the value of 'n' in the formula, and the sum of the interior angles of an octagon = S = (n − 2) × 180° = (8 − 2) × 180° = 6 × 180° = 1080°.

What is the Angle Sum Property of a Pentagon?

As per the angle sum property of a pentagon , the sum of all the interior angles of a pentagon is 540°. In order to find the sum of the interior angles of a pentagon, we multiply the number of triangles in it by 180°. This is expressed by the formula: S = (n − 2) × 180°, where 'n' represents the number of sides in the polygon. In this case, 'n' = 5. Therefore, the sum of the interior angles of a pentagon = S = (n − 2) × 180° = (5 − 2) × 180° = 3 × 180° = 540°.

How to Find the Third Angle in a Triangle?

We know that the sum of the angles of a triangle is always 180°. Therefore, if we know the two angles of a triangle, and we need to find its third angle, we use the angle sum property. We add the two known angles and subtract their sum from 180° to get the measure of the third angle. For example, if two angles of a triangle are 70° and 60°, we will add these, 70 + 60 = 130°, and we will subtract it from 180°, which is the sum of the angles of a triangle. So, the third angle = 180° - 130° = 50°.

How to Find the Exterior Angle of a Polygon?

The exterior angle of a polygon is the angle formed between any side of a polygon and a line that is extended from the adjacent side. In order to find the measure of an exterior angle of a regular polygon, we divide 360 by the number of sides 'n' of the given polygon. For example, in a regular hexagon, where 'n' = 6, each exterior angle will be 60° because 360 ÷ n = 360 ÷ 6 = 60°. It should be noted that the corresponding interior and exterior angles are supplementary and the exterior angles of a regular polygon are equal in measure.

The Triangle Sum Theorem

Old and new proofs concerning the sum of interior angles of a triangle. (more on the hidden depths of triangle qualia.) aaron sloman http://www.cs.bham.ac.uk/~axs.

Triangle Sum Theorem (TST): The interior angles of a triangle add up to a straight line, or half a rotation (180 degrees).

Definition: Two straight lines L1 and L2 are parallel if and only if they are co-planar and have no point in common, no matter how far they are extended. Postulate: Given a straight line L in a plane, and a point P in the plane not on L, there is exactly one line through P that is in the plane and parallel to L. (That was not Euclid's formulation, but is perhaps intuitively the clearest formulation.)

COR: Corresponding angles are equal: If two lines L1, L2 are parallel and a third line L3 is drawn from any point P1 on L1 to a point P2 on L2 and continued beyond P2, then the angle that L1 makes with the line L3 at point P1, and the angle L2 makes with the line L3 at point P2 (where the angles are on the same side of both lines) are equal. ALT: Alternate angles are equal: If two lines L1, L2 are parallel and a third line L3 is drawn from any point P1 on L1 to a point P2 on L2, then the angle L1 makes with the line L3 at point P1, and the angle L2 makes with the line L3 at point P2 (on the opposite sides of both lines) are equal. For more on transversals and relations between the angles they create, see http://www.mathsisfun.com/geometry/parallel-lines.html That page teaches concepts with some interactive illustrations, but presents no proofs. Standard Euclidean proofs of COR and ALT are presented here: https://proofwiki.org/wiki/Parallelism_implies_Equal_Corresponding_Angles https://proofwiki.org/wiki/Parallelism_implies_Equal_Alternate_Interior_Angles

The "standard" proofs of the "Triangle Sum Theorem"

Mary pardoe's proof of the triangle sum theorem.

Note: In the original publication reporting this proof I mistakenly referred to the author as Mary Ensor, her name as a student. I think she was already Mary Pardoe at the time she visited me.

http://www.cs.bham.ac.uk/research/projects/cogaff/misc/p-geometry.html presents a more detailed, but still incomplete, discussion, of the geometrical prerequisites for some of the above reasoning. It introduces the idea of P-geometry, which is intended to be Euclidean geometry without the Axiom of Parallels (Euclid's Axiom 5), but with time and motion added, including translation and rotation of rigid line-segments.

Is the Pardoe proof valid?

Aaron Sloman, 2008, Kantian Philosophy of Mathematics and Young Robots, in Intelligent Computer Mathematics , Eds. Autexier, S., Campbell, J., Rubio, J., Sorge, V., Suzuki, M., and Wiedijk, F., LLNCS no 5144, pp. 558-573, Springer, http://www.cs.bham.ac.uk/research/projects/cosy/papers#tr0802 (This paper referred to Mary Ensor.) Aaron Sloman, 2010, If learning maths requires a teacher, where did the first teachers come from?, In Proceedings Symposium on Mathematical Practice and Cognition, AISB 2010 Convention, De Montfort University, Leicester http://www.cs.bham.ac.uk/research/projects/cogaff/10.html#1001 And in talks on mathematical cognition and philosophy of mathematics here: http://www.cs.bham.ac.uk/research/projects/cogaff/talks/

NOTE ADDED 6 Oct 2012 (Asperti and Scott):

Andrea Asperti, Proof, Message and Certificate, in AISC/MKM/Calculemus , 2012, pp. 17--31, Online: http://www.cs.unibo.it/~asperti/PAPERS/proofs.pdf http://dx.doi.org/10.1007/978-3-642-31374-5_2

http://www.cs.unibo.it/~asperti/SLIDES/message.pdf

"The proof is fine and really is the same as the classical proof. To see this, translate (by parallel translation) all the three angles of the triangle up to the line through the top vertex of the triangle parallel to the lower side."

"I should have commented in my explanation of the proof that if you translate the line on which the base of the triangle sits along each of the sides up to the vertex, then both actions result in the same line - the unique parallel."

If this "proof" is taught to students as a full, valid proof, then I do not see how the teacher will be able to explain to those students where the hell Euclid's fifth postulate (or the parallels axiom) is used here, or even what is the connection between the theorem and parallel lines.

Is it really the same as the classical proof?

What is the cognitive function of a mathematical proof added 16 aug 2018, more on p-geometry, an earlier discovery of mary pardoe's proof added 5 apr 2018.

The proof of the angle sum of a triangle that you attribute to Mary Pardoe was first published by Bernhard Friedrich Thibaut (1775-1832) in the second edition of his Grundriss der reinen Mathematik , published in Goettingen by Vandenhoek und Ruprecht in 1809 (see page 363). It is not valid without assuming an equivalent of the parallel postulate. In Euclidean geometry, the composition of three rotations by (directed) angles adding up to an integer multiple of a full turn is a translation; but this fails to be true without the parallel postulate. Thibaut had put forward the proof as part of an attempted proof of the parallel postulate; his attempted proof is discussed in Roberto Bonola's Non-Euclidean geometry: a critical and historical study of its development (page 63), in William Barrett Frankland's Theories of parallelism: an historical critique (page 37), and in Jean-Claude Pont's L'aventure des paralleles histoire de la geometrie non-Euclidenne: Precurseurs et attardes (pages 240-244). Thus the proof is only valid for plane geometry where the plane is assumed to have the properties that it does in Euclid's Elements ; it does not hold for the hyperbolic plane of Bolyai and Lobachevsky (which satisfies all those properties bar the parallel postulate). (This is likely why the objection about the surface of a sphere was raised to you.) The objection that the surface of a sphere provides a counterexample is also over a century old, going back to Olaus Henrici's criticism of Thibaut's proof in "The axioms of geometry", published in Nature , Volume 29, 1884, pp.453-454 and 573.

Added 9 Feb 2013: Another proof of the sum theorem, by Kay Hughes Modified 4 Mar 2013:

Note : There is a related "visual" proof posted here https://twitter.com/thingswork/status/1121857148068065280 (drawn to my attention by Ron Chrisley), based on what happens if a polygon shrinks to a point. This version applies only to polygons and does does not generalise (smoothly) into a proof that a tangent arrow moving around any simple closed curve, back to its starting point must have a resultant rotation of 360 degrees.

A + a = 180 therefore a = 180 - A B + b = 180 therefore b = 180 - A C + c = 180 therefore c = 180 - A

Nerlich on geometry and metaphysics Added 22 Jul 2018

Graham Nerlich, 1991, How Euclidean Geometry Has Misled Metaphysics, The Journal of Philosophy, 88, 4, Apr, 1991 pp. 169--189, http://www.jstor.org/stable/2026946

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The sum of the interior angles of any triangle

I can demonstrate and prove that in a triangle, the sum of the interior angles is 180°.

Lesson details

Key learning points.

• By considering a number of different triangles you can demonstrate facts about the angles in triangles.
• The interior angles of any triangle sum to 180°
• The angle sum of triangles can be proved using angles in parallel lines.

Common misconception

Pupils may struggle with mathematical proof, especially using other knowledge within it.

Explain to pupils that there are many different styles of mathematical proof but all are showing that a particular fact holds true for all.

Alternate angles - a pair of angles both between or both outside two line segments that are on opposite sides of the transversal that cuts them.

Corresponding angles - Corresponding angles are a pair of angles at different vertices on the same side of a transversal in equivalent positions.

Co-interior angles - Co-interior angles are on the same side of the transversal line and in between the two other lines.

This content is © Oak National Academy Limited ( 2024 ), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).

Starter quiz

6 questions.

scalene triangle -  

All three edges and angles are different to each other.

isosceles triangle -  

At least two edges and two angles are equal to each other.

equilateral triangle -  

All three edges and angles are equal to each other.

right-angled triangle -  

One of the angles is 90°.

∠OAB is equal to -  

∠ABC as they are equal alternate angles

∠DAC is equal to -  

∠ACB as they are equal alternate angles

∠OAB + ∠BAC + ∠DAC = 180° -  

as angles on a line at a point sum to 180°

∠ABC + ∠BAC + ∠ACB = 180° -  

as angles in a triangle sum to 180°

Triangle Sum Theorem

In these lessons, we will learn

• the Triangle Sum Theorem (with worksheets)
• how to prove the Triangle Sum Theorem
• how to solve problems using the Triangle Sum Theorem

Related Pages Angles In A Triangle More Geometry Lessons Geometry Worksheets

The following diagram shows the Triangle Sum Theorem. Scroll down the page for examples and solutions using the Triangle Sum Theorem.

The Triangle Sum Theorem states that

The sum of the three interior angles in a triangle is always 180°.

The Triangle Sum Theorem is also called the Triangle Angle Sum Theorem or Angle Sum Theorem.

Example: Find the value of x in the following triangle.

Solution: x + 24° + 32° = 180° (sum of angles is 180°) x + 56° = 180° x = 180° – 56° = 124°

Worksheet 1 , Worksheet 2 using Triangle Sum Theorem

Proof of the Triangle Sum Theorem

How to prove that the sum of the angles of a triangle is 180 degrees (Triangle Sum Theorem)? This is a two-column proof that involves parallel lines and the alternate angle theorem.

How to prove the Triangle Sum Theorem using right triangles? We will start with right triangles, and then expand our proof later to include all triangles.

Solving problems using the Triangle Sum Theorem

The following videos show more examples of how to solve problems related to the triangle sum theorem using Algebra.

How to Find the Missing Angle in a Triangle Using the Triangle Sum Theorem? Step 1. Write out the equation by adding all the angles and making them equal to 180° Step 2. Solve for x. Step 3: Substitute to find the missing angles.

Use the triangle sum theorem to find the base angle measures given the vertex angle in an isosceles triangle

Solve for angles using the Triangle Sum Theorem

Example: Using the diagram shown, find the value of x and the measure of each missing angle in the triangle.

Triangle Properties

Example: Find the values of x and y for a given triangle problem. This problem uses the Triangle Sum Theorem and the Corollary to the Triangle Sum Theorem.

Example: Find the values of x and y for a given triangle problem. This problem involves linear pair angles as well as the Triangle Sum Theorem.

Solve for angles using the Triangle Sum Theorem and ratios

Example: The measures of the angles of a triangle are in the ratio 2:5:8. Find the measure of each angle.

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Lesson Explainer: Interior Angles of a Polygon Mathematics • First Year of Preparatory School

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In this explainer, we will learn how to find the sum of the measures of the interior angles of a polygon given the number of its sides and the measure of an angle in a regular polygon.

Definition: Polygons

A polygon is a simple two-dimensional closed shape made up of straight line segments called sides. Each point where two sides of a polygon meet is called a vertex (the plural is “vertices”).

The number of sides and the number of interior angles in a polygon are equal, and this number is generally used to classify the shape.

We may already be familiar with the names of simple polygons. For example, a three-sided polygon is called a triangle. The table below shows the general names for 𝑛 -sided polygons for 2 𝑛 ≤ 1 0 .

We may note that a polygon with less than three sides cannot be formed, since this shape would require curved line segments!

Before we can work through some examples, we need a few more definitions to help us describe and classify different polygons.

Definition: Interior and Exterior Angles

An interior angle is an angle inside a polygon at one of its vertices. An exterior angle is an angle outside the polygon; it is formed by a side and the extension of an adjacent side.

At any given vertex, the measures of the interior and exterior angles sum to 1 8 0 ∘ .

Definition: Convex and Concave Polygons

A convex polygon is a polygon with all interior angles measuring less than 1 8 0 ∘ .

A straight line drawn through a convex polygon will intersect its sides exactly twice.

A concave polygon is a polygon with one or more interior angles measuring more than 1 8 0 ∘ .

A straight line drawn through a concave polygon will intersect its sides more than twice.

Let us start with a simple example.

Example 1: Identifying a Polygon as Concave or Convex

Is this polygon concave or convex?

Recall that a convex polygon is a polygon with all interior angles measuring less than 1 8 0 ∘ , while a concave polygon is a polygon with one or more interior angles measuring more than 1 8 0 ∘ .

From the diagram, we see that the interior angles of this polygon have measures that range from 9 0 ∘ up to 1 4 0 ∘ . Therefore, all the interior angles measure less than 1 8 0 ∘ , so we conclude that this polygon is convex.

For the remainder of this explainer, we will be working with convex polygons.

Our next step is to derive a formula for 𝑆  , the sum of the interior angle measures of a polygon with 𝑛 sides.

Let us start with the simplest polygon, which is a triangle, so 𝑛 = 3 . We should be familiar with the fact that the sum of the measures of the interior angles in a triangle is always 1 8 0 ∘ .

For polygons with more than three sides, to find the sum of the interior angles, we split the polygon into triangles as shown below.

We know that the measures of the interior angles in each of the separate triangles must sum to 1 8 0 ∘ .

In addition, the number of triangles within an 𝑛 -sided polygon is 2 less than its number of sides, that is, 𝑛 − 2 . We can see that this is true for the polygon above, which has 5 sides and 5 − 2 = 3 triangles. It is easy to draw further examples to check this.

Therefore, the sum of the interior angles of a polygon with 𝑛 sides will be equal to ( 𝑛 − 2 ) 1 8 0 .

Formula: The Sum of the Measures of the Interior Angles of a Polygon

The sum, 𝑆  , of the measures of the interior angles of a polygon with 𝑛 sides is given by the formula 𝑆 = ( 𝑛 − 2 ) 1 8 0 . 

Let us now look at some examples of applying this formula.

Example 2: Finding the Sum of the Interior Angles of a Hexagon

What is the sum of the interior angles of a hexagon?

Recall that 𝑆  , the sum of the interior angle measures of a polygon with 𝑛 sides, is given by the formula 𝑆 = ( 𝑛 − 2 ) 1 8 0 . 

A hexagon is a polygon with six sides and six vertices, so in this case, we have 𝑛 = 6 .

Hence, we can substitute this value into the formula and simplify to find the sum of the interior angle measures: 𝑆 = ( 6 − 2 ) × 1 8 0 = 4 × 1 8 0 = 7 2 0 .  ∘

We conclude that the sum of the interior angles of a hexagon is 7 2 0 ∘ .

As the formula enables us to calculate the sum of the interior angle measures of an 𝑛 -sided polygon, this means that if we are given an 𝑛 -sided polygon and 𝑛 − 1 of its interior angle measures, we can always work backward from the formula to find the missing angle. Here is an example.

Example 3: Finding the Measure of an Angle of a Pentagon given the Measures of the Other Angles

In the figure, if 𝑚 ∠ 𝐴 = 1 3 7 ∘ , 𝑚 ∠ 𝐵 = 7 8 ∘ , 𝑚 ∠ 𝐶 = 1 1 3 ∘ , and 𝑚 ∠ 𝐸 = 1 3 1 ∘ , find 𝑚 ∠ 𝐷 .

The diagram above shows a pentagon with the measures of four of the five angles given. We know that the sum of the interior angle measures for this shape, 𝑆  , can be expressed as follows: 𝑆 = 𝑚 ∠ 𝐴 + 𝑚 ∠ 𝐵 + 𝑚 ∠ 𝐶 + 𝑚 ∠ 𝐷 + 𝑚 ∠ 𝐸 . 

Using information from the question, we can substitute for all of the interior angle measures except 𝑚 ∠ 𝐷 to give 𝑆 = 1 3 7 + 7 8 + 1 1 3 + 𝑚 ∠ 𝐷 + 1 3 1 = 4 5 9 + + 𝑚 ∠ 𝐷 .  ∘ ∘ ∘ ∘ ∘

A pentagon is a polygon with five sides and five vertices, and we can therefore substitute 𝑛 = 5 into the formula: 𝑆 = ( 5 − 2 ) × 1 8 0 = 3 × 1 8 0 = 5 4 0  ∘

We have now obtained two separate equations for 𝑆  and can equate them to get 4 5 9 + 𝑚 ∠ 𝐷 = 5 4 0 . ∘ ∘

Finally, we solve for 𝑚 ∠ 𝐷 by subtracting 459 from both sides, so 𝑚 ∠ 𝐷 = 5 4 0 − 4 5 9 = 8 1 . ∘ ∘ ∘

We have found that 𝑚 ∠ 𝐷 = 8 1 ∘ .

Now that we know how to find the sum of the interior angle measures of a polygon, let us see how this technique can be applied to regular polygons to find the individual angle values.

Definition: Regular and Irregular Polygons

A polygon is considered to be regular when all its angles are of equal measure and all its sides are equal in length. In any other case, the polygon is considered to be irregular.

The diagram below shows a regular hexagon in comparison to an irregular hexagon ( 𝑛 = 6 ) .

For a regular polygon, we can find the measure of each interior angle, 𝐴  , by dividing the sum of the interior angles, 𝑆  , by the number of angles: 𝐴 = 𝑆 𝑛 .  

Note that this is only true for a regular or equiangular polygon, since all 𝑛 angles are equal in measure.

We have already shown that the sum of the measures of the interior angles for any 𝑛 -sided polygon can be found using the formula 𝑆 = ( 𝑛 − 2 ) 1 8 0 . 

We can therefore substitute for 𝑆  to find the formula for the measure of each interior angle in a regular polygon (in terms of 𝑛 ): 𝐴 = ( 𝑛 − 2 ) 1 8 0 𝑛 . 

Formula: Interior Angles of a Regular Polygon

• The measure, 𝐴  , of each interior angle of a regular 𝑛 -sided polygon is given by the formula 𝐴 = ( 𝑛 − 2 ) 1 8 0 𝑛 . 

It is worth noting that the above relationship implies that the value of 𝐴  is necessarily less than 1 8 0 ∘ for a regular polygon. Any solutions greater than this value should be double-checked, and we should make sure that the shape in question is indeed a regular polygon.

This fact can be seen by imagining a regular arrangement of angle measures greater than 1 8 0 ∘ that form a closed shape. In such a case, the angles would lie outside of the shape and hence would not be classified as the “interior angles”!

For an irregular polygon, interior angle measures may be greater than 1 8 0 ∘ , but this is outside the scope of this explainer.

As a final note, two line segments that form an angle of 1 8 0 ∘ cannot be considered as two sides of a polygon, since they cannot be distinguished from a single segment. For this reason, 1 8 0 ∘ itself is not counted as an angle in this situation.

Let us now look at a question that uses this formula for the interior angles of a regular polygon.

Example 4: Finding the Number of Sides of a Polygon given the Measures of Its Interior Angles

Each interior angle of a regular polygon is 1 7 9 ∘ . How many sides does it have?

Recall that the measure, 𝐴  , of each interior angle of a regular 𝑛 -sided polygon is given by the formula 𝐴 = ( 𝑛 − 2 ) 1 8 0 𝑛 . 

For this question, we are given the measure of each interior angle of a regular polygon ( 𝐴 )  and are asked to find the number of sides and, therefore, the number of angles ( 𝑛 ) .

In order to find 𝑛 , we can substitute the known value 𝐴 = 1 7 9  into the formula for each interior angle in a regular polygon, giving 1 7 9 = ( 𝑛 − 2 ) 1 8 0 𝑛 .

We can now solve this equation by first multiplying both sides by 𝑛 and then grouping and simplifying the terms as follows: 1 7 9 𝑛 = ( 𝑛 − 2 ) 1 8 0 1 7 9 𝑛 = 1 8 0 𝑛 − 3 6 0 .

Adding 360 to both sides, we get 1 7 9 𝑛 + 3 6 0 = 1 8 0 𝑛 , and subtracting 1 7 9 𝑛 from both sides gives 3 6 0 = 𝑛 , which is the same as 𝑛 = 3 6 0 .

We have found that the number of sides (and angles) in this shape is 360.

Even if the above question had not mentioned the regularity of the polygon, we could have still applied the same formula. This is because the measure of each interior angle of a regular 𝑛 -sided polygon is the same as the measure of each interior angle of an equiangular 𝑛 -sided polygon. However, it is unusual to meet examples of this type.

Next, we look at the exterior angles. An interesting property worth noting is that the exterior angle measures of a polygon sum to 3 6 0 ∘ .

To show this, we give a sketch for the simplest polygon, which is a triangle ( 𝑛 = 3 ) . We can imagine a general triangle with interior angle measures 𝑎  , 𝑎  , and 𝑎  . By extending each side beyond the vertices, we can form the exterior angles with measures 𝑏  , 𝑏  , and 𝑏  .

Now, consider an arrow at point 𝑃 pointing in the direction parallel to the base of the triangle. Let us imagine this arrow traveling clockwise around the perimeter of the triangle, turning through each of the exterior angles. Upon arriving back at point 𝑃 , the arrow will have completed one complete turn of 3 6 0 ∘ , so 𝑏 + 𝑏 + 𝑏 = 3 6 0 .    ∘

This property of exterior angles is particularly useful when answering questions about regular polygons, as we will see in our next example.

Example 5: Finding the Number of Sides of a Regular Polygon given an Exterior Angle

If a regular polygon has an exterior angle of 9 0 ∘ , find the number of sides it has.

Recall that the measures of the exterior angles of a polygon sum to 3 6 0 ∘ .

In a regular 𝑛 -sided convex polygon, there will be 𝑛 exterior angles, all of which must have the same measure. Therefore, we deduce that 𝑛 × = 3 6 0 . e x t e r i o r a n g l e

As we are told that the regular polygon has an exterior angle measure of 9 0 ∘ , then substituting this value, we get 𝑛 × 9 0 = 3 6 0 .

Finally, dividing both sides by 90 gives 𝑛 = 4 .

Notice that we could have worked out our answer using an alternative method, as follows. Recall that at any given vertex, the measures of the interior and exterior angles sum to 1 8 0 ∘ ; that is, i n t e r i o r a n g l e e x t e r i o r a n g l e + = 1 8 0 .

Substituting the value of the exterior angle and then rearranging, we get i n t e r i o r a n g l e i n t e r i o r a n g l e + 9 0 = 1 8 0 = 1 8 0 − 9 0 = 9 0 .

Now, we know that the measure of each interior angle of a regular 𝑛 -sided polygon is given by the expression ( 𝑛 − 2 ) 1 8 0 𝑛 , so we have 9 0 = ( 𝑛 − 2 ) 1 8 0 𝑛 .

Multiplying both sides by 𝑛 gives 9 0 𝑛 = ( 𝑛 − 2 ) 1 8 0 , and distributing over the parentheses on the right-hand side, we get 9 0 𝑛 = 1 8 0 𝑛 − 3 6 0 .

Then, subtracting 1 8 0 𝑛 from both sides and dividing through by − 9 0 , we obtain 𝑛 = 4 , as before.

We conclude that the regular polygon has 4 sides, which means it must be a square.

Next, we will show how to find the measure of an exterior angle of a regular polygon given the number of sides it has.

Example 6: Finding the Measures of the Interior and Exterior Angles of a Regular Polygon

Find 𝑥 and 𝑦 .

The diagram shows a regular 8-sided polygon (i.e., an octagon) with an exterior angle measure of 𝑥 and an interior angle measure of 𝑦 .

Starting with the exterior angles of a regular octagon, there are 8 of these, all of which must have the same measure, 𝑥 . This implies that 8 𝑥 = 3 6 0 , and dividing both sides by 8 gives 𝑥 = 4 5 ∘ .

Recall also that at any given vertex, the measures of the interior and exterior angles sum to 1 8 0 ∘ . Therefore, in this case, we must have 𝑥 + 𝑦 = 1 8 0 .

Substituting 𝑥 = 4 5 from above and rearranging, we get that the interior angle measure is 𝑦 = 1 8 0 − 4 5 = 1 3 5 . ∘

We have found that 𝑥 = 4 5 ∘ and 𝑦 = 1 3 5 ∘ .

In our final example, we will need to work back from the sum of the measures of the interior angles in a polygon to find the number of sides.

Example 7: Finding the Number of Sides of a Polygon given the Measures of Its Interior Angles

If the measures of two interior angles of a polygon are 1 2 0 ∘ and 4 0 ∘ and the sum of the rest of the angles is 3 8 0 ∘ , find the number of sides.

Recall that the sum, 𝑆  , of the interior angle measures of a polygon with 𝑛 sides is given by the formula 𝑆 = ( 𝑛 − 2 ) 1 8 0 . 

Here, we are given the measures of two of the interior angles, together with the sum of the rest. Therefore, we can sum these angles to give 𝑆  , where 𝑛 is a number yet to be determined: 𝑆 = 1 2 0 + 4 0 + 3 8 0 = 5 4 0 . 

Now that we know the value of 𝑆  , we can substitute it into the formula to get 5 4 0 = ( 𝑛 − 2 ) 1 8 0 .

Lastly, we solve this equation for 𝑛 . Distributing over the parentheses on the right-hand side, we have 5 4 0 = 1 8 0 𝑛 − 3 6 0 . and adding 360 to both sides gives 9 0 0 = 1 8 0 𝑛 .

Dividing through by 180, we get 𝑛 = 5 .

Thus, we have found that the polygon has 5 sides, which means it is a pentagon.

Let us finish by recapping some key concepts from this explainer.

• A convex polygon is a polygon with all interior angles measuring less than 1 8 0 ∘ . A concave polygon is a polygon with one or more interior angles measuring more than 1 8 0 ∘ .
• The sum, 𝑆  , of the interior angle measures of a polygon with 𝑛 sides is given by the formula 𝑆 = ( 𝑛 − 2 ) 1 8 0 . 
• A polygon is considered to be regular when all its angles have equal measures and all its sides are equal in length. In any other case, the polygon is considered to be irregular.
• The interior angle measures of a regular polygon will always be less than 1 8 0 ∘ , whereas the interior angle measures of an irregular polygon may be greater than 1 8 0 ∘ .
• The sum of the exterior angle measures of a polygon is 3 6 0 ∘ .

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Interior Angle Formula

Interior angle formula (definition, examples, sum of interior angles).

If you take a look at other geometry lessons on this helpful site, you will see that we have been careful to mention interior angles, not just angles, when discussing polygons. Every polygon has interior angles and exterior angles, but the interior angles are where all the interesting action is.

Interior angle formula

From the simplest polygon, a triangle, to the infinitely complex polygon with  n  sides, sides of polygons close in a space. Every intersection of sides creates a vertex, and that vertex has an interior and exterior angle.  Interior angles of polygons  are within the polygon.

Though Euclid did offer an exterior angles theorem specific to triangles, no Interior Angle Theorem exists. Instead, you can use a formula that mathematically describes an interesting pattern about polygons and their interior angles.

Sum of interior angles formula

This formula allows you to mathematically divide any polygon into its minimum number of triangles. Since every triangle has interior angles measuring  180° , multiplying the number of dividing triangles times  180°  gives you the sum of the interior angles.

S = sum of interior angles

n = number of sides of the polygon

Try the formula on a triangle:

Well, that worked, but what about a more complicated shape, like a dodecagon? It has 12 sides, so:

How do you know that is correct?

Take any dodecagon and pick one vertex. Connect every other vertex to that one with a straightedge, dividing the space into 10 triangles. Ten triangles, each  180° , makes a total of  1,800° !

Finding an unknown interior angle

The same formula can help you find a missing interior angle of a polygon. Here is a wacky pentagon, with no two sides equal:

The formula tells us that a pentagon, no matter its shape, must have interior angles adding to  540° :

So subtracting the four known angles from  540°  will leave you with the missing angle:

The unknown angle is  100° .

Finding interior angles of regular polygons

Once you know how to find the sum of interior angles of a polygon, finding one interior angle for any regular polygon is just a matter of dividing.

Where  S =the sum of the interior angles and  n =the number of congruent sides of a regular polygon , the formula is:

Here is an octagon (eight sides, eight interior angles).  First, use the formula for finding the sum of interior angles:

Next, divide that sum by the number of sides:

measure of each interior angle = S n \frac{S}{n} n S ​

measure of each interior angle = 1 , 080 ° 8 \frac{1,080°}{8} 8 1 , 080° ​

measure of each interior angle = 135 ° 135° 135°

Each interior angle of a regular octagon is   =  135° .

Finding the number of sides of a polygon

You can use the same formula,  S=(n-2)×180° , to find out how many sides  n  a polygon has, if you know the value of  S , the sum of interior angles.

You know the sum of interior angles is  900° , but you have no idea what the shape is. Use what you know in the formula to find what you do not know:

State the formula:

Use what you know,  S=900°:

Divide both sides by  180°:

No need for parentheses now:

Add 2 to both sides:

The unknown shape was a heptagon!

Lesson summary

Now you are able to identify interior angles of polygons, and you can recall and apply the formula,  S=(n−2)×180° , to find the sum of the interior angles of a polygon. You also are able to recall a method for finding an unknown interior angle of a polygon, by subtracting the known interior angles from the calculated sum.

Not only all that, but you can also calculate interior angles of polygons using  S n \frac{S}{n} n S ​ , and you can discover the number of sides of a polygon if you know the sum of their interior angles. That is a whole lot of knowledge built up from one formula,  S=(n−2)×180° .

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Prove that the sum of the degrees in the interior angles of a polygon with $n$ sides is $180(n – 2)°$.

I would assume this question involves an inductive hypothesis.

• Show $n=1$ is true.
• Assume that if $n$ is replaced by $k$, the sum of the degrees in the interior angles of a polygon with $k$ sides is $180(n-2)$.
• Assuming the assumption is true, I want to show that when $n=k+1$ that the sum of the degrees in the interior angles of a polygon with $k+1$ sides is $180((k+1)-2)$.

Problem I'm having is that what do i set $180((k+1)-2)$ to?

• algebra-precalculus

2 Answers 2

There is some confusion on your induction.

$n=1$ does not make sense, since polygons have at least 3 sides (a triangle).

This step, although not really wrong, is unnecessary. You are simply renaming $n$ as $k$.

This is ok, might you should be careful when renaming $n=k$ in step 2 (I'm being somewhat pedantic here).

Induction works as follows: take a property $P(n)$ about natural numbers (for example, "$n$ is even", or "The sum of the inner angles of any polygon with $n$ sides is $180^o(n-2)$"), which might be true or false. Fix a number $M$. Then we prove

• $P(M)$ is true.
• If $P(n)$ is true for some $n$, then $P(n+1)$ is true.

And the induction principle says that $P(n)$ is true for all $n\geq M$. In your case you need to show.

• The sum of the angles of any 3-sided polygon (a triangle) is $180^o$.
• Assume that the sum of the angles of any polygon of $n$ sides is $180^o(n-2)$. Then we need to show that the sum of the angles of any polygon of $n+1$ sides is $180^o(n-1)$.

To prove $2$, start with a polygon $P$ of $n+1$ sides. You need somehow to use the hypothesis, that is, make a polygon of $n$ sides appear, and calculate the angles somehow. You should do this by yourself.

• $\begingroup$ I've tried using the hypothesis to prove. What I did was using the hypothesis on $n+1$ by substituting $n$ with $180^o(n-2)$. That would leave me with $180^o(n-2)+1$=$180^o(n-1)$. Somehow, I still can't get them to equal. Am I doing something wrong? $\endgroup$ –  nyorkr23 Commented May 1, 2016 at 4:49
• $\begingroup$ @nyorkr23 Let's say you do know that the sum of the angles on a triangle is $180^o$. Only using this , try to prove that the sum of the angles of a square is $360^o$. Then try to do the same with a pentagon, hexagon, etc... You will probably come with a procedure which works on all polygons, where you "simplify" the problem of calculating the sum of angles to calculating the sum of angles on smaller polygons (which you then know how to do). $\endgroup$ –  Luiz Cordeiro Commented May 1, 2016 at 5:21
• $\begingroup$ There are other more direct ways (instead of induction) to prove the claim. $\endgroup$ –  Mick Commented May 1, 2016 at 15:08

My understanding of math induction, is that you first show the initial case is true. Then you show that any generic place chosen after that, automatically leads to the next case in the sequence. It is sort of like dominoes, once the first one falls, all the others must then fall in sequence. So, here is what I am thinking. The initial case is a triangle, n = 3. We need to show that for the general case, the next figure in the sequence is 180 more than the previous, ie: D(n+1) = D(n)+180 = 180(n-2)+180

D(n) = 180(n-2)

D(3) = 180 (3-2) = 180 (1) = 180

D(n) = 180 (n-2)

D(n+1) = 180((n+1)-2) = 180(n+1-2) = 180(n-2+1) = 180(n-2)+180

I have to be honest here, math induction was never my strong suit. Perhaps someone who knows it better can give some input on my work.

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Interior Angles of Polygons

An Interior Angle is an angle inside a shape:

Another example:

The Interior Angles of a Triangle add up to 180°

It works for this triangle

It still works! One angle went up by 10°, and the other went down by 10°

Quadrilaterals (Squares, etc)

(A Quadrilateral has 4 straight sides)

A Square adds up to 360°

It still adds up to 360°

The Interior Angles of a Quadrilateral add up to 360°

Because there are 2 triangles in a square ...

The interior angles in a triangle add up to 180° ...

... and for the square they add up to 360° ...

... because the square can be made from two triangles!

A pentagon has 5 sides, and can be made from three triangles , so you know what ...

... its interior angles add up to 3 × 180° = 540°

And when it is regular (all angles the same), then each angle is 540 ° / 5 = 108 °

(Exercise: make sure each triangle here adds up to 180°, and check that the pentagon's interior angles add up to 540°)

The Interior Angles of a Pentagon add up to 540°

The General Rule

Each time we add a side (triangle to quadrilateral, quadrilateral to pentagon, etc), we add another 180° to the total:

So the general rule is:

Sum of Interior Angles = ( n −2) × 180 °

Each Angle (of a Regular Polygon) = ( n −2) × 180 ° / n

Perhaps an example will help:

Example: What about a Regular Decagon (10 sides) ?

And for a Regular Decagon:

Each interior angle = 1440 ° /10 = 144°

Note: Interior Angles are sometimes called "Internal Angles"

Angle Properties, Postulates, and Theorems

In order to study geometry in a logical way, it will be important to understand key mathematical properties and to know how to apply useful postulates and theorems. A postulate is a proposition that has not been proven true, but is considered to be true on the basis for mathematical reasoning. Theorems , on the other hand, are statements that have been proven to be true with the use of other theorems or statements. While some postulates and theorems have been introduced in the previous sections, others are new to our study of geometry. We will apply these properties, postulates, and theorems to help drive our mathematical proofs in a very logical, reason-based way.

Before we begin, we must introduce the concept of congruency. Angles are congruent if their measures, in degrees, are equal. Note : “congruent” does not mean “equal.” While they seem quite similar, congruent angles do not have to point in the same direction. The only way to get equal angles is by piling two angles of equal measure on top of each other.

We will utilize the following properties to help us reason through several geometric proofs.

Reflexive Property

A quantity is equal to itself.

Symmetric Property

If A = B , then B = A .

Transitive Property

If A = B and B = C , then A = C .

Addition Property of Equality

If A = B , then A + C = B + C .

Angle Postulates

Angle addition postulate.

If a point lies on the interior of an angle, that angle is the sum of two smaller angles with legs that go through the given point.

Consider the figure below in which point T lies on the interior of ?QRS . By this postulate, we have that ?QRS = ?QRT + ?TRS . We have actually applied this postulate when we practiced finding the complements and supplements of angles in the previous section.

Corresponding Angles Postulate

If a transversal intersects two parallel lines, the pairs of corresponding angles are congruent.

Converse also true : If a transversal intersects two lines and the corresponding angles are congruent, then the lines are parallel.

The figure above yields four pairs of corresponding angles.

Parallel Postulate

Given a line and a point not on that line, there exists a unique line through the point parallel to the given line.

The parallel postulate is what sets Euclidean geometry apart from non-Euclidean geometry .

There are an infinite number of lines that pass through point E , but only the red line runs parallel to line CD . Any other line through E will eventually intersect line CD .

Angle Theorems

Alternate exterior angles theorem.

If a transversal intersects two parallel lines, then the alternate exterior angles are congruent.

Converse also true : If a transversal intersects two lines and the alternate exterior angles are congruent, then the lines are parallel.

The alternate exterior angles have the same degree measures because the lines are parallel to each other.

Alternate Interior Angles Theorem

If a transversal intersects two parallel lines, then the alternate interior angles are congruent.

Converse also true : If a transversal intersects two lines and the alternate interior angles are congruent, then the lines are parallel.

The alternate interior angles have the same degree measures because the lines are parallel to each other.

Congruent Complements Theorem

If two angles are complements of the same angle (or of congruent angles), then the two angles are congruent.

Congruent Supplements Theorem

If two angles are supplements of the same angle (or of congruent angles), then the two angles are congruent.

Right Angles Theorem

All right angles are congruent.

Same-Side Interior Angles Theorem

If a transversal intersects two parallel lines, then the interior angles on the same side of the transversal are supplementary.

Converse also true : If a transversal intersects two lines and the interior angles on the same side of the transversal are supplementary, then the lines are parallel.

The sum of the degree measures of the same-side interior angles is 180°.

Vertical Angles Theorem

If two angles are vertical angles, then they have equal measures.

The vertical angles have equal degree measures. There are two pairs of vertical angles.

(1) Given: m?DGH = 131

Find: m?GHK

First, we must rely on the information we are given to begin our proof. In this exercise, we note that the measure of ?DGH is 131° .

From the illustration provided, we also see that lines DJ and EK are parallel to each other. Therefore, we can utilize some of the angle theorems above in order to find the measure of ?GHK .

We realize that there exists a relationship between ?DGH and ?EHI : they are corresponding angles. Thus, we can utilize the Corresponding Angles Postulate to determine that ?DGH??EHI .

Directly opposite from ?EHI is ?GHK . Since they are vertical angles, we can use the Vertical Angles Theorem , to see that ?EHI??GHK .

Now, by transitivity , we have that ?DGH??GHK .

Congruent angles have equal degree measures, so the measure of ?DGH is equal to the measure of ?GHK .

Finally, we use substitution to conclude that the measure of ?GHK is 131° . This argument is organized in two-column proof form below.

(2) Given: m?1 = m?3

Prove: m?PTR = m?STQ

We begin our proof with the fact that the measures of ?1 and ?3 are equal.

In our second step, we use the Reflexive Property to show that ?2 is equal to itself.

Though trivial, the previous step was necessary because it set us up to use the Addition Property of Equality by showing that adding the measure of ?2 to two equal angles preserves equality.

Then, by the Angle Addition Postulate we see that ?PTR is the sum of ?1 and ?2 , whereas ?STQ is the sum of ?3 and ?2 .

Ultimately, through substitution , it is clear that the measures of ?PTR and ?STQ are equal. The two-column proof for this exercise is shown below.

(3) Given: m?DCJ = 71 , m?GFJ = 46

Prove: m?AJH = 117

We are given the measure of ?DCJ and ?GFJ to begin the exercise. Also, notice that the three lines that run horizontally in the illustration are parallel to each other. The diagram also shows us that the final steps of our proof may require us to add up the two angles that compose ?AJH .

We find that there exists a relationship between ?DCJ and ?AJI : they are alternate interior angles. Thus, we can use the Alternate Interior Angles Theorem to claim that they are congruent to each other.

By the definition of congruence , their angles have the same measures, so they are equal.

Now, we substitute the measure of ?DCJ with 71 since we were given that quantity. This tells us that ?AJI is also 71° .

Since ?GFJ and ?HJI are also alternate interior angles, we claim congruence between them by the Alternate Interior Angles Theorem .

The definition of congruent angles once again proves that the angles have equal measures. Since we knew the measure of ?GFJ , we just substitute to show that 46 is the degree measure of ?HJI .

As predicted above, we can use the Angle Addition Postulate to get the sum of ?AJI and ?HJI since they compose ?AJH . Ultimately, we see that the sum of these two angles gives us 117° . The two-column proof for this exercise is shown below.

(4) Given: m?1 = 4x + 9 , m?2 = 7(x + 4)

In this exercise, we are not given specific degree measures for the angles shown. Rather, we must use some algebra to help us determine the measure of ?3 . As always, we begin with the information given in the problem. In this case, we are given equations for the measures of ?1 and ?2 . Also, we note that there exists two pairs of parallel lines in the diagram.

By the Same-Side Interior Angles Theorem , we know that that sum of ?1 and ?2 is 180 because they are supplementary.

After substituting these angles by the measures given to us and simplifying, we have 11x + 37 = 180 . In order to solve for x , we first subtract both sides of the equation by 37 , and then divide both sides by 11 .

Once we have determined that the value of x is 13 , we plug it back in to the equation for the measure of ?2 with the intention of eventually using the Corresponding Angles Postulate . Plugging 13 in for x gives us a measure of 119 for ?2 .

Finally, we conclude that ?3 must have this degree measure as well since ?2 and ?3 are congruent . The two-column proof that shows this argument is shown below.

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Article Contents

Introduction, the role of early heterogeneities, the 8- to 16-cell transition: polarization, compaction, and inner-outer segregation, history of the two models of the first lineage decision, lineage-specific transcription factors and signaling pathways, linking position, polarity, and transcriptional networks, human versus mouse lineage specification, emerging techniques, conclusions, data availability, conflict of interest.

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Revisiting trophectoderm-inner cell mass lineage segregation in the mammalian preimplantation embryo

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Robin M Skory, Revisiting trophectoderm-inner cell mass lineage segregation in the mammalian preimplantation embryo, Human Reproduction , Volume 39, Issue 9, September 2024, Pages 1889–1898, https://doi.org/10.1093/humrep/deae142

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In the first days of life, cells of the mammalian embryo segregate into two distinct lineages, trophectoderm and inner cell mass. Unlike nonmammalian species, mammalian development does not proceed from predetermined factors in the oocyte. Rather, asymmetries arise de novo in the early embryo incorporating cues from cell position, contractility, polarity, and cell–cell contacts. Molecular heterogeneities, including transcripts and non-coding RNAs, have now been characterized as early as the 2-cell stage. However, it’s debated whether these early heterogeneities bias cells toward one fate or the other or whether lineage identity arises stochastically at the 16-cell stage. This review summarizes what is known about early blastomere asymmetries and our understanding of lineage allocation in the context of historical models. Preimplantation development is reviewed coupled with what is known about changes in morphology, contractility, and transcription factor networks. The addition of single-cell atlases of human embryos has begun to reveal key differences between human and mouse, including the timing of events and core transcription factors. Furthermore, the recent generation of blastoid models will provide valuable tools to test and understand fate determinants. Lastly, new techniques are reviewed, which may better synthesize existing knowledge with emerging data sets and reconcile models with the regulative capacity unique to the mammalian embryo.

Lineage segregation requires crosstalk between cell position, polarity, and mechanical signaling and is defined by the activation of core transcriptional networks. ZGA, zygotic genome activation.

At the beginning of life, the zygote bears potential to form all cells of the conceptus, but on day three the first fate decision segregates cells into two distinct paths. The first cell lineages, trophectoderm (TE) and inner cell mass (ICM), then physically segregate in the blastocyst. This is followed by the second fate decision within the ICM, separating cells into primitive endoderm (PE) and epiblast (EPI). Indeed, the importance of the first fate decision has been the focus of developmental biologists for more than a century and has cultivated significant debate ( Hiiragi et al. , 2006 ; Rossant and Tam, 2009 ; White and Plachta, 2019 ).

The mammalian embryo’s unique properties make it a striking, but challenging system to examine lineage specification. It is both simple in its structure, yet complex with self-organization and regulative capacities. Cues within and between blastomeres begin to favor one cell fate over the other, independent of external factors within the Fallopian tube and endometrial cavity in vivo . Furthermore, the mammalian embryo adapts to both internal and external stressors including blastomere isolation and removal in vitro . This developmental plasticity combines with cellular asymmetries and mechanical forces to further refine cellular identity. In this review, the history of our understanding of lineage specification in addition to current models will be reviewed with a discussion of how future and emerging technologies may fully reveal the origins of the first fate decision.

Nonmammalian species such as Xenopus ( Speman, 1938 ), Drosophila ( Driever and Nüsslein-Volhard, 1988a , b ), and Caenorhabditis elegans ( Goldstein and Hird, 1996 ) have maternally derived factors within the egg that determine body axes prior to fertilization. In mammals, no such oocyte-derived determinant has been identified. However, tools such as lineage tracing ( Fujimori et al. , 2003 ; Tabansky et al. , 2013 ), live imaging ( Plachta et al. , 2011 ; White et al. , 2016a ; Lim et al. , 2020 ), and single-cell sequencing ( Biase et al. , 2014 ; Shi et al. , 2015 ; Goolam et al. , 2016 ) have unveiled transcript heterogeneities as early as the 2- to 4-cell stage. It has been proposed that these early asymmetries are not deterministic, but rather bias cells toward a lineage identity. Prior to zygotic genome activation (ZGA), maternal factors may be unequally distributed at the first cleavage division resulting in so-called ‘partitioning errors’ ( Shi et al. , 2015 ). Furthermore, it has been proposed that these small blastomere-to-blastomere differences are amplified by stochastic transcriptional bursts, or ‘transcriptional noise’, following ZGA ( Fig. 1 ) ( Biase et al. , 2014 ; Piras et al. , 2014 ; Shi et al. , 2015 ).

Overview of mouse preimplantation embryo development . Summary of development from the zygote to blastocyst stages in the mouse. Early heterogeneities are detectable at the 2-cell stage coinciding with zygotic genome activation (ZGA). At the 4-cell stage CARM1 asymmetry drives differential transcription factor expression favoring pluripotency over trophectoderm identity. Changes in embryo metabolism, polarization, and compaction occur at the 8-cell stage with distinct inner and outer cells formed at the 16-cell stage. Unique monoastral spindles in some blastomeres favor asymmetric division producing polar–apolar daughters. The asymmetric inheritance of keratin filaments also favors the trophectoderm program. Transcription factor networks specific to TE or ICM cell lineages are summarized.

In addition to differences in mRNAs, asymmetries in the long-noncoding RNA, LincGET have been identified at the 2-cell stage ( Wang et al. , 2016 , 2018 ). Bound to the methyltransferase CARM1 ( Torres-Padilla et al. , 2007 ; Hupalowska et al. , 2018 ) and its DNA binding partner Prdm14, the complex increases methylation levels, leading to increased chromatin accessibility at the 4-cell stage in some, but not all, blastomeres ( Fig. 1 ). Using photoactivatable fluorescence correlation spectroscopy (paFCS), it was found that CARM1 methylation increases the bound fraction of Sox2, leading to higher expression of pluripotency target genes such as Oct4, Nanog ( White et al. , 2016a ), and Sox21 ( Goolam et al. , 2016 ) ( Fig. 1 ). Thus, it has been proposed that differential CARM1 patterns sway cell fate at the 4-cell stage with high activity favoring pluripotency and low activity favoring TE identity ( Torres-Padilla et al. , 2007 ). Moreover, the early heterogeneity model posits that a blastomere’s history of cleavage patterns and gene expression bias lineage fate probabilistically ( Torres-Padilla et al. , 2007 ; Plachta et al. , 2011 ; Zernicka-Goetz, 2013 ; Chen et al. , 2018 ). However, the precise role of early heterogeneities in biasing lineage fate remains debated ( Hiiragi et al. , 2006 ). Others conclude that lineage identity emerges at the 16-cell stage as a result of stochastic processes unrelated to these early heterogeneities ( Dietrich and Hiiragi, 2007 ; Wennekamp and Hiiragi, 2012 ; Strnad et al. , 2016 ). Nevertheless, emerging models of symmetry breaking must integrate the regulative nature of mammalian preimplantation development and the role of early bias remains an open area of investigation.

To understand lineage segregation, it is important to review key stages comprising preimplantation morphogenesis. The first major structural change occurs from the 8- to 16-cell stage, when three important events transform the relatively simple embryo structure into an organized morula: cellular polarization, compaction, and inner-outer segregation ( Fig. 1 ). These key events predispose cells toward differential fates with cells committed to the TE lineage by the 32-cell stage ( Posfai et al. , 2017 ).

At the 8-cell stage, blastomeres establish apicobasal polarity. With the second wave of ZGA, expression of the key transcription factors Tfap2c and Tead4 triggers actin regulators ( Zhu et al. , 2020 ). This initiates formation of the apical domain in regions lacking cell–cell junctions and asymmetric localization of apical and basolateral proteins ( Fig. 1 ). On the exposed apical cell surface, microvilli ( Reeve and Ziomek, 1981 ) accumulate and a protein cap forms enclosed by an actomyosin ring consisting of atypical protein kinase C (aPKC) ( Pauken and Capco, 2000 ), Pard6b ( Vinot et al. , 2005 ), and Par3 ( Plusa et al. , 2005 ). Along the basolateral domains, adherens and tight junctions form with E-cadherin localizing to cell–cell contacts ( Johnson et al. , 1986 ; Vestweber et al. , 1987 ; Fleming et al. , 1989 ) along with catenins ( Ozawa et al. , 1989 ; McCrea et al. , 1991 ; Staddon et al. , 1995 ). Thus, prior to division, blastomeres at the 8-cell stage contain distinct apical and basolateral domains. In addition, the earliest cytoskeletal heterogeneities arise at the 8-cell stage with keratin intermediate filaments assembling in a subset of cells at the apical cortex ( Lim et al. , 2020 ). During division, these keratin filaments are asymmetrically inherited into outer cells, where they stabilize the apical cortex and promote the expression of TE-associated transcription factors ( Fig. 1 ) ( Lim et al. , 2020 ).

Additionally at this stage, compaction takes place, in which the loosely associated blastomeres transform into tightly grouped cells. At the cellular level, blastomere apical surfaces flatten and cell–cell contacts increase resulting in close apposition of neighboring cell membranes, changes visible by light microscopy (reviewed in White et al. , 2016b ). Unique membrane protrusions known as filopodia form on a subset of blastomeres ( Fierro-González et al. , 2013 ), which extend and anchor to neighboring cells, facilitating changes in cell shape ( Fig. 1 ). The exact signaling mechanisms that initiate the synchronous and conserved timing of compaction remains to be resolved but is known to involve β-catenin phosphorylation by PKC ( Winkel et al. , 1990 ; Ohsugi et al. , 1993 ). In addition, E-cadherin plays an important role in excluding actomyosin from cell–cell contacts, facilitating cortical contractility essential for compaction to occur ( Hyafil et al. , 1981 ; Shirayoshi et al. , 1983 ; Dietrich and Hiiragi, 2007 ; Stephenson et al. , 2010 ; Maître et al. , 2015 ).

During the fourth cleavage division, blastomeres are segregated into two distinct populations defined by polarization and spatial domain. At the 16-cell stage, one to three inner cells are completely enclosed within the embryo and remain apolar while outer polarized cells maintain an exposed surface ( Dietrich and Hiiragi, 2007 ; Morris et al. , 2010 ; Anani et al. , 2014 ; Samarage et al. , 2015 ). Cell position becomes of utmost importance as the first cells positioned within the embryo’s interior go on to form cells of the ICM ( Barlow et al. , 1972 ; Graham and Deussen, 1978 ; Domingo-Muelas et al. , 2023 ). By contrast, most outer cells will go on to become TE. The mechanisms underlying inner-outer segregation have been complex to dissect as inner cell formation does not follow stereotypic patterns. Variable between embryos and between blastomeres, the process involves a combination of division orientation and differential contractility. Cell divisions may be either symmetric, producing two outer daughter cells or asymmetric, producing inner-outer daughter cells as determined by cleavage angle (those occurring ≤30°; Fig. 2A ). Furthermore, asymmetric divisions play an important role in inner cell formation ( Johnson and Ziomek, 1981 ; Sutherland et al. , 1990 ). Indeed, asymmetric inheritance of the apical pole has long formed the basis of lineage segregation models ( Johnson and Ziomek, 1981 ). Division orientation and cell position, however, do not solely determine cell fate. Not all outer cells go on to the TE lineage, with some internalizing independent of cell division. Live imaging combined with membrane labeling revealed that division angle itself could not solely predict cell fate ( Watanabe et al. , 2014 ). Further, some outer cells undergo a process of apical constriction, where differences in cortical tension generated by actomyosin networks drive cells to the embryo’s interior ( Samarage et al. , 2015 ; Maître et al. , 2016 ). Thus, 8-cell stage blastomeres segregate into inner and outer positions based on a combination of division orientation, cellular polarization, and contractility.

Cleavage planes and cell position are linked to cell fate at the 8- to 16-cell stage . ( A ) Cleavage angle determines symmetric versus asymmetric divisions. Asymmetric divisions occur when the cleavage angle is ≤30° between lines drawn from the embryo to blastomere center of mass and between the two daughter cells. Symmetric divisions occur when the cleavage angle is equal to 90°, producing polar-polar outer cells. ( B ) Hippo signaling couples position and contractility with lineage-specific transcription. In outer cells, Amot and Lats1/2 kinase are sequestered at the apical domain rendering Hippo signaling ‘off’. Yap is free to shuttle to the nucleus where in combination with Tead4 results in the expression of trophectoderm transcription factors Cdx2 and Gata3. Without an apical domain, inner cells maintain active Hippo signaling with phosphorylated Amot located in the cytoplasm. This is reinforced by the downregulation of Lamin-A/C in response to a decrease in cortical tension when cells internalize. This leads to Yap phosphorylation via Lats1/2 kinase and the rapid degradation of phosphorylated Yap prevents nuclear shuttling, enabling the expression of pluripotency-associated factors Sox2 and Oct4.

Cleavage planes at the 8-cell stage play a key role in inner-outer segregation, yet determinants of spindle orientation remain unclear. Several determinants have been proposed including cell–cell contacts ( Humięcka et al. , 2017 ), nuclear position ( Ajduk et al. , 2014 ), and presence of the apical domain ( Korotkevich et al. , 2017 ). Interestingly, the spindle itself bears unique and heterogeneous characteristics in the early mouse embryo ( Pomp et al. , 2022 ). While most spindles at the 8- to 16-cell stage lack asters in the absence of centrosomes at this embryonic age, distinctive monoastral spindles form in a few cells ( Fig. 1 ). This promotes orthogonal, or asymmetric, divisions resulting in one outer and one inner daughter cell. Thus, it appears that heterogeneities in early spindle properties play a role in biasing cell fate by driving cleavage orientation during the 8- to 16-cell stage.

Experiments detailing the fate of isolated blastomeres proposed two foundational hypotheses that have led to our modern understanding of TE-ICM specification: the inside–outside hypothesis and the cell polarity model. Distinct microenvironments arise at the 8- to 16-cell stage, forming inner and outer cell populations. Results from in vitro culture experiments of isolated blastomeres at the 4- and 8-cell stages led to the proposal that these differences in cell position direct cell fate ( Tarkowski and Wróblewska, 1967 ). When single blastomeres at the 4-cell stage were isolated, they went on to form blastocysts composed of ICM and TE compartments. However, isolated 8-cell stage blastomeres formed trophoblast vesicles due to an absence of inner cells by the onset of lineage restriction. Furthermore, fate could be altered by repositioning cells within the embryo ( Hillman et al. , 1972 ; Kelly, 1977 ). Thus, it was proposed that following inner-outer cell allocation, local signals based on cell position determined fate, or the ‘inside–outside’ hypothesis ( Tarkowski and Wróblewska, 1967 ).

In addition to inner-outer segregation, cell polarity has been proposed as the driving factor of lineage divergence. Studying isolated blastomeres at the 8-cell stage, it was found that the majority of cell divisions were asymmetric, resulting in polar–apolar pairs ( Fig. 2A ) ( Johnson and Ziomek, 1981 ). Furthermore, it has been shown that transplanting the apical domain into apolar cells induced lineage-specific transcription patterns ( Korotkevich et al. , 2017 ). Thus the ‘cell polarity model’ proposed that cell fate was not dictated by cells sensing position, per se , but was driven by the asymmetry of apical and basal regions of the polarized blastomere ( Johnson and Ziomek, 1981 ) and inheritance of the apical domain.

Neither model, however, sufficiently explains all lineage specification. For instance, some outer cells remain apolar and slowly internalize, eventually contributing to the ICM ( Plusa et al. , 2005 ; Anani et al. , 2014 ), suggesting that the inside–outside hypothesis could not act as an exclusive mechanism. On the other hand, advances in live imaging enabled the visualization of cell divisions with high temporal resolution revealing that the apical domain is disassembled just prior to cytokinesis and that apolar cells can spontaneously polarize many hours after division ( Anani et al. , 2014 ; Korotkevich et al. , 2017 ; Zenker et al ., 2018 ). Live imaging also revealed that most cleavage planes are oblique, rather than perpendicular or parallel to the embryo surface ( Sutherland et al. , 1990 ; Yamanaka et al ., 2010 ; McDole et al ., 2011 ). Thus, neither position nor inheritance of the apical domain act as sole drivers of cell fate. Rather, the variability and regulative nature of mammalian embryos necessitate models that encompass reinforcing signals including cell position, polarity, and shape. Indeed, unified models have been proposed wherein signals such as the apical domain drives cell fate, but are influenced and amplified by the sum of molecular heterogeneities, cleavage orientation, and cell position ( Mihajlović and Bruce, 2017 ; Lamba and Zernicka-Goetz, 2023 ).

Work from both developmental and stem cell biology has given us a significant understanding of the transcription factors that either maintain pluripotency or push cells toward TE identity. Cells of the ICM are the precursors of all fetal cell types and must therefore remain in the pluripotent state. This is regulated by a circuit of transcription factors, driven by Sox2, Oct4, and Nanog ( Nichols et al. , 1998 ; Avilion et al. , 2003 ; Mitsui et al. , 2003 ; Niwa et al. , 2005 ; Li et al. , 2007 ) ( Fig. 1 ). Single-cell gene expression analysis identified Sox2 as the earliest marker of inner cells with strong induction beginning at the 16-cell stage ( Avilion et al. , 2003 ; Guo et al. , 2010 ; Wicklow et al. , 2014 ). By the early blastocyst stage (∼64 cell stage), Sox2 forms a complex with Oct4 to drive Nanog expression within inner cells, establishing the pluripotency core required for self-renewal ( Palmieri et al. , 1994 ; Yuan et al. , 1995 ; Rodda et al. , 2005 ; Guo et al. , 2010 ). Furthermore, Sox2 and Oct4 act synergistically to promote their own expression, a mechanism reinforcing pluripotency ( Chew et al. , 2005 ).

Unlike inner cells, outer cells must override the pluripotent program to restrict cells toward the TE lineage. This occurs via the caudal-like transcription factor, Cdx2, which represses Oct4 and Nanog expression ( Strumpf et al. , 2005 ; Chen et al. , 2009 ; Huang et al. , 2017 ) ( Fig. 1 ). Beginning at the 8-cell stage, Cdx2 can be detected in the nucleus, but is not equivalent across all blastomeres ( Dietrich and Hiiragi, 2007 ; Ralston and Rossant, 2008 ). Expression is gradually restricted to outer cells by the late 16-cell stage ( Strumpf et al. , 2005 ), both inhibiting ICM-associated genes and favoring TE-transcription factors including Tead4 and Gata3 ( Yagi et al. , 2007 ; Nishioka et al. , 2008 ; Ralston et al. , 2010 ). Downstream, Cdx2 regulates the T-box transcription factor Eomes, which is first detected in TE at the blastocyst stage ( Russ et al. , 2000 ; Niwa et al. , 2005 ). Moreover, the Ets transcription factor Elf5 reinforces both Cdx2 and Eomes expression, providing positive feedback toward TE identity ( Ng et al. , 2008 ).

Preimplantation development is characterized by a metabolic switch, which is proposed to also play a role in lineage segregation. It has long been known that the embryo first requires pyruvate through the morula stage. Around the time of compaction, progression to the blastocyst stage then becomes glucose-dependent ( Brown and Whittingham, 1991 ; Fig. 1 ). Interestingly, it was found that glucose plays an important role in TE specification. Byproducts of two glycolytic shunts (pentose phosphate pathway and hexosamine biosynthetic pathway), are required for Tfap2c translation ( Chi et al. , 2020 ) and Cdx2 expression, which are necessary for TE specification. Thus, the transcriptional network depends on metabolic state and carries redundant and reinforcing mechanisms which ensure proper development.

Our understanding of the transcriptional regulation of pluripotency versus TE differentiation has established the foundation for pioneering work in embryonic stem cells, embryoid models, and regenerative medicine ( Cockburn and Rossant, 2010 ). However, how are the morphogenetic events of early development linked to these transcriptional networks? The Hippo signaling pathway modulates diverse cellular processes including differentiation, cell size, and proliferation ( Mo et al. , 2014 ) and in development plays a key role in reconciling cell position, polarity, and lineage-specific transcription ( Nishioka et al. , 2009 ; Lorthongpanich et al. , 2013 ; Sasaki, 2017 ).

Angiomotin (Amot), a key mediator of Hippo signaling, couples position, polarization, and actomyosin contractility with fate ( Chan et al. , 2011 ; Hirate et al. , 2013 ; Skory et al. , 2023 ). Differences in cell–cell junctions and polarity determine Amot localization in inner versus outer cells. In apolar inner cells, Amot binds to adherens junction proteins, where it is phosphorylated by Lats1/2 kinase. Lats1/2 kinase also phosphorylates the transcriptional activator Yap1 and its related protein Taz (Wwtr1, together referred to as Yap; Fig. 2B ) ( Nishioka et al. , 2009 ; Sasaki, 2017 ) and is then retained in the cytoplasm where it is quickly degraded ( Wang et al. , 2012 ). Moreover, active Hippo signaling within inner cells prevents TE-specific transcription factors and is required for Sox2 expression ( Wicklow et al. , 2014 ).

Conversely, in polarized outer cells, Amot is sequestered to the apical cortex, preventing downstream Hippo signaling. Amot sequestration depends on the Par-aPKC system, proteins of the apical domain, Rho and F-actin ( Hirate et al. , 2013 ; Leung and Zernicka-Goetz, 2013 ; Anani et al. , 2014 ; Shi et al. , 2017 ). Moreover, without Amot at the junctional complexes, Hippo signaling remains inactivated, which leaves unphosphorylated Yap to shuttle to the nucleus ( Nishioka et al. , 2009 ). Nuclear Yap subsequently interacts with Tead4, promoting the expression of TE-specific genes (Cdx2, Gata3) ( Yagi et al. , 2007 ; Nishioka et al. , 2009 ; Ralston et al. , 2010 ) and in parallel, suppresses the pluripotency-associated factor Sox2 ( Wicklow et al. , 2014 ) ( Fig. 2B ).

Differences in contractility also modulate Yap subcellular localization ( Maître et al. , 2016 ) and Cdx2 expression ( Samarage et al. , 2015 ) suggesting mechanosensitive regulation. Our laboratory recently showed that the nuclear lamina plays an essential role in coupling contractility with Hippo signaling via Lamin-A/C levels leading to differences in cytoplasmic Amot abundance ( Skory et al. , 2023 ). Thus, Amot links compaction, polarization, and inner-outer segregation with lineage specification through differential cell–cell contacts and contractility.

Lineage specification and symmetry breaking have been extensively explored in the mouse. With more human embryo studies, however, we are increasingly learning about divergent patterns in the spatio-temporal expression of lineage markers. Morphologically, mouse and human preimplantation development proceed through the same order of events from fertilization to compaction, cavitation, and blastocyst expansion. However, an additional cell cycle takes place prior to implantation in human, occurring on days post-fertilization 6-8 versus 4-4.5 in mouse ( Niakan and Eggan, 2013 ). The timing of compaction also differs, occurring at the 8-cell stage in mouse ( Johnson and Maro, 1984 ) and at the 12- to 16-cell stage in human ( Nikas et al. , 1996 ; Iwata et al. , 2014 ; Domingo-Muelas et al. , 2023 ).

The origins of lineage specification likely also differ between species due in part to ZGA timing, with the major wave occurring in the late 2-cell stage in mouse and at the 8-cell stage in human ( Braude et al. , 1988 ; Petropoulos et al. , 2016 ; Rossant and Tam, 2017 ). In addition, lineage restriction occurs later in human as demonstrated in reconstitution and single-cell sequencing studies. TE cells isolated from early human blastocysts are capable of regenerating both cell types ( Paepe et al. , 2013 ), while in mice, outer cells become TE-restricted at the late morula stage ( Suwińska et al. , 2008 ). Furthermore, lineage-specific expression of TE and pluripotent markers such as CDX2, OCT4, NANOG, and GATA6 does not occur until the late blastocyst stage in human embryos ( Niakan and Eggan, 2013 ), which is observed prior to cavitation in mouse.

Similar to mouse, initiating the TE program in human appears to share common signaling mediators within the Hippo pathway ( Gerri et al. , 2020 , 2023 ). Furthermore, cellular polarization appears to play a role in human TE specification, albeit an initiating signal in mouse and a reinforcing role in human ( Zhu et al. , 2021 ). An ongoing debate remains regarding the timing and molecular pathways segregating cells to the ICM in human. The emergence of single cell transcriptomic atlases of human embryos has led to three models: (i) concurrent specification of all three lineages (TE, EPI, PE) at the expanded blastocyst stage ( Petropoulos et al. , 2016 ); (ii) a 2-step process with initial TE-EPI segregation during early blastocyst expansion followed by EPI-PE specification ( Meistermann et al. , 2021 ); and (iii) ICM-TE segregation in the early expanded blastocyst ( Radley et al. , 2023 ). Thus, it remains an open question of whether there is a single or two-step differentiation process to EPI and PE lineages with uncertainty regarding the existence of a bona fide ICM state in human preimplantation development.

A deep framework for understanding lineage specification has been established in the mouse and emerging techniques in single cell approaches, live imaging, and embryo modeling will likely challenge and refine existing models. Furthermore, increasing effort and attention toward understanding human development will undoubtedly identify how lineage segregation may differ from the mouse.

The combination of single-cell sequencing with large-scale omics, AI technology, and computational modeling has begun to provide an integrated view of early fate decisions. Atlases of human embryos have now been generated using multi-ome single cell approaches, providing key roadmaps of transcription ( Yan et al. , 2013 ; Shi et al. , 2015 ; Petropoulos et al. , 2016 ; Meistermann et al. , 2021 ) and methylation patterns ( Li et al. , 2018 ; Zhu et al. , 2018 ). Additionally, retrospective lineage tracing based on genome sequencing and somatic variant barcoding has recently been employed in human embryos ( Coorens et al. , 2021 ; Fasching et al. , 2021 ; Park et al. , 2021 ) identifying bias as early as the 2-cell stage.

Advancements in live imaging, including inverted light-sheet microscopy and non-invasive labeling, now enable in toto imaging of mammalian preimplantation embryogenesis ( Strnad et al. , 2016 ; Domingo-Muelas et al. , 2023 ). Higher spatial and temporal resolution will resolve new heterogeneities and provide the opportunity to functionally test their role in early bias. Experiments using live imaging and microinjection of 2-cell human embryos recently enabled prospective lineage tracing, unveiling that the timing of divisions at the 2-cell stage bias toward ICM versus TE fates ( Junyent et al. , 2024 ). In addition, the advancement of functional studies utilizing CRISPR-Cas9 has begun to test the role of genes known to be important in the mouse. The application of CRISPR-Cas9 in 2017 uncovered the role of OCT4 in both ICM and TE lineages in human embryos ( Fogarty et al. , 2017 ), in contrast to its well-characterized role in TE specification in the mouse. Given the ethical concerns of human germline gene editing, future similar studies will continue under extensive regulations notably for research and not clinical application. Nonetheless, such prospective studies of human embryogenesis will be vital to our understanding and likely uncover mechanisms unique to lineage specification in human.

Given the ethical, legal, and technical limitations of human embryo research, several stem-cell based blastocyst-like models (blastoids) have emerged ( Liu et al. , 2021 ; Sozen et al. , 2021 ; Yu et al. , 2021 ; Kagawa et al. , 2022 ). Human blastoids enable unprecedented studies of human development. Importantly, however, they do not fully recapitulate preimplantation development, particularly within the TE ( Posfai et al. , 2021 ). Furthermore, stem-cell based human gastruloids have also been described, which interestingly completely bypass a blastocyst-like stage ( Oldak et al. , 2023 ; Weatherbee et al. , 2023 ). In parallel, embryo culture systems continue to evolve, which are now able to support prolonged nonhuman primate development through gastrulation ( Ma et al. , 2019 ; Niu et al. , 2019 ; Zhai et al. , 2023 ). These prolonged systems enable opportunities to study lineage specification continuously from ICM-TE to TE-PE-EPI within intact nonhuman primate embryos. Thus, advancement of culture systems and stem-cell based embryoid models will require finer deconstruction of the stages before and after the first lineage segregation event.

In conclusion, TE-ICM lineage segregation in the mammalian embryo requires crosstalk between cell position, polarity, mechanical, and molecular signaling. The timing of lineage bias and then restriction is tied to the timing of ZGA, which remains an open area of investigation. Furthermore, the origins, role, and significance of early heterogeneities remain unknown. Addressing these questions may resolve longstanding debates and further reveal the self-organizing and regulative capacity of the mammalian preimplantation embryo.

No new data were generated or analyzed in support of this research.

NICHD RSDP (K12HD849-36) and ASRM to R.M.S.

The author has no conflicts of interest.

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• transcription factor
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