Theorems About Triangles

Consider a triangle with vertices A, B, and C, and one of the exterior angles corresponding to ∠ C.

The diagram shows that ∠ C and ∠ PCA form a linear pair, so the sum of their measures is 180^(∘). Additionally, by the Triangle Angle Sum Theorem, the sum of the angle measures of △ ABC is 180^(∘). m∠ C + m∠ PCA = 180^(∘) & (I) m∠ A + m∠ B + m∠ C = 180^(∘) & (II) Now m∠ C can be isolated in Equation (I). m∠ C + m∠ PCA = 180^(∘) ⇕ m∠ C = 180^(∘)-m∠ PCA Next, the expression of m∠ C can be substituted into Equation (II). It has been proven that the measure of ∠ PCA is equal to the sum of the measures of ∠ A and ∠ B. Therefore, it can be said that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two nonadjacent interior angles.

Consider △ ABC, where D and E are the midpoints of AB and AC, respectively. Let ∠ PCA be one exterior angle of △ ABC.

Now, △ ABC can be rotated 180^(∘) about D. Since a rotation is a rigid motion, the image of △ ABC after the rotation is congruent to △ ABC. Corresponding parts of congruent figures are congruent, so the measures of the angles and the lengths of the sides remain unchanged. Since a 180^(∘)-rotation is equivalent to a reflection, C'A is parallel to BC and C'B is parallel to AC. Therefore, C'ACB is a parallelogram and ∠ C'AC is congruent to ∠ CBC'. Now the parallelogram C'ACB will be rotated 180^(∘) about E. By the Parallelogram Opposite Angles Theorem, ∠ PCA is congruent to ∠ AB''P. Congruent angles have the same measure by the definition. ∠ PCA &≅ ∠ AB''P &⇕ m∠ PCA &= m∠ AB''P Since m∠ AB''P is equal to the sum of m∠ A and m∠ B and because of the Transitive Property of Equality, m∠ PCA is equal to the sum of m∠ A and m∠ B. m∠ PCA=m∠ AB''P m∠ AB''P = m∠ A + m∠ B ⇓ m∠ PCA = m∠ A + m∠ B This completes the proof.

Consider a triangle ABC with two congruent sides, or an isosceles triangle.

In this triangle, let P be the point of intersection of BC and the angle bisector of ∠ A. From the diagram, the following facts about △ BAP and △ CAP can be observed.

Statement	Reason
∠ BAP ≅ ∠ CAP	Definition of an angle bisector
BA ≅ CA	Given
AP ≅ AP	Reflexive Property of Congruence

Therefore, △ BAP and △ CAP have two pairs of corresponding congruent sides and one pair of congruent included angles. By the Side-Angle-Side Congruence Theorem, △ BAP and △ CAP are congruent triangles. △ BAP ≅ △ CAP Corresponding parts of congruent figures are congruent. Therefore, ∠ B and ∠ C are congruent. ∠ B ≅ ∠ C It has been proven that if two sides of a triangle are congruent, then the angles opposite them are congruent.

Consider an isosceles triangle △ ABC.

A line passing through A and the midpoint of BC will be drawn. Let P be the midpoint.

Since BP and PC are congruent, the distance between B and P is equal to the distance between C and P. Therefore, B is the image of C after a reflection across AP. Also, because A lies on AP, a reflection across AP maps A onto itself. The same is true for P.

Reflection Across AP
Preimage	Image
C	B
A	A
P	P

The table shows that the images of the vertices of △ CAP are the vertices of △ BAP. It can be concluded that △ BAP is the image of △ CAP after a reflection across AP. Since a reflection is a rigid motion, this proves that the triangles are congruent. Corresponding parts of congruent figures are congruent, so ∠ B and ∠ C are congruent. ∠ B≅∠ C

This theorem can be proven by placing the triangle on a coordinate plane. For simplicity, vertex B will be placed at the origin and vertex C on the x-axis.

Since B lies on the origin, its coordinates are (0,0). Point C is on the x-axis, meaning its y-coordinate is 0. The remaining coordinates are unknown and can be named a, b, and c. B(0,0) C(a,0) A(b,c) If DE is the midsegment from BA to CA, then by the definition of a midpoint, D and E are the midpoints of BA and CA, respectively.

To prove this theorem, it must be proven that DE is parallel to BC and that DE is half of BC.

BC ∥ DE

If the slopes of these two segments are equal, then they are parallel. The y-coordinate of both B(0, 0) and C(a, 0) is 0. Therefore, BC is a horizontal segment. Next, the coordinates of D and E will be found using the Midpoint Formula.

M(x_1+x_2/2,y_1+y_2/2)
Segment	Endpoints	Substitute	Simplify
BA	B( 0, 0) and A( b,c)	D(0+ b/2,0+c/2)	D(b/2,c/2)
CA	C( a, 0) and A( b,c)	E(a+ b/2,0+c/2)	E(a+b/2,c/2)

The y-coordinate of both D( b2,c2) and E( a+b2,c2) is c2. Therefore, DE is also a horizontal segment. Since all horizontal segments are parallel, it can be said that BC and DE are parallel. BC ∥ DE ✓

DE=1/2BC

Since both BC and DE are horizontal, their lengths are given by the difference of the x-coordinates of their endpoints.

Segment	Endpoints	Length	Simplify
BC	B(0,0) and C(a,0)	BC=a-0	BC=a
DE	D(b/2,c/2) and E(a+b/2,c/2)	DE=a+b/2-b/2	DE=1/2a

Since 12a is half of a, it can be stated that the midsegment DE is half the length of BC. DE=1/2BC ✓ Therefore, a midsegment of two sides of a triangle is parallel to the third side of the triangle and half its length.

This proof will be developed based on the given diagram, but it is valid for any triangle. To prove this theorem, it must be proven that DE is parallel to BC and that DE is equal to half of BC. Each statement will be proven one at a time.

BC ∥ DE

This part can be proven by using rigid motions. First, translate △ ADE along DB so that D is mapped onto B. Since D is the midpoint of AB, A is mapped onto D. Next, it must be proven that the image of E — which is marked as E' — lies on BC. This proof will be done using indirect reasoning. In this method, it is temporarily assumed that the negation of the statement is true. It has been proven that E' lies on BC. Because translations preserve angles, ∠ ADE is congruent to ∠ DBE'.

By the Converse Corresponding Angles Theorem, BC is parallel to DE.

DE=1/2BC

It has been previously obtained that BC and DE are parallel. Now, another rigid motion to △ BDE' will be applied. By the Segment Addition Postulate, the length of BC can be calculated by adding the lengths of smaller segments. BC = BE' + CE' Because corresponding sides of congruent triangles are congruent, DE is congruent to BE' and DE is congruent to CE'. Therefore, DE=BE' and DE=CE'. By the Substitution Property of Equality, BC can be expressed in terms of DE. BC = DE + DE ⇔ BC= 2DE Finally, by the Division Property of Equality, the second statement of the theorem is obtained.