Theorems About Triangles

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.

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.

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.

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.

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.

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

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.

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.

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.

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.

1

Assume That E^' Does Not Lie on BC

expand_more

Assume that E', the image of E after the translation along DB, does not lie on BC. This means that E' lies either above or below BC. The proof will be developed for only one case but is valid for both.

Based on the assumption, let F denote the point of intersection of BE' and EC.

Next, it will be proven that △ ADE is congruent to △ EE'F.

2

Show That △ ADE ≅ △ EE'F

expand_more

It is given that AD= DB because D is the midpoint of AB. Additionally, since △ ADE is translated along DB, it can be concluded that DB=EE'. Then, by the Transitive Property of Equality, AD = EE'.

Recall that AE = DE'. Additionally, DE is the common side of △ ADE and △ E'ED. All pieces of information can now be summarized.

• AD = EE'
• AE = DE'
• DE is the common side of △ ADE and △ E'ED.

Using all the information, △ ADE is congruent to △ E'ED by the Side-Side-Side Congruence Theorem. △ ADE ≅ △ E'ED Because corresponding angles of congruent triangles are congruent, ∠ ADE is congruent to ∠ DEE'.

Since translations preserve angles, DE is parallel to BF. By the Alternate Interior Angles Theorem, ∠ DEE' is congruent to ∠ EE'F. ∠ ADE ≅ ∠ DEE' and ∠ DEE' ≅ ∠ EE'F Therefore, by the Transitive Property of Congruence, ∠ ADE ≅ ∠ EE'F.

Since DE is parallel to BF and the image of E is translated in the same direction as the image of D, ∠ BDE' is congruent to ∠ E'EF. Additionally, ∠ BDE' is congruent to ∠ DAE. ∠ E'EF ≅ ∠ BDE' and ∠ BDE' ≅ ∠ DAE One more time, by the Transitive Property of Congruence, ∠ E'EF ≅ ∠ DAE.

Summarize the obtained information about the triangles △ ADE and △ EE'F.

• AD=EE'
• ∠ ADE ≅ ∠ EE'F
• ∠ DAE ≅ ∠ E'EF

Therefore, by the Angle-Side-Angle Congruence Theorem, △ ADE is congruent to △ EE'F.

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.

1

Rotate △ BDE' Around the Midpoint of DE'

expand_more

Rotate △ BDE' counterclockwise about the midpoint of DE' so that E' is mapped onto D. It can be noted that the triangle is rotated 180^(∘). Since rotations preserve angles and lengths, this rotation maps B onto E. Therefore, BD is mapped onto EE'.

As a result of the rotation, it can be concluded that △ BDE' and △ B''D''E'', or △ EE'D, are congruent triangles.

2

Rotate △ DED'' Around the Midpoint of D''E

expand_more

Rotate △ DED'' counterclockwise about the midpoint of D''E so that E is mapped onto D''. It can be noted that the triangle is again rotated 180^(∘). Since rotations preserve angles and lengths, this rotation maps D onto C. Therefore, DD'' and DE are mapped onto CE and CE', respectively.

As a result of the rotation, it can be concluded that △ DED'' and △ EE'C'' are congruent triangles.

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.