From the definition of an isosceles triangle as one in which two sides are equal, we proved the Base Angles Theorem – the angles between the equal sides and the base are congruent.
Now we’ll prove the converse theorem – if two angles in a triangle are congruent, the triangle is isosceles.
Problem
In triangle ΔABC, the angles ∠ACB and ∠ABC are congruent. Prove that ΔABC is isosceles, i.e. that AB=AC.
Strategy
As with other converse theorems, we’ll try to apply the same strategy we used to prove the original one – the Base Angles Theorem. There, we drew a line from A to the base BC and proved the resulting triangles are congruent.
As a result, the base angles were congruent. We’ll do the same here, prove the triangles are congruent relying on the fact that the base angles are congruent. And as a result, the corresponding sides, AB and AC, will be equal.
And just like in the original theorem, we have a choice of which line to draw. We can draw either the altitude to the base, and use the fact that it creates a linear pair of equal right angles. Or, draw the angle bisector of A, and use the fact that it creates a pair of equal angles at A.
We’ll provide a proof for both.
Proof
First, we’ll draw AD, the height to the base:
(1) ∠ACB ≅ ∠ABC //Given
(2) AD = AD // Common side to both triangles, reflexive property of equality
(3) m∠ADC= m∠ADB=90° //construction
(4) ∠ADC≅∠ADB //(3), definition of congruent angles
(5) △ABD≅△ACD //(1), (4), (3), Angle-Angle-Side postulate
(6) AB = AC // Corresponding sides in congruent triangles (CPCTC)
Now let’s repeat it, using AD, the angle bisector:
(1) ∠ACB ≅ ∠ABC //Given
(2) AD = AD // Common side to both triangles, reflexive property of equality
(3) ∠CAD≅∠BAD // Given, AD is the bisector
(5) △ABD≅△ACD //(1), (3), (2), Angle-Angle-Side postulate
(6) AB = AC // Corresponding sides in congruent triangles (CPCTC)