From A draw the altitude AH intersecting BC in H
Let's prove that triangle ABH is congruent to triangle ACH
The above 2 triangles are right triangles due to the altitude AH
Angle B= Angle C (given)
Angle AHB = Angle AHC =90° (since AH is the altitude)
Then angle BAH = CAH (both complementary to B & C respectively
And AH is a common side
Now Tri. ABH = Tri. ACH because ASA, hence AB=AC
