Question

Proof that angles opposite to equal sides of an iscolelus triangle are equal

Answer 1

in triangle ABC drop a perpendicular from a on bc and name it d as a point on bc ab = bc (sides of an isosceles triangle) ad = ad (common side) angleADB = angleADC =90° hence triangle ADB is congruent to triangle ADC thererefore angleB = angleC (c.p.c.t)