Question

Prove that in an isosceles triangle the angle opposite to equal sides are equal.
In ΔABC with Angle B=90°, AB=BC. Using above theorem, find Angle A​

Answer 1

Answer:

. Take a triangle ABC, in which AB=AC. Construct AP bisector of angle A meeting BC at P. Hence proved that angles opposite to equal sides of a triangle are equal.

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Answer 2

Answer:

in isoceles triangle we know that two side are equal than two angle are equal Angle B is 90° and angle A and B are also equal A+B+C =180° then 180-90 = 90°. two angle are equal then 90/2 = Angle A = 45° and angle B= 45° thank you