Question

show that in an isosceles triangle angle opposite to equal sides equal

Answer 1

Take a triangle ABC, in which AB=AC.
Construct AP bisector of angle A meeting BC at P.
In ∆ABP and ∆ACP 
AP=AP[common]
AB=AC[given]
angle BAP=angle CAP[by construction]
Therefore, ∆ABP congurent ∆ACP[S.A.S]
This implies, angle ABP=angleACP[C.P.C.T]
Hence proved that angles opposite to equal sides of a triangle are equal.
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Answer 2

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