prove that angles opposite to equal sides of an isosceles triangle are equal
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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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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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