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6. Prove that: Angles opposite to equal sides of an isosceles triangle are equal. |
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Answer» 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. *PLEASE rate it brainliest* Step-by-step explanation: |
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