Question

In an isosceles triangle ABC, AB= AC, angle bisector AD is drawn from < A to BC. Prove, giving reasons, that triangle ADB is congruent to triangle ADC.
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Answer 1

Step-by-step explanation:

In ΔABP and ΔACP,

∠APB=∠APC (Both equal to 90

0

)

AB=AC [∵ΔABC is an isosceles triangle.]

AP=AP (Common Side)

By R.H.S. criterion of congruence,

△ABP≅△ACP

⇒∠B=∠C (CPCT)

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