Question

In the figure ABC is a triangle in which AB = AC and AD is an
altitude on BC. Prove that AD bisects ∠A

Answer 1

Hey mate here is your answer:

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→ABC is an isosceles △.

→AB=AC ,

→AD is altitude.

→∠ADB=∠ADC=90° .

→To prove:-  

→(I) AD bisects BC, i.e., BD=CD ,

→(ii) AD bisects ∠A, i.e., ∠BAD=∠CAD,

→Proof:-

→In △ADB and △ADC,

→∠ADB=∠ADC[Each 90°] ,

→AB=AC[Given] ,

→AD=AD[Common] ,

→By R.H.S congruency,

→△ADB≅△ADC ,

→By C.P.C.T.

→BD=CD ,

→∠BAC=∠CAD.

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Hope it helps you//