In the adjoining figure ABC is an isosceles triangle in which AB = AC. Also, D is a point such that BD = CD. Prove that AD bisects angle A and angle D.

Answer :

we form our diagram from given information , As  :




Now In $∆$ ABD and $∆$ ACD 

AB =  AC                                           (  Given )

BD  =  CD                                          ( Given D is  mid point of BC )

And

$\angle$ ABD  =  $\angle$ ACD                      (  We know base angle of isosceles triangle are equal , So $\angle$ ABC  =  $\angle$ ACB  )

Hence

$∆$ ABD $\cong$ $∆$ ACD                    ( By SAS rule )

So,

$\angle$ BAD  =  $\angle$ CAD       and  $\angle$ ADB =  $\angle$ ADC                         (  BY CPCT )

So,

We can say that AD bisect both $\angle$ A and $\angle$ D    .                        (  Hence proved )