in triangle ABC , AD is the bisector of such that AD is perpendicular to BC. Prove that triangle ABC is an isoceles triangle

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Please find below the solution to the asked query:

We form diagram from given information , As :


In $∆$ ABD and   $∆$ ACD 

$\angle$ BAD =  $\angle$ CAD                                        ( Given AD is angle bisector )

AD  = AD                                                       ( Common side )

And

$\angle$ BDA =  $\angle$ CDA = 90$°$                             ( Given AD is also perpendicular on BC )

So,

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

Then,

 $\angle$ ABD  =  $\angle$ ACD                                      ( By CPCT )                   --- ( 1 )

We know from base angle theorem  ( As $\angle$ ABC  =  $\angle$ ACB  from equation 1 )  that AB  =  AC  , So

$∆$ ABC is a isosceles triangle .                                                                   ( Hence proved )


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