If the bisector of an angle of a triangle bisects the opposite side, prove that the triangle is isosceles.

Answer :

We are given a point D on side BC of a $∆$ ABC such that

$\angle$ BAD = $\angle$ CAD and BD = CD

We want to prove that AB =AC.

Now we produce our line AD to E , As AD  = DE , Then join CE  , Now we have our figure , As :



Now In $∆$ ABD and $∆$ ECD

BD  =  CD  ( Given  )

AD  =  ED  ( By construction )

$\angle$ ADB  =  $\angle$  EDC  ( Vertically opposite angles )

Hence

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

So,

AB  =  EC  -------------------- ( 1 )  ( BY CPCT )
And
$\angle$  BAD  =  $\angle$  CED   -------------------- ( 2 )  ( BY CPCT )
So we get

$\angle$ CED  =  $\angle$ CAD  ( As given $\angle$ BAD  =  $\angle$ CAD  )

So from base angle theorem we get

AC  =  EC 
So,

AB  =  AC  ( As we know from equation 1 , AB  = EC )

Hence

In $∆$ ABC , AB  = AC , So $∆$ ABC is a isosceles triangle .  ( Hence proved )