In ABC,We have

<B=2<C Or,<B=2Y,Where <C=y

AD is the bisector of <BAC.So let <BAD=<CAD=x

let BP be the bisector of <ABC.JOIN PD.

In BPC We have

<CBP=<BCP=y BP=PC

In ABP and DCP,We have

<ABP=<DCP=y

AB=DC                        [Given]

and, BP=PC                             [As proved above]

So, by SAS congruence criterion,we have

<BAP=<CDP and AP=DP

<CDP=2x and <ADP=DAP=x                 [ <A=2X]

In ABD,We have

<ADC=<ABD+<BAD x+2x=2y+xx=y

In ABC, We have

2x+2y+y=          

5x=           [ X=Y]

Hence,<BAC=2X=