ABC is a triangle in which $\displaystyle \angle B = 2 \angle C.$ If D is a point on BC such that AD bisects $\displaystyle \angle BAC and AB =CD$Prove that $\displaystyle \angle BAC = 72^{\circ}$

Can we go this way :

Let $\displaystyle \angle A = 2t ; \angle B = 2x ; \angle C = x$[/latex] ( as $\displaystyle \angle B = 2\angle C$ Let $\displaystyle \angle ADC = m ; \angle ADB =n$ such that $\displaystyle \angle m + \angle n = 180^{\circ}$

Now in $\displaystyle \triangle ADB ; 2x+n+t =180^{\circ}$( since $\displaystyle \angle A = 2t$ and D is the bisector of $\displaystyle \angle BAC$ also $\displaystyle \angle m + \angle x + \angle DAC = 180^{\circ}$

$\displaystyle \angle m = \angle n + \angle 2x$( as m is external angle which is equal to sum of the opposite interior angles)