# Geometry - triangle problem

• Mar 14th 2013, 07:36 PM
sachinrajsharma
Geometry - triangle problem
ABC is a triangle in which $\angle B = 2 \angle C.$ If D is a point on BC such that AD bisects $\angle BAC and AB =CD$Prove that $\angle BAC = 72^{\circ}$

Can we go this way :

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

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

$\angle m = \angle n + \angle 2x$( as m is external angle which is equal to sum of the opposite interior angles)
• Mar 14th 2013, 11:38 PM
ibdutt
Re: Geometry - triangle problem
• Mar 15th 2013, 05:03 AM
sachinrajsharma
Re: Geometry - triangle problem
thanks a lot...