# Geometry - triangle problem

• Mar 14th 2013, 07:36 PM
sachinrajsharma
Geometry - triangle problem
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)
• 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...