ABC is a triangle in which $ \angle B = 2 \angle C$ D is a point on BC such that AD bisects $\angle BAC$ and AB = CD. Prove that $\angle BAC =72^{\circ}$
Here $\angle BAD = \angle CAD$ AB = DC
Can we go this way :
Let $\angle A = 2t ; \angle B = 2x ; \angle C = x $( 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)
Best Answer
$\angle ACB$ is common. And $AB=CD$, $AC=AC$(Common).
Therefore the triangles are similar (Two sides have lengths in the same ratio, and the angles included between these sides have the same measure).
Which means,
$\frac{AB}{AC}=\frac{AC}{AB}$, and the angle between them is common( $\angle ACB$)
Now,
$\angle CAD = \angle ACB =c$
Thus, $a=c$
Therefore,
$5a=180^0$. $a=36^0$ and $\angle BAC=72^0$.