geometry - triangle

Question

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)

Answer 1

Observe Triangles $ACB$ and $ADC$.

$\angle ACB$ 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 $\angle BAC=72^0$.

Answer 2

Hint:

This is rather nice and easy problem. All you need:

• properties of inscribed angles,
• properties of parallel lines,
• properties of isosceles triangles.

Good luck!

Answer 3

Answer cannot be 108 degrees because 3x must be smaller than 180 degrees.

Edit 1:

The solution can not be competed without the trigonometry.