# geometry - triangle

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)

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is this homework? –  user4140 Mar 15 '13 at 3:48
INMO question, right? –  user63477 Mar 15 '13 at 3:49
In fact we have $\,m=t+2x\;,\;\;n=t+x\,$ , so your last equality is wrong. –  DonAntonio Mar 15 '13 at 3:50
@SachinSharmaa: It was asked in one of the regional math olympiad(India)! –  user63477 Mar 15 '13 at 7:39

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!

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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 $\angle BAC=72^0$.

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How AD = AD (common ) of triangles ABC and ACD. –  Sachin Sharmaa Mar 17 '13 at 14:48
Oh sorry! That was a sure typo. –  user63477 Mar 18 '13 at 4:16