3
$\begingroup$

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)

$\endgroup$
7
  • $\begingroup$ is this homework? $\endgroup$
    – Asinomás
    Mar 15, 2013 at 3:48
  • $\begingroup$ INMO question, right? $\endgroup$
    – Inceptio
    Mar 15, 2013 at 3:49
  • $\begingroup$ In fact we have $\,m=t+2x\;,\;\;n=t+x\,$ , so your last equality is wrong. $\endgroup$
    – DonAntonio
    Mar 15, 2013 at 3:50
  • $\begingroup$ in which olympiad it was asked @Inceptio... please suggest and also suggest the source of solutions of Olympiad questions... please help.. $\endgroup$ Mar 15, 2013 at 5:30
  • $\begingroup$ @SachinSharmaa: It was asked in one of the regional math olympiad(India)! $\endgroup$
    – Inceptio
    Mar 15, 2013 at 7:39

3 Answers 3

3
$\begingroup$

enter image description here 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 $\angle BAC=72^0$.

$\endgroup$
2
  • $\begingroup$ How AD = AD (common ) of triangles ABC and ACD. $\endgroup$ Mar 17, 2013 at 14:48
  • $\begingroup$ Oh sorry! That was a sure typo. $\endgroup$
    – Inceptio
    Mar 18, 2013 at 4:16
0
$\begingroup$

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!

$\endgroup$
0
-1
$\begingroup$

A High level of critical thinking needed

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

Edit 1:

The solution can not be competed without the trigonometry.

$\endgroup$
1
  • $\begingroup$ i didnt under stand it can you dowithout trinometry $\endgroup$
    – user206291
    Jan 7, 2015 at 14:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .