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In triangle $ABC$, angle $C$ is an obtuse angle. The bisectors of the exterior angles at $A$ and $B$ meet $BC$ and $AC$ produced at $D$ and $E$ respectively. If $AB = AD = BE$, then the value of angle $ACB$ is::

$(i) 105$

$(ii) 108$

$(iii) 110$

$(iv) 135$

I have been trying to solve this question. But, I am just not able to get a clear diagram. Even if I take the angle $C$ as an obtuse angle , then the length of the sides $AB$ , $AD$ and $BE$ dont seeem to be equal in my diagram.

I need some serious help here please. And, if there is any trick to slve such questions , then please share it with me.enter image description here

  • 0
    What have you done besides try to get the right diagram? You can solve it with the diagram you posted, although it is not to scale. – 2017-02-16
  • 0
    Add some parallel lines. Make good use of alternate angles. You will find $\angle D = \angle E$. – 2017-02-16

1 Answers 1

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Let's call $\beta=\angle CBE$ and $\theta=\angle CAD$

Once $BE=BA$ then $\angle BEC=\angle A$ and once $AD=AB$ then $\angle ADC=\angle B$, so

$1)$ $\angle C=\angle A+ \beta=\angle B+\theta\to \angle A-\angle B=\ \theta-\beta$

$2)$ $\angle B+2\beta=\angle A+2\theta=180°\to \angle A-\angle B=2(\beta-\theta)$

Using $(1)$ and $(2)$ we conclude that $\theta=\beta$ and $\angle A=\angle B$.

$3)$ we also have that $\angle C+2\angle A=180°$, by triangle $ABC$,

$4)$ $\angle C=\angle A+\theta=A+\frac{180°-\angle A}{2}\to \angle A=2\angle C-180°$

Now using $(3)$ and $(4)$ we get

$$\angle C+2(2\angle C-180°)=180° \to \angle C=108°$$