In ΔABC, ∠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 ∠ACB = ?

Answer: Option B

According to question,

Let ∠CAB = x and ∠CBA = y
$$\begin{array}{rl}& \Rightarrow \angle CAD=\frac{180-x}{2}=90-\frac{x}{2}\\ & \text{and}\\ & \Rightarrow \angle EBC=\frac{180-y}{2}=90-\frac{y}{2}\\ & \text{also}\angle AEB=\angle EAB=x\end{array}$$
(∵ AB = EB ⇒ ABE is an isosceles triangle)
and ∠ADB = ∠ABD = y
(∵ AB = AD ⇒ ADB is an isosceles triangle)
In ΔAEB,
∠AEB + ∠ABE + ∠BAE = 180°
x + x + y + 90 - $$\frac{y}{2}$$ = 180°
⇒ 4x + y = 180°
Similarly in ΔADB
4y + x = 180°
⇒ 4y + x + 4x + y = 180 + 180
⇒ 5x + 5y = 360°
⇒ x + y = 72°
In triangle ABC,
∠ACB + x + y = 180°
⇒ ∠ACB = 180 - 72
⇒ ∠ACB = 108°