A geometry problem by vishwash kumar

Const : $\text{Angle bisector of }\angle B \text{ meeting AC at E}$

Soln :

$\text{Let } \angle BAD = \angle CAD = x, \angle ABE = \angle CBE = y \\ \text{Given } \angle B = 2 \angle C \\ \implies \angle C = \dfrac{\angle B}{2} = y \\ \implies BE = EC ~ (\text{Sides opp to equal angles are equal}) \\ \text{In } \Delta ABC \\ \angle A + \angle B + \angle C = 180 \\ \boxed{2x + 3y = 180 ~ \rightarrow ~ 1} \\ \text{Now, in } \Delta ABC \text{ and } \Delta AEB \\ \angle ACB = \angle BEA = y \\ \angle BAE = \angle BAC = 2x ~(\text{common}) \\ \implies \Delta ABC ~ \Delta AEB ~ (\text{AA}) \\ \text{by cpst, } \dfrac{AB}{AE} = \dfrac{BC}{EB} = \dfrac{AC}{AB} \\ \dfrac{BC}{EB} = \dfrac{AC}{AB} \\ \dfrac{BC}{AC} = \dfrac{EB}{AB} \\ \dfrac{BC}{AC} = \dfrac{EC}{CD} ~ (\text{AB = CD, given. EC = BE, proved})\\ \text{In }\Delta CDA \text{ and } \Delta CEB \\ \dfrac{BC}{AC} = \dfrac{EC}{CD} \\ \angle C = \angle C ~ (\text{common}) \\ \implies \Delta CDA ~ \Delta CEB \\ \implies \angle CAD = \angle CBE \\ \implies x = y \\ \therefore 2x + 3y = 180 \\ 5y = 180 ~~ \text{and} ~~ 5x = 180 \\ y = x = 36 \\ \boxed{\therefore \angle BAC = 2x = 2 \times 36 = 72 \circle}$