This is great

$\text{Let BE be the angle bisector of angle B. Angles are shown in the sketch.} \\ Applying~ Sin~ Law,~\dfrac a b =\dfrac{SinA}{SinB},~ we~ get:-\\ In~~\Delta~ABD~~\angle~ADB=X+Y, internal ~angles~of~\Delta~ADC.\\ \dfrac{AB}{ AD}= ~~\dfrac{ Sin(X+Y)}{Sin(2X)}.\\ In~~\Delta~ADB~~\dfrac{ DC=AB}{Sin(AD)}= ~~\dfrac{Sin(Y)}{Sin(X)}.\\ \therefore~\dfrac{ Sin(X+Y)}{Sin(2X)}=\dfrac{Sin(Y)}{Sin(X)}\\ \implies~~\dfrac{ Sin(X+Y)}{Sin(Y)}=\dfrac{Sin(2X)}{Sin(X)}\\ \therefore~Sin(X)Cot(Y)+Cos(X) = 2Cos(X), \text{expanding Sin of sum.}\\ \therefore~Cot(Y)=Cot(X),~~~\implies~Y=X.\\ But~180=2Y+3X=5Y,~~\implies~2Y=\Large~~~~\color{#D61F06}{72^o}$

Let $\angle ABC =2\alpha$ and $\angle BAD = \beta$ . If we draw a line segment $\overline{DF}$ such that $\angle CDF=\alpha$ we have that triangles $ABD$ and $ADF$ are congruent, so $\overline{AB}=\overline{AF}$ and $\overline{BD}=\overline{DF}$ . Also we have that triangle $CDF$ is isosceles, so $\overline{DF}=\overline{CF}$ . As $\overline{CD}=\overline{AF}$ and $\overline{BD}=\overline{CF}$ , triangle $ABC$ is isosceles, so $\beta=\alpha=36^{o} \rightarrow \angle A = 72^{o}$