But every thing yields the same result!

Let $\angle C = \theta$ , then $\space \angle B = 2\theta$ , $\space \angle BAC = \angle A = 180^\circ - 3\theta$ , $\space \angle DAB = \angle DAC = 90^\circ - \dfrac {3\theta}{2}\space$ and $\angle BDA = 180^\circ - 90^\circ + \dfrac {3\theta}{2} - 2\theta = 90^\circ - \dfrac {\theta}{2}$ .

Using Sine Rule, we have:

$\dfrac {CD}{AD} =\dfrac {\sin{\left(90^\circ - \frac {3\theta}{2}\right)}}{\sin{\theta}} \space$ and $\space \dfrac {AB}{AD} =\dfrac {\sin{\left(90^\circ - \frac {\theta}{2}\right)}}{\sin{2\theta}}$ .

Since $\space AB = CD$ ,

$\begin{aligned} \Rightarrow \dfrac {\sin{\left(90^\circ - \frac {3\theta}{2}\right)}}{\sin{\theta}} & = \dfrac {\sin{\left(90^\circ - \frac {\theta}{2}\right)}}{\sin{2\theta}} \\ \dfrac {\cos{\frac {3\theta}{2}}}{\sin{\theta}} & = \dfrac {\cos{\frac {\theta}{2}}}{\sin{2\theta}} \\ \cos{\frac {3\theta}{2}} & = \dfrac {\cos{\frac {\theta}{2}}}{2\cos{\theta}} \\ 2\cos{\theta}\cos{\frac {3\theta}{2}} & = \cos{\frac {\theta}{2}} \\ 2\cos{\theta} \left(4\cos^3{\frac {\theta}{2}} - 3\cos{\frac {\theta}{2}} \right) & = \cos{\frac {\theta}{2}} \\ 2\cos{\theta} \left(4\cos^2{\frac {\theta}{2}} - 3 \right) & = 1 \\ 2\cos{\theta} \left(2\cos{\theta} -1 \right) & = 1 \\ 4\cos^2{\theta} - 2\cos{\theta} -1 & = 0 \\ \Rightarrow \cos{\theta} & = \begin{cases} \frac{1+\sqrt{5}}{4} & \Rightarrow \theta = 36^\circ \\ \frac {1- \sqrt{5}}{4} & \Rightarrow \theta = 108^\circ \quad \text{unacceptable} \end{cases} \end{aligned}$

Therefore, $\space \angle BAC = 180^\circ - 3\theta = 180^\circ - 3\times 36^\circ = \boxed{72^\circ}$