A geometry problem by Aly Ahmed

Let the height of triangle ABC is h, and DC=b, so BD= h√2. ΔABC is similar to ΔBDC : BC/DC = 2DC/BC, BC= b√2. [BDC] = hb/2 = ½ (h√2. b√2. sin x), six x= ½, x=30

Apply sine rule on $\triangle BDC$ . We have $\dfrac{DC}{\sin~x}=\dfrac{BC}{\sin~135}$ $\implies$ $\dfrac{DC}{BC}=\dfrac{\sin~x}{\sin~135}=\sqrt{2} \sin~x$

Apply sine rule on $\triangle ABC$ . We have $\dfrac{BC}{\sin~x}=\dfrac{AC}{135}$ $\implies$ $\dfrac{BC}{AC}=\dfrac{\sin~x}{\sin~135}=\sqrt{2} \sin~x$

Since $2DC=AC$ , we have

$2\sqrt{2}\sin~x(BC)=\dfrac{BC}{\sqrt{2}\sin~x}$ or

$\sin~x)^2=\dfrac{1}{4}$

$\sin~x=\dfrac{1}{2}$

$x= \sin^{-1} \left(\dfrac{1}{2}\right)$

$x=\dfrac{\pi}{6}$

or

$\color{#D61F06}\large \boxed{x=30^\circ}$

Using sine rule , we have:

$\begin{aligned} \frac {BD}{\color{#3D99F6}AD} & = \frac {BD}{\color{#3D99F6}AD} & \small \color{#3D99F6} \text{Note that }AD = DC \\ \frac {\sin \angle BAD}{\sin \angle ABD} & = \frac {\sin \angle BCD}{\sin \angle CBD} \\ \frac {\sin x}{\sin (135^\circ - x)} & = \frac {\sin (45^\circ - x)}{\sin x} \\ \sin^2 x & = \sin (45^\circ - x) \sin (135^\circ - x) \\ \frac 12 \left(1-\cos 2x \right) & = \left(\frac 1{\sqrt 2}\cos x - \frac 1{\sqrt 2}\sin x \right) \left(\frac 1{\sqrt 2}\cos x + \frac 1{\sqrt 2}\sin x \right) \\ \frac 12 \left(1-\cos 2x \right) & = \frac 12 (\cos^2x - \sin^2 x) \\ 1-\cos 2x & = \cos 2x \\ \implies \cos 2x & = \frac 12 \\ 2x & = 60^\circ \\ \implies x & = \boxed{30}^\circ \end{aligned}$