Finding an angle in a triangle 
I want to find angle x in this picture.

And this is what I've done so far. Without loss of generality, assume $\overline{\rm BC}=1$
then, $$\overline{\rm BD}= 2\sin{\frac{x}{2}}$$, $$\overline{\rm BH}= 4\sin^2{\frac{x}{2}}= 2(1-\cos{x}), \quad \overline{\rm CH} = 2\cos{x}-1$$
$$\overline{\rm CE}=\frac{2\cos{x}-1}{\sqrt{2-2\cos{x}}}$$
Let $\overline{\rm DE}=y$,
since $\bigtriangleup DCE = \bigtriangleup HCE$,
$$\frac{1}{2}y\sin{50^{\circ}}=\frac{1}{2}\sin{x}\frac{(2\cos{x}-1)^2}{2-2\cos{x}}$$
Then by applying law of cosines to $\bigtriangleup DEC$,
$$y^2+1-2y\cos{50^{\circ}}=\frac{(2\cos{x}-1)^2}{2-2\cos{x}}$$
So we have a system of equations
$$\begin{cases}y\sin{50^{\circ}}=\sin{x}\frac{(2\cos{x}-1)^2}{2-2\cos{x}}\\y^2+1-2y\cos{50^{\circ}}=\frac{(2\cos{x}-1)^2}{2-2\cos{x}} \end{cases}$$
But it's too messy to solve since 50 is not special angle.
How can I solve this problem?


Draw a perpendicular on CD from D., it touches AC at F. Extend CD. Draw a circle on A and D tangent on CD at D. For this draw perpendicular bisector of AD  it intersects the perpendicular on ED at O which is the center of circle.This circle intersects  extension of CD at H. Since DF is perpendicular on DH at D , the points F, O and H are on diameter of the circle. In this way $\angle FAH=90^o$  and we have:
$\angle FAH=90^o$
DE is tangent on the circle and $Arc (HD)=2\times 50=100^o$. Hence $\angle DAH=\frac {100}2=50^o$ which results in $\angle BAC=40^o$.
