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 DC from D that intersect AC at E. Draw a circle on C, D and E. Extend BA equal it's measure to get point A'. Connect A' to C. Clearly $\angle A'CB=90^o$. A'C intersect the circle at G. So OG is diameter of the circle. Extend DF to touch the circle at H. Connect H to E. Clearly OG bisects EH and the arc EGH, because:
Let intersection of circle and BC be I, then $arc(EG)=arc (IC)$
$\rightarrow (CE=180)-EG=(GI=180)-GH\rightarrow arc (IH)=arc (CG)\rightarrow\angle GEC=\angle KEI$,also $\angle EIG=\angle ECG$.
Triangle EGC and KEI have two equal angles so their third angles are equale, Sonce $\angle EGC=90^o$ therefore $\angle EKI=90^i$ that is OG is perpendicular on chord EH, so it bisects EH and it's arc EGH, hence G is midpoint of arc EH.
This results in:
$\angle EDG=\angle GDH=\angle ECG$
But $\angle EDH=40^o$, therefore :
$\angle EDG=\angle GDH=\angle ECG=20^o$
Triangle AA'C is isosceles and we have:
$\angle AA'C=\angle ACA'=20^o$
Hence $\angle BAC=40^o$
