# geometry – 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, $$overline{rm BD}= 2sin{frac{x}{2}}$$, $$overline{rm BH}= 4sin^2{frac{x}{2}}= 2(1-cos{x}), quad overline{rm CH} = 2cos{x}-1$$
$$overline{rm CE}=frac{2cos{x}-1}{sqrt{2-2cos{x}}}$$
Let $$overline{rm DE}=y$$,

since $$bigtriangleup DCE = bigtriangleup HCE$$,
$$frac{1}{2}ysin{50^{circ}}=frac{1}{2}sin{x}frac{(2cos{x}-1)^2}{2-2cos{x}}$$
Then by applying law of cosines to $$bigtriangleup DEC$$,
$$y^2+1-2ycos{50^{circ}}=frac{(2cos{x}-1)^2}{2-2cos{x}}$$
So we have a system of equations
$$begin{cases}ysin{50^{circ}}=sin{x}frac{(2cos{x}-1)^2}{2-2cos{x}}\y^2+1-2ycos{50^{circ}}=frac{(2cos{x}-1)^2}{2-2cos{x}} end{cases}$$
But it’s too messy to solve since 50 is not special angle.
How can I solve this problem?