Find "x" in the right triangle $ABC$, right angle at $B$. 
My progress:

I extended the lines and formed an equilateral triangle but an equation is still missing
$x+\theta = 60^\text{o}\\
x+ \alpha = 50^\text{o}\\
\ldots\ldots$
 A: Here's a possible path with Euclidean geometry, also following in part your construction.


*

*Construct equilateral triangle $\triangle CEF$.

*Produce $AC$ to $G$ so that $\measuredangle CFG = 20^\circ$.

*Observe that $\triangle AFG$ is isosceles, and so is $\triangle CGF$.

*By SAS criterion, show that $\triangle AGH \cong \triangle AFH$ and $\triangle CEH \cong\triangle CFH$.

*Conclude that $\triangle GHF$ and $\triangle EHF$ are isosceles and furthermore $\triangle GHF \cong \triangle EHF$ (SSS criterion).

*Thus $FH$ bisects $\angle GFA$.

*Conclude that $\measuredangle HEC = 20^\circ$.

A: By the law of sines in $\triangle EHC$, $EH=\frac{HC}{2\sin x}$. Since $\angle CEB=60$, $EH=\frac{BH}{\sin(60-x)}$. Thus $HC\sin(60-x)=2BH\sin x$. Hence
$$BC=HC\left(1+\frac{\sin(60-x)}{2\sin x}\right)$$
But notice that $AB\tan 10=BH$. Thus
$$BC=AB\tan 20=BH\frac{\tan 20}{\tan 10}=HC\frac{\sin(60-x)\tan 20}{2\sin x\tan 10}$$
Equating the two expressions for $BC$, we get that
$$\frac{2\sin x+\sin(60-x)}{2\sin x}=\frac{\sin(60-x)\tan 20}{2\sin x\tan 10}$$
$$2\sin x\tan 10+\sin(60-x)\tan 10=\sin(60-x)\tan 20$$
Can you finish from here? (All angles are in degrees)
