Solve for $x$ in the $80^\circ$-$80^\circ$-$20^\circ$ triangle I've been solving this for days but I still I couldn't solve this. Can I know how to solve $x$ ?
I fail to see the importance and difficulty of Langley’s Problem mentioned by Justin Benfield in his answer here apart perhaps some historical mention or maybe some interesting applications. In effect, here I give an EASY generalization of the problem proposed by the OP. I would appreciate any substantiated comment against (am I wrong?).

Let $\triangle{ABC}$ not necessarily isosceles. Given side $a$ and angles $\alpha,\beta,\gamma,\delta$ determine the angles $\phi_1,\phi_2,\phi_3,x,y$ as well as the segments $BE,BD,CD$ and the  said difficulty would be to determine $x$ or $y$ since $\phi_1,\phi_2,\phi_3$  are  immediate to calculate



We have
$$BE=\frac{\sin \beta}{\sin \phi_1}\cdot a\\BD=\frac{\sin \phi_3}{\sin \phi_2}\cdot a\\CD=\frac{\sin \alpha}{\sin \phi_2}\cdot a\\ED=BE\cdot\frac{\sin \delta}{\sin x}=\frac{\sin \beta\sin\delta}{\sin \phi_1\sin x}\cdot a $$ It follows the easy system
$$\begin{cases}x+y=\phi_3\\\space\\\dfrac{\sin (\phi_1+\phi_3-x)}{\sin x}=\dfrac{\sin\phi_1\sin\phi_3}{\sin\beta\sin\phi_2}\end{cases}$$ This gives an elementary equation of the form $$A\sin x+B\cos x=C$$ one of the quite popular way to solve is the following:
$$A\sin x+B\cos x=C\iff\frac{A\sin x}{\sqrt{A^2+B^2}}+\frac{B\cos x}{\sqrt{A^2+B^2}}=\frac{C}{\sqrt{A^2+B^2}}$$ so $$x+\theta= \arcsin{\frac{C}{\sqrt{A^2+B^2}}}\space\text{where } \space \theta=\arccos{\frac{A}{\sqrt{A^2+B^2}}}$$
Hey I got $\angle x=20°$

I did it by constructing triangle as per your requirement

(Not the correct way)