Find $\angle DCF$. 
Let $ABC$ an isoscel triangle with $AB=AC$ with $\angle{BAC}=100^{\circ}$ and $D\in (BC)$ s.t. $AC=DC$ and $F$ on $(AB)$ with $DF||AC$.
Find $\angle DCF$.

I tried a lot of constructions, to find an inscriptible quadrilateral but I didn't succeed.

Let $\angle DCF =x$ and $AB =AC=CD=1$. The given $FD || AC$ leads to similar triangles BDF and BCA, and 
$\frac{AF}{AC} = \frac 1{2\sin50}$. Then, apply the sine rule to the triangle FCA
$$\frac{\sin\angle ACF}{\sin\angle AFC}=\frac{AF}{AC}
\implies\frac{\sin(40-x)}{\sin(40+x)}=\frac1{2\sin50}$$
or 
\begin{align}
\sin x &=\cos(10+x)-\sin(40+x) \\
&=\cos(10+x)-\cos(50-x)=\sin(20-x)
\end{align}
which yields the solution $\angle DCF =x=10^\circ$.
