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.

Draw equilateral triangle $DCE$. Then $\angle ACE = 20^{\circ}$ and since triangle $ACE$ isosceles we have $\angle CAB =80^{\circ}$ (so $B,A,E$ are collinear) and $\angle FED = 20^{\circ}$. Now $\angle DFE = \angle FDE = 80^{\circ}$ we have $EF = ED (=EC)$ so $E$ is a center of circle through points $C,D,F$ and thus $$\angle DCF ={1\over 2}\angle DEF =10^{\circ}$$

