Find $\angle DCF$.

Question

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.

Answer 1

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 $$\angle DCF ={1\over 2}\angle DEF =10^{\circ}$$

Answer 2

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 $\angle DCF =x=10^\circ$.