# What is the length of $x$ in this pentagon diagram?

ABCDE is a regular pentagon. $$\angle AFD = \angle EKC$$

$$|FH|=1$$ cm; $$|AH|=3$$ cm

What is $$|DK|?$$ I know that triangles $$EFA$$ and $$DEK$$ are similar and that $$|EK|=4$$ cm. Also because this is a regular pentagon each one of the interior angles are $$108^o$$. Naming similar angles inside the pentagon, I tried to find an isosceles triangle, but I couldn't. I can't progress any further from here.

How can I solve this problem?

Answer: $$x=2$$.
Since $$\angle EFA=\angle DKE$$, $$\angle AEF=\angle EDK$$ and $$AE=ED$$ we obtain $$\triangle AEF=\triangle EDK$$ (they are similar and corresponding sides are equal). Hence, $$EF=DK=x$$ and $$\angle FEH=\angle DEK=\angle EAF$$. Therefore, trinagles $$\triangle FEH$$ and $$\triangle FAE$$ are similar, so $$\frac{FE}{FH}=\frac{FA}{FE}.$$ It means that $$x^2=FE^2=FA\cdot FH=4\cdot 1=4$$. Thus, $$x=2$$.
Let $$\measuredangle FEH=\measuredangle EAF=\alpha.$$
Thus, by your work and by law of sines we obtain: $$\frac{x}{\sin{\alpha}}=\frac{4}{\sin108^{\circ}}$$ and $$\frac{x}{\sin108^{\circ}}=\frac{1}{\sin\alpha},$$ which gives $$x^2=4$$ and $$x=2.$$