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

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

What is $|DK|?$

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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?

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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, $x=2$.

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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 $$x=2.$$

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