Let $\triangle ABC$ be a right-angled triangle with $\angle B = 90^\circ$. Let $\overline{AD}$ be the bisector
of $\angle A$ with $D$ on $\overline{BC}$. 

Let the circumcircle of $\triangle ACD$ intersect $\overline{AB}$
again in $E$, and let the circumcircle of $\triangle ABD$ intersect $\overline{AC}$ again in
$F$. 

Let $K$ be the reflection of $E$ with respect to the line $\overline{BC}$. 

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> Prove that $|\overline{FK}| = |\overline{BC}|$.
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[![enter image description here](https://i.stack.imgur.com/VCCMy.png)](https://i.stack.imgur.com/VCCMy.png)