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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 $\overline{BC}$.

Prove that $|\overline{FK}| = |\overline{BC}|$.

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Let $AB=b$ and $BC=a$.

Thus, since $AD$ is a bisector of $\Delta ABC$, we obtain: $$\frac{BD}{DC}=\frac{AB}{AC}$$ or $$\frac{BD}{a-BD}=\frac{b}{\sqrt{a^2+b^2}}$$ or $$BD=\frac{ab}{b+\sqrt{a^2+b^2}}.$$ Now, $$BE\cdot AB=BD\cdot BC$$ or $$BE\cdot b=\frac{a^2b}{b+\sqrt{a^2+b^2}}$$ or $$BE=\frac{a^2}{b+\sqrt{a^2+b^2}}$$ or $$BE=\sqrt{a^2+b^2}-b,$$ which says that $BE=FC$ and from here $BK=FC$.

Thus, since $BD=DF$ and $$\measuredangle KBD=\measuredangle CFD=90^{\circ},$$ we obtain $$\Delta BDK\cong\Delta FDC,$$ which says $D\in KF$ and $BC=KF$.

Done!

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  • $\begingroup$ Can you send me the figure of it $\endgroup$ – Dwip Dalal Oct 7 '17 at 2:56

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