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$\triangle ABC$ is a right-angled triangle at $A$. Circle centre $D$, diameter $AB$ cuts $BC$ and $CD$ respectively at $E$ and $F$. $M$ is the midpoint of $BE$. $d$ is a line that passes through point $A$ and is perpendicular to $CD$ at point $H$. $d$ cuts $BC$ and $DM$ respectively at $K$ and $I$. $FK \cap (D) = J$ ($J \not\equiv F$). Prove that $IJ$ passes through the reflection of point $F$ in point $D$.

That means $\widehat{IJK} = 90^\circ \Longleftarrow IJMK$ or $IJHF$ is a cyclic quadrilateral $\Longleftarrow \left [ \begin{align} \widehat{JIM} = \widehat{JKM}\\ \widehat{JIH} = \widehat{JFH} \end{align} \right.$.

Neither of which have I figured the way to solve the problem.

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$$\frac{KJ}{KI}=\frac{FJ-FK}{KI}=\frac{2r\frac{FH}{FK}-FK}{KI}=\frac{2r\frac{r-DH}{FK}-FK}{KI}=\frac{2r^2-2rDH-FK^2}{FK\cdot KI}=\frac{2r^2-2rDH-HF^2-HK^2}{FK\cdot KI}=\frac{2r^2-2rDH-(r-DH)^2-HK^2}{FK\cdot KI}=\frac{r^2-DH^2-HK^2}{FK\cdot KI}=\frac{AH^2-HK^2}{FK\cdot KI}=\frac{CH\cdot DH-HK^2}{FK\cdot KI}=\frac{HK\cdot HI-HK^2}{FK\cdot KI}=\frac{HK(HI-HK)}{FK\cdot KI}=\frac{HK}{FK}$$

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  • $\begingroup$ I'm sorry for my over-the-top editing. Hope you can understand. Your answer is great. $\endgroup$ – Lê Thành Đạt Apr 30 '19 at 4:10

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