# Prove that $IJ$ passes through the reflection of point $F$ in point $D$.

$$\triangle ABC$$ 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.

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

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