# Let $K$ be the midpoint of $DJ \perp HI$ $(J \in HI)$. Prove that $K \in (ABC)$.

$$(A, AD)$$ intesects $$(ABC)$$ at $$E$$ and $$F$$ where $$AD$$ is the altitude of $$\triangle ABC$$. Let $$D'H \parallel EF$$ and $$D'$$ is the reflection of $$D$$ in $$G$$, which is the foot of the perpendicular line to $$EF$$ from $$D$$ $$(H \in BC)$$. $$I$$ is the reflection of $$D$$ in point $$A$$. Let $$K$$ be the midpoint of $$DJ \perp HI$$ $$(J \in HI)$$. Prove that $$K \in (ABC)$$.

Because $$A$$ is the midpoint of $$DI$$, we might need to prove that $$AK \parallel IJ \implies AK \perp DK$$

$$\implies K, D$$ and $$L$$ are collinear where $$AL$$ is the diameter of $$(ABC)$$.

Further than that, I don't know how.

• Did you mean $(A,AD)$ intersects $(ABC)$ at $E$ and $F$? – steven gregory Sep 18 at 11:55
• Yup. It was a typo. – Lê Thành Đạt Sep 18 at 11:59

Let $$d$$ denote the dilation about the center $$D$$ with factor $$+1/2$$. Then $$d(D')=G$$, $$d(I)=A$$, $$d(J)=K$$. In particular $$d(\overline{IH})=\overline{AK}$$ and $$d(\overline{D'H})=\overline{EF}$$, where $$\overline{XY}$$ is the straight line passing through $$X$$ and $$Y$$. Consequently $$\overline{AK}$$ meets $$\overline{EF}$$ at the point $$H'$$ which is the midpoint of $$DH$$.
Now let $$i$$ be the inversion about the circle centered at $$A$$ with radius $$|AD|$$. Then $$i$$ sends the circle $$\mathcal{C}$$ circumscribing $$\triangle ABC$$ to $$\overline{EF}$$ and vice versa. Particularly, as $$H'$$ is on $$\overline{EF}$$ we know that $$i(H')$$ must lie on $$\mathcal{C}$$. Since $$\triangle AKD \sim \triangle ADH'$$ we get $$\frac{|AK|}{|AD|}=\frac{|AD|}{|AH'|}$$ so that $$|AK|\cdot |AH'|=|AD|^2$$. This means $$K=i(H')$$ so $$K$$ is on $$\mathcal{C}$$.