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$(A, AD)$ 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 $K \in (ABC)$.

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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.

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

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    $\begingroup$ This answer is just so neat. Thanks for that. $\endgroup$ – Lê Thành Đạt Sep 21 at 15:26

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