Question 140
Let $ABC$ be an acute triangle and $(O)$ be its circumcircle. Denote by $H$ its orthocenter and $I$ the midpoint of $BC$. The lines $BH$, $CH$ intersect $AC$, $AB$ at $E$, $F$ respectively. The circles $(IBF)$ and $(ICE)$ meet again at $D$.
1. Prove that $D$, $I$, $A$ are collinear and $HD$, $EF$, $BC$ are concurrent.
2. Let $L$ be the foot of the angle bisector of $\angle BAC$ on the side $BC$. The circle $(ADL)$ intersects $(O)$ again at $K$ and intersects the line $BC$ at $S$ out of the side $BC$. Suppose that $AK$, $AS$ intersect the circle $(AEF)$ again at $G$, $T$ respectively. Prove that $TG = TD$.
1. Prove that $D$, $I$, $A$ are collinear and $HD$, $EF$, $BC$ are concurrent.
2. Let $L$ be the foot of the angle bisector of $\angle BAC$ on the side $BC$. The circle $(ADL)$ intersects $(O)$ again at $K$ and intersects the line $BC$ at $S$ out of the side $BC$. Suppose that $AK$, $AS$ intersect the circle $(AEF)$ again at $G$, $T$ respectively. Prove that $TG = TD$.