Let $A D, B F$ and $C E$ be the altitudes of $\triangle A B C$. A line passing through $D$ and parallel to $A B$ intersects the line $E F$ at the point $G$. If $H$ is the orthocenter of $\triangle A B C$, find the angle $\widehat{C G H}$.
We can see easily that points $C, D, H, F$ lie on a circle of diameter $[C H]$.

Take $\left\{F, G^{\prime}\right\}=\odot(C H F) \cap E F$. We have $\widehat{E F H}=\widehat{B A D}=\widehat{B C E}=\widehat{D F H}$ since the quadrilaterals $A E D C, A E H F, C D H F$ are cyclic. Hence $[F B$ is the bisector of $\widehat{E F D}$, so $H$ is the midpoint of the arc $D G^{\prime}$. It follows that $D G^{\prime} \perp C H$ since $[C H]$ is a diameter. Therefore $D G^{\prime} \| A B$ and $G \equiv G^{\prime}$. Finally $G$ lies on the circle $\odot(C F H)$, so $\widehat{H G C}=90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_b2c1a517537edbece730g-3.jpg?height=468&width=537&top_left_y=360&top_left_x=765)

## Solution 2

The quadrilateral $A E H F$ is cyclic since $\widehat{A E H}=\widehat{A F H}=90^{\circ}$, so $\widehat{E A D} \equiv \widehat{G F H}$.

But $A B \| G D$, hence $\widehat{E A D} \equiv \widehat{G D H}$. Therefore $\widehat{G F H} \equiv \widehat{G D H} \Rightarrow D F G H$ is cyclic.

Because the quadrilateral $C D H F$ is cyclic since $\widehat{C D H}=\widehat{C F H}=90^{\circ}$, we conclude that the quadrilateral $C F G H$ is cyclic, which gives that $\widehat{C G H}=\widehat{C F H}=90^{\circ}$.