Solution. Let $F$ be the midpoint of segment $AB$.

![](https://cdn.mathpix.com/cropped/2024_05_02_22edec424ec968fe1a03g-1.jpg?height=542&width=542&top_left_y=220&top_left_x=781)

Since triangle $ABC$ is half of a square, $CF = AF$, and triangles $ACF$ and $BCF$ are isosceles right triangles. $\angle CAP = \angle PCG$, because $CAP$ and $PCG$ form a pair of complementary angles (both are acute angles).

$\angle HAF = \angle GCF = 45^{\circ} - \angle PCG$. Triangles $GCF$ and $HAF$ are congruent because two adjacent sides and the included angle are equal. These two triangles are $90^{\circ}$ rotations of each other, because $CF$ is perpendicular to $AF$ and $CG$ is perpendicular to $AH$, so $FG$ and $FH$ are also perpendicular to each other.

This means that segment $GH$ is seen at a right angle from the midpoint of segment $AB$.
In the isosceles right triangle $ABC$, the right angle is at $C$. Let $P$ be any point on the side $BC$, and let $G$ be the orthogonal projection of point $C$ onto $AP$. Let $H$ be the point on segment $AP$ such that $AH = CG$. At what angle is the segment $GH$ seen from the midpoint of $AB$?