8.3. Given a triangle $A B C$, a point $D$ is chosen on side $B C$ and a point $H$ on side $A C$. Additionally, $D K$ is the angle bisector of triangle $B D A$. It turns out that angles $C H D$ and $H D K$ are right angles. Find $H C$, if $A C=2$.
Answer: $H C=1$.

Solution: Since angles $C H D$ and $H D K$ are right angles, line $K D$ is parallel to line $A C$. Therefore, angles $H A D$ and $A D K$ are equal as alternate interior angles, and angles $D C A$ and $B D K$ are equal as corresponding angles. Since $D K$ is the angle bisector, angles $D A C$ and $D C A$ are equal, hence triangle $A D C$ is isosceles, and $H D$ is the altitude in an isosceles triangle, which means it is also the median. Therefore, $H C=A C / 2=1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_246cf2059f4db554afa0g-1.jpg?height=474&width=897&top_left_y=2016&top_left_x=1025)

Criteria: only the answer - 0 points.

Proved that lines $K D$ and $A C$ are parallel - 1 point.

Proved that $A D C$ is isosceles or that $D H$ is the angle bisector - 4 points.