8.2. In a convex quadrilateral $\mathrm{ABCD}$, the bisector of angle $\mathrm{B}$ passes through the midpoint of side $\mathrm{AD}$, and $\angle \mathrm{C}=\angle \mathrm{A}+\angle \mathrm{D}$. Find the angle $\mathrm{ACD}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_b8f46b2c2d4464c00b2dg-1.jpg?height=274&width=423&top_left_y=2213&top_left_x=291)
Solution. Let $\mathrm{E}$ be the midpoint of side AD, and $\mathrm{F}$ be the intersection point of $\mathrm{BE}$ and AC. From the condition, we have: $\angle \mathrm{B}=360-2(\angle \mathrm{A}+\angle \mathrm{D})$, from which $\angle \mathrm{AEB}=180-\angle \mathrm{A}-\angle \mathrm{B} / 2=\angle \mathrm{D}$. Therefore, $\mathrm{BE} \| \mathrm{CD}$, and $\mathrm{EF}$ is the midline of triangle $\mathrm{ACD}$, meaning $\mathrm{AF}=\mathrm{FC}$. Thus, $\mathrm{BF}$

is the bisector and median of triangle $\mathrm{ABC}$, and therefore its altitude. Consequently, line $\mathrm{CD}$, parallel to $\mathrm{BF}$, is also perpendicular to $\mathrm{AC}$, from which the answer follows.

Answer. Angle ACC is 90.

Criteria. Only the answer: 0 points.