Blinkov A.D:

On the side $AB$ of rectangle $ABCD$, a point $M$ is chosen. Through this point, a perpendicular to the line $CM$ is drawn, which intersects the side $AD$ at point $E$. Point $P$ is the foot of the perpendicular dropped from point $M$ to the line $CE$. Find the angle $APB$.
Since

$$
\angle M A E=\angle M P E=\angle C P M=\angle M B C=90^{\circ}
$$

(see figure), quadrilaterals $A E P M$ and $B M P C$ are cyclic. Therefore, $\angle B P A=\angle B P M+\angle M P A=\angle B C M+\angle M E A=$ $=180^{\circ}-\angle E M A-\angle C M B=\angle E M C=90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_b0b755a355dbd05a96cdg-11.jpg?height=760&width=600&top_left_y=1052&top_left_x=729)

## Answer

$90^{\circ}$.