10. (20 points) The diagonals $A C$ and $B D$ of a convex quadrilateral $A B C D$ intersect at point $P$. Find $\angle A P B$, if $\angle B A C=2 \alpha, \angle B C A=2 \beta, \angle C A D=90^{\circ}-\alpha$ and $\angle A C D=90^{\circ}-\beta$.
# Solution:

![](https://cdn.mathpix.com/cropped/2024_05_06_f48d5baeb4f84f12ed51g-10.jpg?height=914&width=1034&top_left_y=431&top_left_x=611)

It is clear that

$$
\angle L C D=180^{\circ}-\left(90^{\circ}-\beta\right)-2 \beta=90^{\circ}-\beta=\angle A C D
$$

and

$$
\angle K A D=180^{\circ}-\left(90^{\circ}-\alpha\right)-2 \alpha=90^{\circ}-\alpha=\angle C A D
$$

We obtain that $A D$ and $C D$ are the external bisectors of triangle $A B C$. Therefore, $B D$ is the bisector of $\angle A B C$. Hence,

$$
\angle A B P=\angle P B C=\frac{180^{\circ}-2 \alpha-2 \beta}{2}=90^{\circ}-\alpha-\beta
$$

Then the required angle can be found as the external angle of triangle $B P C$

$$
\angle B P A=\angle P B C+\angle B C P=90^{\circ}-\alpha-\beta+2 \beta=90^{\circ}-\alpha+\beta
$$

Answers:

| Option | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Answer | $70^{\circ}$ | $85^{\circ}$ | $72^{\circ}$ | $80^{\circ}$ | $74^{\circ}$ | $65^{\circ}$ | $75^{\circ}$ | $75^{\circ}$ | $80^{\circ}$ | $81^{\circ}$ |