8.6. In a convex quadrilateral $A B C D$, side $B C$ is half the length of $A D$. Diagonal $A C$ is perpendicular to side $C D$, and diagonal $B D$ is perpendicular to side $A B$. Find the larger acute angle of this quadrilateral, given that the smaller one is $36^{\circ}$.
Answer: $84^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_402f7649344ac77c021fg-8.jpg?height=363&width=602&top_left_y=1880&top_left_x=727)

Solution. Let point $M$ be the midpoint of side $A D$. Since angles $A B D$ and $A C D$ are right angles, angles $B$ and $C$ of quadrilateral $A B C D$ are obtuse, and angles $A$ and $D$ are acute, as angles of right triangles $A B D$ and $A C D$. Let $\angle A=\alpha$ be the given angle, and $\angle D=\beta$ be the angle to be found.

Triangles $A B D$ and $A C D$ are right triangles, so their medians to $A D$ are equal to half the hypotenuse, thus $A M=B M=C M=D M=B C$. From this, it follows that (1) $\angle A B M=\alpha$ (since $\triangle A M B$ is isosceles), (2) $\angle D C M=\beta$ (since $\triangle C M D$ is isosceles), (3) $\angle C B M=\angle B C M=60^{\circ}$ (since $\triangle B M C$ is equilateral).

The sum of the angles of quadrilateral $A B C D$ is $360^{\circ}$, so the equation $\alpha+\left(\alpha+60^{\circ}\right)+\left(\beta+60^{\circ}\right)+\beta=360^{\circ}$ holds, from which it follows that $\alpha+\beta=120^{\circ}$ or $\beta=120^{\circ}-\alpha$.

In this case, if $\alpha=36^{\circ}$, then $\beta=120^{\circ}-\alpha=120^{\circ}-36^{\circ}=84^{\circ}$.