VII-3. Given is a parallelogram ABCD. The diagonal AC divides $\angle B A D$ into two parts that differ by $20^{\circ}$. The height of the parallelogram drawn from vertex $D$ to side $B C$ intersects diagonal $A C$ at point $H$ and at the same time $\angle A H D=50^{\circ}$.

Determine the angles of the parallelogram.
Solution: The opposite angles in a parallelogram are equal, and so are the parts into which the diagonal divides them. If $\varphi$ is the required angle, then $\alpha=\varphi+\varphi+20^{\circ}=2 \varphi+20^{\circ}$.

We will consider three cases:
a) $\alpha<90^{\circ}$, then $D_{1}$ is on the extension of $B C$. $\triangle D_{1} C H$ is a right triangle, so $\angle D_{1} C H=\varphi=40^{\circ}$, i.e., the angles of the parallelogram are $100^{\circ}$ and $80^{\circ}$ (Fig. 2).

![](https://cdn.mathpix.com/cropped/2024_06_05_4e46949c190613f78f8fg-2.jpg?height=413&width=235&top_left_y=320&top_left_x=1209)

Fig. 2