Let $A B C D$ be a parallelogram such that $\angle B A D=60^{\circ}$. Let $K$ and $L$ be the midpoints of $B C$ and $C D$, respectively. Assuming that $A B K L$ is a cyclic quadrilateral, find $\angle A B D$.
Let $\angle B A L=\alpha$. Since $A B K L$ is cyclic, $\angle K K C=\alpha$. Because $L K \| D B$ and $A B \| D C$, we further have $\angle D B C=\alpha$ and $\angle A D B=\alpha$. Let $B D$ and $A L$ intersect at $P$. The triangles $A B P$ and $D B A$ have two equal angles, and hence $A B P \sim D B A$. So

$$
\frac{A B}{D B}=\frac{B P}{A B}
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_8b636f8a6b85b9bd036bg-09.jpg?height=357&width=462&top_left_y=1164&top_left_x=1294)

The triangles $A B P$ and $L D P$ are clearly similar with similarity ratio $2: 1$. Hence $B P=$ $\frac{2}{3} D B$. Inserting this into (1) we get

$$
A B=\sqrt{\frac{2}{3}} \cdot D B
$$

The sine theorem applied to $A B D$ (recall that $\angle D A B=60^{\circ}$ ) immediately gives

$$
\sin \alpha=\frac{A B}{B D} \sin 60^{\circ}=\sqrt{\frac{2}{3}} \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{2}}{2}=\sin 45^{\circ} .
$$

So $\angle A B D=180^{\circ}-60^{\circ}-45^{\circ}=75^{\circ}$.