Let $A B C D$ be a rhombus, with $\angle B A D>\angle A B C$, and $P$ and $Q$ points on sides $A B$ and $A D$, respectively, such that the triangle $P C Q$ is equilateral, with side equal to the side of the rhombus. Find the measures of the angles of the rhombus.
Solution

Since $\angle A B C=\angle Q D C=\alpha$ and $B C=P C=Q C=C D$, it follows that $\angle B C P=\angle D C Q=180^{\circ}-2 \alpha$. Thus,

$$
\begin{aligned}
180^{\circ}-\alpha & =\angle B C D \\
& =\left(180^{\circ}-2 \alpha\right)+60^{\circ}+\left(180^{\circ}-2 \alpha\right) \\
& =420^{\circ}-4 \alpha .
\end{aligned}
$$

Therefore, $240^{\circ}=3 \alpha$, that is, $\alpha=80^{\circ}$. Hence, the angles of the rhombus are $80^{\circ}, 80^{\circ}, 100^{\circ} \text{ and } 100^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_42ab6d7ce4e674d231dcg-15.jpg?height=394&width=677&top_left_y=1844&top_left_x=761)