The rhombus $ABCD$ is divided into two equilateral triangles by its diagonal $BD$. A point $P$ is given on the segment $AD$, and a point $Q$ is on the segment $CD$ such that $\angle PBQ = 60^{\circ}$. What are the other two angles of the triangle $PBQ$?
I. solution. Since $A B P \angle + P B D \angle = 60^{\circ}$ and $P B D \angle + D B Q \angle = 60^{\circ}$, it follows that $A B P \angle = D B Q \angle$. Therefore, the triangles $A B P$ and $D B Q$ are congruent, as one of their sides and two angles are equal. Thus, $P B = Q B$, and since the included angle is $60^{\circ}$, the triangle $P B Q$ is equilateral, so $Q P B \angle = P Q B \angle = 60^{\circ}$.

II. solution. Rotate the equilateral triangle $A B D$ around $B$ by $60^{\circ}$ in the direction of the clock, as shown in the figure!

![](https://cdn.mathpix.com/cropped/2024_05_02_cacdbec78acdc38656c3g-1.jpg?height=366&width=521&top_left_y=430&top_left_x=802)

The rotated image of the triangle $A B D$ is clearly the triangle $D B C$. Since $P B Q \angle = 60^{\circ}$, the image of the point $P$ after rotation must lie on the line $B Q$. However, since $P$ is on the segment $A D$, its image must be on the segment $D C$, so it can only be the intersection of $B Q$ and $D C$, which is $Q$. Therefore, $P B = B Q$, and the triangle $P B Q$ is also equilateral, meaning all its angles are $60^{\circ}$.