3. On the sides $A B$ and $B C$ of an equilateral triangle $A B C$, points $P$ and $Q$ are chosen such that $A P: P B = B Q: Q C = 2: 1$. Find $\angle A K B$, where $K$ is the intersection point of segments $A Q$ and $C P$.
Answer: $90^{\circ}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_66368caef28b3de8e403g-26.jpg?height=340&width=1042&top_left_y=2170&top_left_x=560)

Solution 1. Let $B H$ be the height and median of triangle $A B C$. Draw a line through vertex $B$ parallel to $A C$ and denote the point of its intersection with line $C P$ as $D$ (see the left figure). Triangles $B P D$ and $A P C$ are similar with a coefficient of $\frac{1}{2}$, hence $D B = \frac{1}{2} A C = A H$. Therefore, $A D B H$ is a rectangle, which means $\angle A D B = 90^{\circ}$. Note that triangles $A B Q$ and $C A P$ are congruent by two sides and an angle. Then

$$
\angle B D K = \angle D C A = \angle B A K.
$$

Thus, quadrilateral $A D B K$ is cyclic, from which $\angle A K B = 180^{\circ} - \angle A D B = 90^{\circ}$.

Solution 2. Draw the height $C H$ in triangle $A B C$ (see the right figure). Since $B H = \frac{1}{2} A B$, we get $\frac{B P}{B H} = \frac{2}{3} = \frac{B Q}{B C}$. Therefore, triangles $B P Q$ and $B H C$ are similar, from which $\angle B P Q = \angle B H C = 90^{\circ}$. Note now that $B Q = A P$, $A B = C A$, and $\angle A B Q = \angle C A P = 60^{\circ}$. Then triangles $A B Q$ and $C A P$ are congruent by two sides and an angle. Since $\angle A Q B = \angle C P A = 180^{\circ} - \angle C P B$, quadrilateral $B P K Q$ is cyclic, from which

$$
\angle A K B = 180^{\circ} - \angle B K Q = 180^{\circ} - \angle B P Q = 90^{\circ}
$$