A regular hexagon $A B C D E F$ is inscribed in a circle. Points $P$ and $Q$ are chosen on the tangents to this circle at points $A$ and $D$ respectively, such that the line $P Q$ is tangent to the smaller arc $E F$ of this circle. Find the angle between the lines $P B$ and $Q C$.
Let $T$ be the point of tangency of $P Q$ with the circle, $M, N$ be the midpoints of segments $A T, D T$. Since $P B$ and $C Q$ are symmedians of triangles $A B T$ and $C D T$ respectively (see problem $\underline{\text { 56983) }}$), then $\angle A B P=\angle M B T, \angle D C Q=\angle N C T$. Since $M N$ is the midline of triangle $A D T$, then $M N=A D / 2=B C$ and $M N \| B C$ (see the figure). Therefore, $B C N M$ is a parallelogram, and the angle between $P B$ and $Q C$ is $\angle P B M+\angle N C Q=\angle A B M+\angle N C D-\angle M B T-\angle T C N=1 / 2(-A D-$ $2-F E)=30^{\circ}$

## Answer

$30^{\circ}$.

| Problem 65955 Topics: | [ Properties and criteria of isosceles triangles. |  |
| :---: | :---: | :---: |
|  | Properties and criteria of parallelograms | Difficulty: 4 - Grades: 8,9 |
|  | Congruent triangles. Criteria for congruence |  |
|  | Equilateral (equilateral) triangle |  |
|  | [ Isosceles, inscribed, and circumscribed trapezoids ] |  |
|  | Midline of a trapezoid |  |

On the lateral sides $A B$ and $A C$ of isosceles triangle $A B C$, points $K$ and $L$ are marked such that $A K=C L$ and $\angle A L K+\angle L K B=60^{\circ}$.

Prove that $K L=B C$.

## Solution

First method. Extend the given triangle to form parallelogram $A B C D$ and draw segment $K N$ parallel to $B C$ (see left figure). Then $C N=C D-D N=A B-A K=B C-C L=A L$. Triangles $N C L$ and $L A K$ are congruent by two sides and the included angle. Therefore, $N L=L K$. Moreover, $\angle N L A=\angle L K B$, as these are corresponding exterior angles in congruent triangles.

Thus, $\angle N L K=\angle A L K+\angle N L A=\angle A L K+\angle L K B=60^{\circ}$, meaning triangle $K L N$ is equilateral. Hence, $K L=K N=B C$.
![](https://cdn.mathpix.com/cropped/2024_05_06_6409e68af5a1ba5e989dg-49.jpg?height=572&width=1226&top_left_y=1166&top_left_x=424)

Second method. Draw lines through points $K$ and $L$ parallel to $B C$, intersecting $A C$ and $A B$ at points $M$ and $N$ respectively (see right figure). Then $K M L N$ is an isosceles trapezoid, so $M N=K L$ and $\angle M N K=\angle M L K$.

Let $P$ be the intersection of the diagonals of the trapezoid, then

$\angle N P L=\angle P N K+\angle N K P=\angle A L K+\angle L K B=60^{\circ}$.

Draw a line through point $M$ parallel to $K L$, intersecting line $N L$ at point $Q$. Then in triangle $M N Q$, $M N=M Q$ and $\angle Q M N=60^{\circ}$, so this triangle is equilateral. Therefore, $M Q=N Q$, thus $K L=N L+L Q=N L+K M$.

Draw the midline $E F$ of triangle $A B C$, which will also be the midline of trapezoid $K M L N$ (since $A M=A K=C L=B N$). Then

$K L=N L+K M=2 E F=B C$.