Problem 11.6. Point $M$ is the midpoint of the base $B C$ of trapezoid $A B C D$. A point $P$ is chosen on the base $A D$. Ray $P M$ intersects ray $D C$ at point $Q$. The perpendicular to base $A D$, drawn through point $P$, intersects segment $B Q$ at point $K$.

It is known that $\angle K Q D=64^{\circ}$ and $\angle K D Q=38^{\circ}$. How many degrees does the angle $K B C$ measure?

![](https://cdn.mathpix.com/cropped/2024_05_06_9703912434d201fc4f33g-05.jpg?height=503&width=591&top_left_y=318&top_left_x=431)
Answer: $39^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_9703912434d201fc4f33g-05.jpg?height=450&width=711&top_left_y=942&top_left_x=369)

Fig. 11: to the solution of problem 11.6

Solution. From the condition, it follows that $\angle B K D=\angle K Q D+\angle K D Q=64^{\circ}+38^{\circ}=102^{\circ}$.

Let the lines $Q B$ and $A D$ intersect at point $X$ (Fig. 11). Triangles $Q X P$ and $Q B M$ are similar with a coefficient of $Q P / Q M$, and triangles $Q D P$ and $Q C M$ are similar with the same coefficient. Then, from the equality $M B=M C$, it follows that $P X=P D$. Therefore, point $K$ lies on the perpendicular bisector of segment $X D$, so $\angle K X D=\angle K D X=\frac{1}{2}\left(180^{\circ}-102^{\circ}\right)=39^{\circ}$. Since $B C \| A D$, we get that $\angle K B C=\angle K X D=39^{\circ}$.

Remark. The equality $P X=P D$ could also have been established using a homothety with center $Q$, which maps triangle $Q B C$ to triangle $Q X D$.