Problem 10.5. On the side $AB$ of the rectangle $ABCD$, a circle $\omega$ is constructed with $AB$ as its diameter. Let $P$ be the second intersection point of the segment $AC$ and the circle $\omega$. The tangent to $\omega$ at point $P$ intersects the segment $BC$ at point $K$ and passes through point $D$. Find $AD$, given that $KD=36$.

![](https://cdn.mathpix.com/cropped/2024_05_06_62ad5020cf8c9b2474f9g-3.jpg?height=321&width=410&top_left_y=1321&top_left_x=518)
Answer: 24.

Solution. Note that triangles $A D P$ and $C K P$ are similar (the equality $\angle D A P=\angle K C P$ and $\angle A D P=\angle C K P$ is ensured by the parallelism $A D \| B C$). Moreover, triangle $A D P$ is isosceles, as $A D$ and $D P$ are segments of tangents. Therefore, triangle $C K P$ is also isosceles, meaning $C K=K P$.

On the other hand, $K P=K B$, as these are also segments of tangents. We get $K C=K B$, so $K$ is the midpoint of side $B C$.

Then the similarity ratio of triangles $A D P$ and $C K P$ is the ratio $A D: K C$, which is 2. Therefore, $P K=\frac{1}{2} P D$, from which $A D=P D=\frac{2}{3} K D=24$.