$\left[\begin{array}{l}\text { Isosceles, Inscribed, and Circumscribed Trapezoids } \\ {[\quad \text { Auxiliary Similar Triangles }}\end{array}\right]$

The center of a circle with radius 6, which touches the sides $A B, B C$, and $C D$ of the isosceles trapezoid $A B C D$, lies on its larger base $A D$. The base $B C$ is equal to 4. Find the distance between the points where the circle touches the lateral sides $A B$ and $C D$ of this trapezoid.
Let $O$ be the center of the circle, $Q, M$, and $P$ be the points of tangency with sides $A B, B C$, and $C D$, respectively. Let $L$ be the projection of vertex $C$ onto the base $A D$, and $K$ be the intersection point of segments $P Q$ and $M O$.

## Solution

Let $O$ be the center of the circle, $Q, M$, and $P$ be the points of tangency with sides $A B, B C$, and $C D$, respectively. Let $L$ be the projection of vertex $C$ onto the base $A D$, and $K$ be the intersection point of segments $P Q$ and $M O$.

Since $C O$ is the bisector of angle $B C D$, we have $\angle C O D = \angle O C B = \angle O C D$. Therefore, triangle $O D C$ is isosceles: $C D = O D$.

Let $O D = x$. Then $L D = O D - O L = O D - M C = x - 2$. By the Pythagorean theorem, $C D^2 = C K^2 + K D^2$, or $x^2 = 36 + (x - 2)^2$. From this, we get $x = 10$.

From the similarity of right triangles $O K P$ and $D L C$, we find that $K P : C L = O P : C D$. Therefore, $K P = O P \cdot \frac{C L}{C D} = 6 \cdot \frac{6}{10} = \frac{18}{5}$.

Thus, $P Q = 2 K P = \frac{36}{5}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_f7f9d90261bfc416bd71g-03.jpg?height=440&width=531&top_left_y=452&top_left_x=772)

## Answer

## 7.2.