$[\underline{C}$  Tangent Circles

Two circles touch each other externally at point $A$. Their common tangent touches the first circle at point $B$ and the second circle at point $C$. A line passing through points $A$ and $B$ intersects the second circle at point $D$. It is known that $A B=5$ cm, $A D=4$ cm. Find $C D$.
Prove that $C A$ is the altitude of the right triangle $B C D$, drawn from the vertex $C$ of the right angle.

## Solution

Let the common tangent, drawn to the given circles at point $A$, intersect the segment $B C$ at point $K$. Then $B K=K A=K C$. Therefore, $\angle B A C=90^{\circ}$. This means that the chord $C D$ of the second circle is seen from point $A$ of this circle at a right angle. Therefore, $C D$ is the diameter of the second circle, and since $B C$ is a tangent, $\angle B C D=90^{\circ}$. Thus, the segment $A D$ is the projection of the leg $C D$ of the right triangle $B C D$ onto the hypotenuse $B D$. Therefore,

$$
C D^{2}=A D \cdot B D=4 \cdot 9=36 \Rightarrow C D=6
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_8027e864227276d12944g-39.jpg?height=443&width=732&top_left_y=1202&top_left_x=663)

## Answer

6.