[ Auxiliary similar triangles ] [ Angle between a tangent and a chord $\quad$ ]

On the sides of the acute angle $A B C$, points $A$ and $C$ are taken. One circle is tangent to the line $A B$ at point $B$ and passes through point $C$. The second circle is tangent to the line $B C$ at point $B$ and passes through point $A$. Point $D$ is the second common point of the circles. It is known that $A B=a, C D=b, B C=c$. Find $A D$.
Triangles $D B C$ and $D A B$ are similar.

## Solution

By the theorem on the angle between a tangent and a chord, $\angle D B C = \angle B A D, \angle A B D = \angle B C D$, therefore triangles $D B C$ and $D A B$ are similar by two angles. Thus, $B D: C D = A B: B C$. From this, $B D = A B \cdot C D / B C = a b / c$, and since $A D: B D = A B: B C$, then $A D = B D \cdot A B / B C = a b / c \cdot a / c = b\left(\frac{a}{c}\right)^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-24.jpg?height=643&width=577&top_left_y=1342&top_left_x=749)

## Answer

$b\left(\frac{a}{c}\right)^{2}$.