$\left.\begin{array}{ll}\text { [Sine Theorem } \\ \text { [Inscribed, circumscribed, and exscribed circles; their radii] }\end{array}\right]$

A circle of radius $r$ passes through vertices $A$ and $B$ of triangle $ABC$ and intersects side $BC$ at point $D$. Find the radius of the circle passing through points $A, D$, and $C$, if $AB=c$ and $AC=b$.
$\sin \angle A D C=\sin \angle A D B$

## Solution

Since

$$
\angle A D C+\angle A D B=180^{\circ}
$$

then

$$
\sin \angle A D C=\sin \angle A D B=\frac{c}{2 r}
$$

If $R$ is the radius of the circle passing through points $A, C$, and $D$, then $b=2 R \sin \angle A D C$. From this, we find that

$$
R=\frac{b}{2 \sin \angle A D C}=\frac{b}{2 \cdot \frac{c}{2 r}}=\frac{b r}{c} .
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-03.jpg?height=409&width=652&top_left_y=1596&top_left_x=703)

## Answer

$\frac{b r}{c}$.

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