8,9 [ Inscribed and Circumscribed Circles ]

The diagonals of a quadrilateral $A B C D$ inscribed in a circle of radius $R$ intersect at point $M$. It is known that $A B=B C=a, B D=m$.

Find the radius of the circumscribed circle of triangle $B C M$.

#
Triangles $B C M$ and $B D C$ are similar.

## Solution

Inscribed angles $B D C$ and $B D A$ subtend equal arcs, so they are equal. Therefore, $\angle B C M = \angle C A B = \angle B D C$. Triangles $B C M$ and $B D C$ are similar by two angles with a similarity ratio of ${ }^{B C} /{ }_{B D}=a / m$. Thus, the radius of the circumscribed circle of triangle $B C M$ is $a / m \cdot R$.

![](https://cdn.mathpix.com/cropped/2024_05_06_95d66af6ee6f6b4b6717g-10.jpg?height=497&width=515&top_left_y=0&top_left_x=777)

## Answer

$a R / m$.