[Parallel lines, properties and signs. Secants] [The ratio in which the bisector divides the side]

Avoor: Rozhkov M.

Given an isosceles triangle $ABC$, where $BC = a$, $AB = AC = b$. On the side $AC$, a triangle $ADC$ is constructed outward, where $AD = DC = a$. Let $CM$ and $CN$ be the bisectors in triangles $ABC$ and $ADC$ respectively. Find the radius of the circumscribed circle of triangle $CMN$.
Let $K$ be a point on the segment $A C$ such that $M K \| B C$. Then $\angle M C A=\angle M C B=\angle C M K$, so $M K=K C$; moreover, by symmetry $K C=M B$. According to Thales' theorem and the property of the angle bisector, $C K: A K=B M: A M=a: b=$ $D N: A N$. Therefore, $K N \| C D$; similarly, $K N=K C$. Thus, $K$ is the center of the circumcircle of triangle $C M N$, and its radius $K C=M B=a b / a+b$ (see figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_0e468145bb05e34a17aag-08.jpg?height=532&width=649&top_left_y=1652&top_left_x=727)

## Answer

$a b / a+b$.