\left.\begin{array}{l}\text { [ Properties of medians. Centroid of a triangle. ]} \\ {[\quad \text { Inscribed quadrilaterals (other). ] }}\end{array}\right]

Medians $A M$ and $B E$ of triangle $A B C$ intersect at point $O$. Points $O, M, E, C$ lie on the same circle. Find $A B$, if $B E = A M = 3$.
Prove that triangle $ABC$ is equilateral.

## Solution

Since $OE=\frac{1}{3} BE$ and $OM=\frac{1}{3} AM$, then $OE=OM$. Therefore, $CO$ is the bisector of angle $ECM$.

From the equality of medians $BE$ and $AM$, it follows that triangle $ABC$ is isosceles. Therefore, $EC=CM$. Then triangles $CEO$ and $CMO$ are equal, and since

$$
\angle CEO + \angle CMO = 180^{\circ}
$$

then

$$
\angle CEO = \angle CMO = 90^{\circ}
$$

i.e., medians $AM$ and $BE$ are altitudes. Therefore, triangle $ABC$ is equilateral. Consequently,

$$
AB = \frac{AM}{\sin 60^{\circ}} = 2 \sqrt{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-08.jpg?height=417&width=506&top_left_y=2421&top_left_x=776)

## Answer

## Problem