The median $A M$ of triangle $A B C$ is equal to $m$ and forms angles $\alpha$ and $\beta$ with sides $A B$ and $A C$ respectively. Find these sides.

#
On the extension of the median $A M$ beyond point $M$, lay off the segment $M K$ equal to $A M$, and apply the Law of Sines to triangle $A C K$.

## Solution

On the extension of the median $A M$ beyond point $M$, lay off the segment $M K$ equal to $A M$. Then the quadrilateral $A B K C$ is a parallelogram, so $\angle A K C = \angle B A M = \alpha$. By the Law of Sines in triangle $A C K$, we find that

$$
\frac{C K}{\sin \beta} = \frac{A K}{\sin (180^\circ - \alpha - \beta)}
$$

from which

$$
A B = C K = \frac{A K \sin \beta}{\sin (\alpha + \beta)} = \frac{2 m \sin \beta}{\sin (\alpha + \beta)}
$$

Similarly, we find that

$$
A C = \frac{A K \sin \alpha}{\sin (\alpha + \beta)} = \frac{2 m \sin \alpha}{\sin (\alpha + \beta)}
$$

## Answer

$\frac{2 m \sin \beta}{\sin (\alpha + \beta)}, \frac{2 m \sin \alpha}{\sin (\alpha + \beta)}$