10.021. The median to the hypotenuse of a right triangle is equal to $m$ and divides the right angle in the ratio 1:2. Find the sides of the triangle.
Solution.

By the condition $C O=m, A O=B O, \frac{\angle B C O}{\angle A C O}=\frac{2}{1}, A B=2 m(O-$ is the center of the circumscribed circle) (Fig. 10.21). Let $\angle A C O=x, \angle B C O=2 x$, $x+2 x=90^{\circ}, x=30^{\circ}, \quad \angle A C O=\angle C A O . B \quad \triangle A B C \quad \angle A C B=90^{\circ}$, $B C=2 m \cdot \sin 30^{\circ}=m, A C=\sqrt{4 m^{2}-m^{2}}=m \sqrt{3}$.

Answer: $\quad m ; m \sqrt{3}, 2 m$.