The legs of a right triangle are equal to $a$ and $b$. Find the length of the bisector drawn from the vertex of the right angle.
Let $x$ be the desired length. The sum of the areas of the triangles into which the bisector divides the given triangle is equal to its area, that is, $a x \sin 45^{\circ}+b x \sin 45^{\circ}=a b$. From this, $x=\frac{a b \sqrt{2}}{a+b}$.

## Answer

$\frac{a b \sqrt{2}}{a+b}$.