In a right-angled triangle, the angle bisector of the right angle and the median to the hypotenuse enclose an angle of $16^{\circ}$. What are the angles of the triangle?
In the right-angled triangle $ABC$, let $D$ be the midpoint of the hypotenuse $AC$. Then the triangles $BCD$ and $BAD$ are isosceles, because $D$ is the center of the circle circumscribed around triangle $ABC$.

Thus,

$$
\angle BAC = \angle ABD = 45^{\circ} + 16^{\circ} = 61^{\circ}
$$

and

$$
\angle BCA = \angle CBD = 45^{\circ} - 16^{\circ} = 29^{\circ}
$$

(Jakab Bauer, Zenta.)