5. In triangle $ABC$ (angle B is obtuse), the altitude $BH$ and the bisector $AK$ are drawn. Find the angle $AKB$, if the angle $KHC$ is $45^{\circ}$.
Answer: $45^{\circ}$.

Solution. Point K is equidistant from lines AB and AC (since AK is the bisector), and from lines HB and HC (since HK is the bisector of the right angle). Therefore, point K is equidistant from lines AB and BH. This means that BK is the bisector of the angle external to angle ABH. Therefore, angle HBC is half of this external angle: $90^{\circ}$ - $C=\frac{1}{2}\left(A+90^{\circ}\right)$, from which $C+\frac{1}{2} A=45^{\circ}$. But this is precisely the angle AKC we are looking for (it is the external angle in triangle AKC).

Grading. Full solution - 7 points. For a "trigonometric" solution - full credit. For a correctly set up but unsolved equation - 2 points. For the correct answer (without solution) - 0 points.