8.4. Given a triangle $\mathrm{ABC}$ with angle $\mathrm{BAC}$ equal to $30^{\circ}$. In this triangle, the median $\mathrm{BD}$ was drawn, and it turned out that angle $\mathrm{BDC}$ is $45^{\circ}$. Find angle $\mathrm{ABC}$.
Answer: $45^{\circ}$

Solution: Draw the height $C H$. Then $H D=A D=C D$ as the median to the hypotenuse. Moreover, $\angle H C D=\angle C H A-\angle H A C=60^{\circ}$, so triangle $C H D$ is equilateral, which means $\angle H D C=60^{\circ}$ (from which it follows, in particular, that $H$ lies between $A$ and $B$). Then $\angle H D B=\angle H D C-\angle B D C=15^{\circ}$. Furthermore, $\angle A B D=\angle B D C-\angle B A D=15^{\circ}$. Therefore, triangle $H B D$ is isosceles, and $H B=H D=H C$, which implies that $B H C$ is an isosceles right triangle, and thus $\angle A B C=45^{\circ}$.

Criteria: The height $C H$ is drawn - 1 point.

It is additionally proven that $\angle H D B=15^{\circ}$ - another 2 points.

No points are deducted for the lack of proof that $H$ lies between $A$ and $B$.