8.3. In triangle $A B C$ with angle $C$ equal to $30^{\circ}$, median $A D$ is drawn. Angle $A D B$ is equal to $45^{\circ}$. Find angle $B A D$.

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# Answer: $30^{\circ}$.

Solution. Draw the height $B H$ (see the figure). In the right triangle $B H C$, the leg $B H$ lies opposite the angle $30^{\circ}$, so $B H=\frac{B C}{2}=B D$. The angle $H B C$ is $180^{\circ}-90^{\circ}-30^{\circ}=60^{\circ}$. Triangle $B H D$ is isosceles with an angle of $60^{\circ}$, so it is equilateral, and $B H=B D=H D$. Let's find the angles of triangle $A D H$: $\angle A D H=60^{\circ}-45^{\circ}=15^{\circ} ; \quad \angle A H D=90^{\circ}+60^{\circ}=150^{\circ} ; \quad \angle D A H=$ $180^{\circ}-150^{\circ}-15^{\circ}=15^{\circ}$. Therefore, $A H=H D$. Then $A H=B H$, and

![](https://cdn.mathpix.com/cropped/2024_05_06_62b90d6eaa4fe8177ba8g-2.jpg?height=363&width=651&top_left_y=241&top_left_x=1316)
triangle $A B H$ is isosceles. Since it is a right triangle, $\angle B A H=45^{\circ}$. Then $\angle B A D=$ $\angle B A H-\angle D A H=45^{\circ}-15^{\circ}=30^{\circ}$.

Comment. A correct and justified solution - 7 points. A correct solution with some non-obvious and unexplained transitions - 5-6 points. If the solution is not completed, points are given for proving useful auxiliary statements - 2-3 points. Only the answer is provided - 0 points. The problem is not solved or solved incorrectly - 0 points.