Problem 4. The median $\mathrm{AA}_{0}$ of triangle $\mathrm{ABC}$ is laid off from point $\mathrm{A}_{0}$ perpendicular to side $\mathrm{BC}$ to the outside of the triangle. Denote the second endpoint of the constructed segment as $\mathrm{A}_{1}$. Similarly, points $\mathrm{B}_{1}$ and $\mathrm{C}_{1}$ are constructed. Find the angles of triangle $\mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1}$, if the angles of triangle $\mathrm{ABC}$ are $30^{\circ}, 30^{\circ}$, and $120^{\circ}$.
Answer._ $\triangle \mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1}$ is equilateral, all angles are $60^{\circ}$.

## Solution.

Since $\triangle \mathrm{ABC}$ is isosceles, $\mathrm{BB}_{0}$ is the perpendicular bisector of the base AC. Therefore, $\mathrm{B}_{1}$ lies on this perpendicular and $\mathrm{CB}_{0} \perp \mathrm{BB}_{1}$. In $\Delta \mathrm{BCB}_{1}$, point $\mathrm{B}_{0}$ is the foot of the altitude and median on side $\mathrm{BB}_{1}$, hence $\mathrm{BC}=\mathrm{B}_{1} \mathrm{C}$, and since

![](https://cdn.mathpix.com/cropped/2024_05_06_259088e5783b98240440g-2.jpg?height=520&width=629&top_left_y=837&top_left_x=1385)
$\angle \mathrm{B}_{1} \mathrm{BC}=1 / 2 \angle \mathrm{ABC}=60^{\circ}$. Therefore, $\Delta \mathrm{B}_{1} \mathrm{BC}$ is equilateral, and $\mathrm{A}_{0}$, being the midpoint of side $\mathrm{BC}$, is the foot of the altitude in triangle $\mathrm{B}_{1} \mathrm{BC}$. This means that points $\mathrm{A}_{1}$ and $\mathrm{B}_{1}$ lie on the perpendicular bisector of segment $\mathrm{BC}$. Similarly, $\mathrm{C}_{1}$ and $\mathrm{B}_{1}$ lie on the perpendicular bisector of segment AB. Therefore, $\angle \mathrm{C}_{1} \mathrm{~B}_{1} \mathrm{~A}_{1}=180^{\circ}-\angle \mathrm{ABC}=60^{\circ}, \mathrm{B}_{1} \mathrm{~A}_{1}=\mathrm{B}_{1} \mathrm{~A}_{0}$ $+\mathrm{A}_{0} \mathrm{~A}_{1}=\mathrm{B}_{1} \mathrm{~A}_{0}+\mathrm{A}_{0} \mathrm{~A}=\mathrm{B}_{1} \mathrm{C}_{0}+\mathrm{C}_{0} \mathrm{C}=\mathrm{B}_{1} \mathrm{C}_{0}+\mathrm{C}_{0} \mathrm{C}_{1}=\mathrm{B}_{1} \mathrm{C}_{1}$, i.e., $\Delta \mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1}$ is isosceles with the vertex angle $60^{\circ}$.