## Task B-4.5.

Point $M$ is the midpoint of the height of a regular tetrahedron $A B C D$ drawn from vertex $A$ to the base $B C D$. Determine the measure of angle $B M C$.


## First solution.

Let $E$ be the foot of the perpendicular from vertex $A$ to the base $BCD$.

![](https://cdn.mathpix.com/cropped/2024_05_30_d4daff78382705a7c34eg-26.jpg?height=534&width=536&top_left_y=544&top_left_x=226)

Let the edges of the regular tetrahedron $ABCD$ be of length $a$.

Then, $|BE|=|EC|=\frac{2}{3} \cdot \frac{a \sqrt{3}}{2}=\frac{a \sqrt{3}}{3}$.

From the right triangle $AEB$, we have: $|AE|^2 = a^2 - \left(\frac{a \sqrt{3}}{3}\right)^2 \Rightarrow |AE| = \frac{a \sqrt{6}}{3}$, and thus $|EM| = \frac{1}{2}|AE| = \frac{a \sqrt{6}}{6}$.

From the right triangle $MEB$, we have: $|MB|^2 = \left(\frac{a \sqrt{3}}{3}\right)^2 + \left(\frac{a \sqrt{6}}{6}\right)^2 \Rightarrow |MB| = \frac{a \sqrt{2}}{2}$, and thus $|MC| = |MB| = \frac{a \sqrt{2}}{2}$.

In the isosceles triangle $MBC$, the third side is of length $a$, so by the converse of the Pythagorean theorem, the angle $\angle BMC$ is a right angle.