1.2. Given an equilateral triangle $A B C$. Point $D$ is such that $\angle B D C=90^{\circ}$ and $D$ and $A$ lie in different half-planes relative to the line $B C$. Point $M$ is the midpoint of side $A B$. Find the angle $B D M$.
1.2. Solution. Let's construct a circle on segment $BC$ as its diameter (see Fig. 17). Notice that point $D$ lies on it,

![](https://cdn.mathpix.com/cropped/2024_06_04_b1380f2021d62732a78bg-19.jpg?height=1276&width=991&top_left_y=53&top_left_x=538)

Fig. 17: to the solution of problem 1.2

as does point $M$. Angle $BDM$ subtends segment $BM$, which is equal to the radius of the circumscribed circle. This means that the required angle is $30^{\circ}$.