7. On the sides $\mathrm{AB}$ and $\mathrm{AC}$ of the right triangle $\mathrm{ABC}\left(\angle B A C=90^{\circ}\right)$, right triangles АВТ and АСК are constructed externally such that $\angle A T B=\angle A K C=90^{\circ}$, $\angle A B T=\angle A C K=30^{\circ}$. On the side $\mathrm{BC}$, a point М is chosen such that $\mathrm{BM}=\mathrm{MC}$. Determine the degree measure of angle КМТ.
Solution. Mark points P and O at the midpoints of sides AB and AC, respectively. Connect point P with points M and T, and point O with points K and M.

![](https://cdn.mathpix.com/cropped/2024_05_06_9fc9f23f10a928994848g-08.jpg?height=600&width=514&top_left_y=1536&top_left_x=634)

Then: 1) $\triangle T P M = \Delta K O M$, by two sides and the angle between them, since

$$
\begin{gathered}
A O = \frac{1}{2} A C = K O = P M \\
A P = \frac{1}{2} A B = T P = O M \\
\angle T P M = \angle T P A + \angle A P M = \angle A O K + \angle A O M = \angle K O M
\end{gathered}
$$

Thus, TM = MK, $\angle P M T = \angle M K O$, and $\angle P T M = \angle K M O$.

2) Find the sum of angles $\angle P M T$ and $\angle K M O$:

$$
\angle P M T + \angle K M O = \angle M K O + \angle K M O = 180^{\circ} - \angle M O K.
$$

In turn, $\angle M O K = 360^{\circ} - \angle K O C - \angle M O C = 360^{\circ} - 120^{\circ} - \angle B A C = 240^{\circ} - \angle B A C$. 3) $\angle K M T = \angle P M O - (\angle P M T + \angle K M O) = \angle B A C - (180^{\circ} - \angle M O K) = \angle K M T = \angle B A C - 180^{\circ} + 240^{\circ} - \angle B A C = 60^{\circ}$.

Answer: 60.