Problem 3. From point $M$ which side $A B$ of the equilateral triangle $A B C$ divides into segments of length $6 \mathrm{~cm}$ and $2 \mathrm{~cm}$, perpendiculars are drawn to the other sides. Determine the distance from point $C$ to the drawn perpendiculars.
Solution. The triangles AMD and $M B E$ are right-angled with angles $30^{0}$ and $60^{\circ}$, so

$$
\begin{aligned}
& \overline{A D}=\frac{1}{2} \overline{A M}=3 \mathrm{~cm} \text { and } \\
& \overline{B E}=\frac{1}{2} \overline{M B}=1 \mathrm{~cm}
\end{aligned}
$$

Since the side of triangle $A B C$ is $8 \mathrm{~cm}$, the required distances from $C$ to the normals are $5 \mathrm{~cm}$ and $7 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_1cb7beb611b796adc950g-07.jpg?height=521&width=586&top_left_y=1090&top_left_x=894)