Consider an equilateral triangle $ABC$ where
$P$ is a point on $AB$,
$Q$ is a point on $BC$.Suppose we draw perpendiculars from $P$ to other sides. Let $s_1$ be the sum of the length of these perpendiculars.
Suppose we draw perpendiculars from $Q$ to other sides. Let $s_2$ be the sum of the length of these perpendiculars.
Then, will $s_1$ be $s_2$?
This doubt came while going to through the solution given in this website here, According to the solution there, $s_1$ must be equal to $s_2$.
My attempt to see whether $s_1$ is equal to $s_2$
Using Viviani's_theorem, sum of perpendicular drawn from all sides to any arbitrary point is equal to altitude.
I tried to see if this is any possible extension of this theorem. But, finally came to the conclusion that we cannot use Viviani's_theorem for this because, the point lies on a side and we are calculating sum of perpendiculars from each side.
I tried to split the triangle and use Pythagorean theorem. But all my efforts did not succeed.
I am also not sure whether this is true (i.e., $s_1=s_2$).
The triangles $PAM,PLB$ are similar. Both are half an equilateral triangle. So $\frac{PM}{PA}=\frac{PL}{PB}=\frac{\sqrt3}{2}$.
Note that if you reflect $PAM$ in the line $PM$ you get an equilateral triangle side $PA$. So $AM$ is half $PA$ and you can then get $PM$ by Pythagoras' Theorem. Hence the $\frac{\sqrt3}{2}$.
So $PM+PL=\frac{\sqrt3}{2}(PA+PB)=\frac{\sqrt3}{2}AB$ or if you prefer the height of the triangle.