Q:

Sum of distances from interior point to sides of equilateral triangle

Let $P$ be an arbitrary point within an equilateral triangle. Let $h_{1}, h_{2} $ and $h_{3}$ be the shortest distances from $P$ to each of the three sides. Show that the sum $h_{1} + h_{2} + h_{3}$ is independent of where point $P$ is placed in the triangle.
I've found a few geometric proofs in similar questions to mine, but am interested in a neat algebraic proof that provides a solution to the problem given above. Can anyone help please?

A:

Let the triangle be $ABC$ with side length $s$ and height $H$. Then, we have
 $$[ABC] = \dfrac{s \cdot H}{2} = [PAB] + [PBC] + [PAC] = \dfrac{s \cdot h_1}{2} +\dfrac{s \cdot h_2}{2} +\dfrac{s \cdot h_3}{2}
$$
Therefore, we have $h_1 + h_2 + h_3 = H$.
Note: $[ABC]$ denotes the area of triangle $ABC$.
P.S. This proof is rather related to the proof that the area of a triangle $ABC$ is equal to $rs$, where $r$ is the radius of its incircle, and $s$ is half its perimeter. Try showing this fact as well!

