Question

Show that the sum of the lengths of the perpendiculars drawn from an interior point of an equilateral triangle on to the sides of the triangle is independent of the point chosen, but depends only on the triangle.

Answer 1

Given: an equilateral triangle $ABC$ with side $a$, an interior point $P$ and the perpendiculars $PD$, $PE$, $PF$ respectively on to the sides $BC$, $CA$, $AB$.
To prove: $PD+PE+PF$ is independent of the point $P$, but depends only on $△ABC$.
Observe
area$\left(ABC\right)=$ area$\left(PBC\right)+$ area$\left(PCA\right)+$ area$\left(PAB\right)$.
But we know
area$\left(PBC\right)=\frac{1}{2}BC\times PD$,
area$\left(PCA\right)=\frac{1}{2}CA\times PE$,
area$\left(PAB\right)=\frac{1}{2}AB\times PF$.
But we know $BC=CA=AB=a$, since $ABC$ is equilateral. Hence
area$\left(ABC\right)=\frac{1}{2}a(PD+PE+PF)$.
We obtain
$PD+PE+PF=\frac{2area\left(ABC\right)}{a}$.
Hence $PD+PE+PF$ depends only on the triangle $ABC$, but not on the point $P$.