91Ó°ÊÓ

An equilateral triangle such that x, y and z are the perpendicular distances from any point inside the triangle to each side.

The triangle drawn below is an equilateral triangle.

Let P be any point inside the triangle and x, y and z are the perpendicular distances from this point to each side.

Draw a line from to each of A, B, and C forming three triangles$PAB$ , $PBC$, and $PCA$.

The formula to find the area of triangle is:

$\text{Area}=\frac{1}{2}×\text{base}×\text{height}$

So,

$\begin{array}{c}\text{´¡°ù±ð²¹â€‰o´Ú }\mathrm{Δ}PAC=\frac{a.z}{2}\\ \text{´¡°ù±ð²¹â€‰o´Ú }\mathrm{Δ}PAB=\frac{a.x}{2}\\ \text{´¡°ù±ð²¹â€‰o´Ú }\mathrm{Δ}PCB=\frac{a.y}{2}\end{array}$

So, the area of$\mathrm{Δ}ABC$ can be calculated as:

$\begin{array}{c}\mathrm{Δ}ABC=\mathrm{Δ}PAC+\mathrm{Δ}PAB+\mathrm{Δ}PCB\\ \frac{a.h}{2}=\frac{a.z}{2}+\frac{a.x}{2}+\frac{a.y}{2}\\ x+y+z=h\end{array}$

Observation is that the sum of the distances from any interior point to the sides of the equilateral$\mathrm{Δ}ABC$ is equal to the length of triangle’s altitude.

Therefore, the equilateral triangle is shown in Figure and the value of $x+y+z=h$.