Accurately draw or construct a large equilateral triangle. Choose any point inside the triangle and carefully measure the distance x, y, z and h. Then find$x+y+z$.

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:

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

So,

$\begin{array}{c}\text{Area\hspace{0.17em}of\hspace{0.17em}}\Delta PAC=\frac{a.z}{2}\\ \text{Area\hspace{0.17em}of\hspace{0.17em}}\Delta PAB=\frac{a.x}{2}\\ \text{Area\hspace{0.17em}of\hspace{0.17em}}\Delta PCB=\frac{a.y}{2}\end{array}$

So, the area of$\Delta ABC$ can be calculated as:

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

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

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