Let AX ⊥ BC of an equilateral triangle ABC. Then the sum of the perpendicular distances of the sides of ΔABC from any point inside the triangle is:

Answer: Option D

Let side = 2 units
Side = $$\frac{2}{\sqrt{3}}$$(PT + QT + TR)
2 = $$\frac{2}{\sqrt{3}}$$(PT + QT + TR)
∴ PT + QT + TR = $$\sqrt{3}$$
& AX $$=\frac{\sqrt{3}}{2}\times 2=\sqrt{3}$$
So it is equal to AX