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:

Solution (By Examveda Team)

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