If two medians BE and CF of a triangle ABC, intersect each other at G and if BG = CG, ∠BGC = 60° and BC = 8 cm then area of the triangle ABC is

In Δ BGC

∠BGC = 60° and BG = GC

$$\therefore$$ ΔBGC is an isosceles triangle

∠GBC = ∠GCB

As, we know sum of angles of a triangle = 180°

$$\therefore$$ ∠GBC + ∠GCB + ∠BGC = 180°

=> 2∠GBC = 180° - 60°

=> ∠GBC = 120°/2 = 60°

∠GBC = ∠GCB = ∠BGC = 60°

$$\therefore$$ ΔGBC is equilateral triangle

Area of equilateral triangle = $$\frac{\sqrt{3}}{4} * side^2$$

Area of $$\triangle$$GBC = $$\frac{\sqrt{3}}{4} * 8^2$$

= 16$$\sqrt{3}$$

Median of triangle divides the triangle into two parts of equal area.

3 medians of a triangle divide the triangle into six parts of equal area.

$$\therefore$$ Area of ΔGBC = 2 × area of ΔGDC

Area of ΔABC = 6 × area of ΔGDC

= 3 × 2 × ΔGDC

= 3 × area of ΔGBC

= 3 × 16$$\sqrt{3}$$

= 48$$\sqrt{3} cm^2$$