A (4, -6), B (3, -2) and C (5, 2) are the vertices of a Δ ABC and AD is its median. Prove that the median AD divides Δ ABC into two triangles of equal areas.

Given A (4, -6), B (3, -2) and C (5, 2) are the vertices of Δ ABC.

Mid- point of BC, D = (4, 0)

We know that area of triangle = 1/2 |x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂) |

Consider Δ ABD,

⇒ Area = 1/2 [4 (-2 – 0) + 3 (0 – (-6)) + 4 (-6 – (-2))]

= 1/2 |-8 + 18 – 16|

Area of Δ ABD = 3 square units … (1)

Now Δ ADC,

⇒ Area = 1/2 [4 (0 – 2) + 4 (2 – (-6)) + 5 (-6 – 0)]

= 1/2 |-8 + 32 – 30|

Area of Δ ADC = 3 square units … (2)

From (1) and (2),

ar (Δ ABD) = ar (Δ ADC)

∴ Median AD divides Δ ABC into two triangles of equal area.