Answer :

Given A (4, 2), B (7, 6) and C (1, 4) are the vertices of Δ ABC.

Mid- point of BC, D = (4, 5)

We know that area of triangle = 1/2 |x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2}) |

Consider Δ ABD,

⇒ Area = 1/2 [4 (6 – 5) + 7 (5 – 2) + 4 (2 – 6)]

= 1/2 [4 + 21 – 16]

Area of Δ ABD = 9/2 square units … (1)

Now Δ ADC,

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

= 1/2 [4 + 8 – 3]

Area of Δ ADC = 9/2 square units … (2)

From (1) and (2),

ar (Δ ABD) = ar (Δ ADC)

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

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