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A(4, - 6), B(3,- 2) and C(5, 2) are the vertices of a 8 ABC and AD is its median. Prove that the median AD divides Δ ABC into two triangles of equal areas.

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#### Solution

Solution:

Let the co-ordinates of D be D (x,y) and D is midpoint of BC

x = (3+5)/2 = 4; y = (2-2)/2 = 0

Now Area of triangle ABD = `1/2 [4(-2-0) +3(0+6)+4(-6+2)]` = `3 `

and Area of triangle ACD `=1/2 [5(-6-0) + 4(0 - 2) + 4(2+6)] = 3 `

Hence, the median AD divides triangle ABC into two triangle of equal area.

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