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

Given: AD be the median of ΔABC
BD = DC
Since D is the mid-point of BC
Using the mid point theorem to find the coordinates of D,
D=[x2+x3/2, y2+y3/2]
D=[3+5/2, -2+2/2]=[4,0]
Area of ΔABD=1/2|[(x1(y2-y4)+x2(y4-y1)+x4(y1-y2))]|
=1/2|[(4(-2-0)+3(0-(-6))+4(-6-(-2)))]|
=1/2|[-8+18-16]|
=1/2|[-6]|=3 Sq.units
Area of ΔADC=1/2|[(x1(y4-y3)+x4(y3-y1)+x3(y1-y4))]|
=1/2|[(4(0-2)+4(2-(-6))+5(-6-0))]|
=1/2|[-8+32-30]|
=1/2|[-6]|=3 Sq.units
Area of ΔABD=ΔADC
Hence proved that the median of triangle AD divides the triangles into equal areas.