If Δ1 is the area of the triangle formed by the centroid and two vertices of a triangle, Δ2 is the area of the triangle formed by the mid-points of the sides of the same triangle, then Δ1:Δ2=

Let Ax1,y1,Bx2,y2 and Cx3,y3 be the vertices of a ∆ ABC, and let G be its centroid. Then,Δ1= Area of ΔGBC⇒Δ1=12x1+x2+x33y1+y2+y331x2y21x3y31⇒Δ1=16x1+x2+x3y1+y2+y33x2y21x3y31 Applying R1→R1(3)⇒Δ1=16x1    y1    1x2    y2    1x3    y3    1  Applying R1→R1−R2−R3⇒Δ1=Δ3 where ∆is the area of ∆ABCΔ2=Area of triangle formed by the mid-points of the sides⇒Δ2=x1+x22y1+y221x2+x32y2+y321x3+x12y3+y1211⇒Δ2=18x1+x2y1+y21x2+x3y2+y31x3+x1y3+y11 Applying C1→C1(2),C2→C2(2)⇒Δ2=182x1+x2+x32y1+y2+y33x2+x3y2+y31x3+x1y3+y11 Applying R1→R1+R2+R3⇒Δ2=28x1+x2+x3y1+y2+y332x2+x3y2+y31x3+x1y3+y11  Applying R1→R1(1/2)⇒Δ2=14x1+x2+x3y1+y2+y33/2−x1−1/1−1/2−x2−1/2−1/2 R2→R2−R1R3→R3−R1⇒ Δ2=18x1+x2+x3y1+y2+y33−x1−y1−1−x2−y2−1  Applying C3→C3⇒ Δ2=18x3    y3    1−x1    −y1    −1−x2    −y2    −1  Applying R1→R1+R2+R3⇒ Δ2=14Δ∴ Δ1:Δ2=4:3