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 Three Centroids (Posted on 2015-05-28)

Let parallelogram ABCD have point P on side BC.
Let K, L, and M be the centroids of triangles
PBA, PAD, and PDC respectively.

Prove that [BCMK] = [AKL] + [DLM], where

[·] denotes area of polygon.

 See The Solution Submitted by Bractals No Rating

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 Simplified solution & more area relationships | Comment 1 of 2
Shearing a parallelogram back to a rectangle doesn't change the relative sizes of any of these areas.
Compressing the rectangle to a square doesn't change the relative sizes of any of these areas.

So we just need to prove this for a square.  For further convenience
A=(0,6)
B=(0,0)
C=(6,0)
D=(6,6)
P=(3p,0)
so that
K=(p,2)
L=(p+2,4)
M=(p+4,2)

BCMK is a trapezoid so
[BCMK] = .5*2*(6+4) = 10

[AKL] + [DLM] = [AKMD] - [KLM] - [ADL]
= .5*4*(6+4) - .5*2*4 - .5*6*2
= 20 - 4 - 6 = 10

This completes the proof.  There are other correspondences between areas as can be seen from the above.  The last two triangles areas also sum to
[ABK] + [DCM]
=.5*6*p + .5*6*(2-p)
=3p + 6-3p = 6 = [ADL]

 Posted by Jer on 2015-05-29 09:00:05
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