Let A', B', C', and D' be the centroids of triangles BCD, CDA, DAB, and ABC respectively.
Let P, Q, R, and S be the midpoints of sides AB, BC, CD, and DA respectively.
If ^ denotes intersection, then
A' = BR ^ DQ
B' = CS ^ AR
C' = DP ^ BS
D' = AQ ^ CP
From triangles BCD and CDA we get
|RA'| |RB'| 1
------- = ------ = ---
|RB| |RA| 3
Therefore, from triangle ARB we have A'B' is parallel to AB and |A'B'| = |AB|/3.
A similar argument shows that B'C', C'D', and D'A' are parallel to and 1/3 the lengths of BC, CD, and DA respectively.
Therefore, A'B'C'D' and ABCD are homothetic and AA', BB', CC', and DD' are concurrent at their homothetic center.
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