Medians Theorem
The medians of a triangle are concurrent.
Probably the shortest proof is as follows:
If AQ, BR, CP are the medians of ΔABC, then
(BQ/QC)(CR/RA)(AP/PB) = 1,
since each of the ratios on the left is 1. The Converse of Ceva's Theorem
shows that the medians are concurrent.
Many of the known proofs involve the fact that the centroid X has the property
that (AX/XQ) = (BX/XR) = (CX/XP) = 2. This does not emerge in the proof
given above.
This property is a special case of the following result :
Van Obel's Theorem If the point X does not lie on any side of ΔABC, and AB meets CX in P, BC meets AX in Q and CA meets BX in R, then (AX/XQ) = (AP/PB) + (AR/RC), (BX/XR) = (BP/PA) + (BQ/QC), (CX/XP) = (CQ/QB) + (CR/RA).
Proof
Corollary
These follow at once since AQ, BR, CP are the medians,


Theorem 2 If the point X lies inside ΔABC, then AB meets CX in P, BC meets AX in Q and CA meets BX in R, such that (AX/XQ) + (BX/XR) + (CX/XP) ≥ 6. Equality occurs if and only if X is the centroid of ΔABC.
Proof

Suppose that x > 0. Then x + 1/x 2 = (x^{2}+12x)/x = (x1)^{2}/x ≥ 0 with equality if and only if x = 1. 