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Statement
The medians of a triangle pass through the same point which divides each of the medians in the ratio 2:1.

Given:
In DABC, medians BE and CF intersects at G. AG is joined and produced to meet BC in D.
To prove:
i) AD is the median.
i.e., BD = DCii) AG : GD = BG : GE = CG : GF = 2 : 1
Construction:
Produce AD to K such that AG = GK. Join KB and KC.
Proof:
i) AD is the median
i.e., BD = DCIn DABK,
F is the mid-point of AB . (given)G is the mid-point of AK. (by construction)
FG || BK
or GC || BK ...(1)
By theorem on line joining the mid-points of any twosides of a triangle.
Similarly in DACK,GE || CK or GB || CK
From (1) and (2), BKCG is a parallelogram.
Now in BKCG, BC and GK are diagonals.The diagonals of a parallelogram bisect each other.
BD = DC or AD is the median from A.
(ii) GK and BC are diagonals of GCKB.
GD = DK

Similarly, BG : GE = 2 : 1 and
CG : GF = 2 : 1Hence the theorem is proved.

