Theorem 4

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 = DC

ii) 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 = DC

In 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 two

sides 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

i.e., AG : GD = 2 : 1

Similarly, BG : GE = 2 : 1 and

CG : GF = 2 : 1

Hence the theorem is proved.