Question

Prove that the centroid divides the median in the ratio 2:1​

Answer 1

Simple Proof

Reflect the triangle along AC,

ABCB1 is a parallelogram.

BEB1 is a straight line .

Since CD = AD1 and CD // AD1,

DCD1A is a parallelogram. (opposite sides equal and parallel.)

\ DG // CG1

Since BD = DC and DG // CG1 \ BG = GG1 (intercept theorem)

BG : GG1 = 1 : 1

Since GE = EG1 , BG : GE = 2 : 1.

diagram acha nhi h

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Answer 2

THE CENTROID OF A TRIANGLE DIVIDES EACH MEDIAN IN THE RATIO 2:1

The theorem

D, E, F are the mid points BC, CA, AB.

AD, BE, and CF are medians.

The medians cut each other are centroid G.

We need to show that :

AG:GD = BG:GE = CG:GF = 2:1

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SIMPLE PROOF

Reflect the triangle along AC, you can get a diagram.

ABCB^1 is a parallelogram.

BEB^1 is a straight line.

DCD^1A is a parallelogram (opposite sides are equal and parallel.)

\ DG // CG^1.

Since BD = DC and DG // CG^1 \ BG = GG^1 (intercept theorem).

BG : GG^1 = 1 : 1

Since GE = EG^1, BG : GE = 2:1

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