If the medians of a ` triangle A B C` intersect at `G ,` show that `a r( triangle A G B)=a r( triangle A G C)=a r( triangle B G C)=1/3a r( triangle A B C)`. GIVEN : `triangle A B C` such that its medians `A D ,B E` and `C F` intersect at `G`. TO PROVE : `a r(triangle A G B)=a r( triangle B G C)=a r(triangle A G C)=1/3a r(triangle A B C)`

Explanation

`In/_ABC`, AD is median

`ar(/_ABD)=ar(/_ACD)-(1)`

`In/_GBC`,GD is median

`ar(/_GBD)=ar(/_GCD)-(2)`

subtracting equation 2 from 1

`ar(/_ABG)=ar(/_ACG)-(3)`

`ar(/_AGB)=ar(/_BGC)-(4)`

`ar(/_ABG)=ar(/_BGC)=ar(/_AC C_1)`

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