If the medians of a triangle ABC intersect at G, prove that:
ar.(ΔAGB)=ar.(ΔAGC)=ar(ΔBGC)=13ar(ΔABC)
Given,
AM, BN and CL are medians.
To prove:
ar(ΔAGB)=ar(ΔAGC)=ar(ΔBGC)
=13.ar(ΔABC)
Proof:
To ΔAGB & ΔAGC
AG is the median
∴ ar.ΔAGB=ar.ΔAGC
Similarly
BG is the median
∴ar.ΔAGB=ar.ΔBGC
So we can say that
ar.ΔAGB=ar.ΔAGC=ar.ΔBGC
Now,
ΔAGB+ΔAGC+ΔBGC=ar.ΔABC
13.ΔAGB+13ΔAGC+13ΔBGC=ar.ΔABC
(they are in equal area)
⇒13(ΔAGB+ΔAGC+ΔBGC)=ar.ΔABC
⇒ΔAGB+ΔAGC+ΔBGC=13.arΔABC
Hence proved.