Question

If the medians of a triangle ABC intersect at G, prove that area of triangle BCG = 1/3 area of triangle ABC.

Answer 1

given ,
AM , BN & CL are medians
to prove -ar. ΔAGB = ar.ΔAGC , ar.ΔAGB = ar. ΔBGC & ΔAGB= 1/3ar.ΔABC
proof ,
in 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 = ΔBGC
now ,
ΔAGB + ΔAGC + ΔBGC = ar. ΔABC
1/3ΔAGB +1/3ΔAGC + 1/3ΔBGC ( they are equal in area ) = ar. ΔABC
so we can say that ,
ΔAGB = ar.ΔAGC = ΔBGC = ar 1/3 ΔABC