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Question

If the median of a ABC intersect at G. show that ar (AGC) = ar (AGB) = ar (BGC) = 13 ar (ABC)

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Solution

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
( PROVED )


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