Question

If the median of a $△ABC$ intersect at G. show that ar ($△AGC$) = ar ($△AGB$) = ar ($△BGC$) = $\frac{1}{3}$ ar ($△ABC$)

Answer 1

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 )