Question

In figure, medians of the ∆ABC intersect at point G. show that:
ar(∆AGB)=⅓ar(∆ABC)
( use ratio property of centroid that is centroid divides median in a ratio of 2:3)​

Answer 1

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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  

  ( PROVED )