Question

the medians of a triangle intersect at g, show that area of triangle agb=1÷3area of triangleabc
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Answer 1

Step-by-step explanation:

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 )