Question

medians of triangle ABC intersect at G show that triangle AGC = BGA=BGC=
1/3 of triangle ABC

Answer 1

Hey there,

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 )