Question

4. Medians BE and CF of a A ABC intersect at a point G. Prove that the
(i) ar(triangleGBC) = ar(quadrilateralAFGE)
(ii) ar(triangleAGB) = ar(triangleAGC) = ar(GBC) = 1/3ar(triangleABC)​

Answer 1

Answer:

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 )

Answer 2

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