Question

If the medians of a ΔABC intersect at G show that ar (AGB) = ar (AGC) =ar (BGC) = 1/3 ar (ABC)

Answer 1

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

Answer:

hey

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AD is the median of ∆ABC

ar(∆ABD)=ar(∆ADC).....(1)

GD is median of ∆GBC

ar(∆GBD)=ar(∆GDC)......(2)

eqn(1)-eqn(2)

ar(∆ABD)-ar∆GBD)=ar(∆ADC)-ar(∆GDC)

ar(∆ABG)=ar(∆AGC)

BE is median of ∆ABC

ar(∆ABE)=ar(∆BEC).....(3)

GE is median of ∆AGC

ar(∆AGE)=ar(∆GEC)......(4)

subtracting eqn (3) -eqn(4)

ar(∆ABE)-ar(∆AGE)=ar(∆BEC)-ar(∆GEC)

ar(∆AGB)=ar(∆BGC)

now ,area of all triangles are equal

ar(∆AGB)=ar(∆AGC)=ar(∆BGC)

ar(∆AGB)+ar(∆AGC)+ar(BGC)=ar(∆ABC)

ar(∆AGB)+ar(∆AGB)+ar(∆AGB)=ar(ABC)

3×(ar∆AGB)=ar(∆ABC)

•ar(∆AGB)=1/3ar(∆ABC)

prooved

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hope helped

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