maths

Asked on December 26, 2019 by **Apoorva Christina**

$ar.(ΔAGB)=ar.(ΔAGC)=ar(ΔBGC)=31 ar(ΔABC)$

Given,

AM, BN and CL are medians.

To prove:

$ar(ΔAGB)=ar(ΔAGC)=ar(ΔBGC)$

$=31 .ar(ΔABC)$

Proof:

To $ΔAGB$ & $ΔAGC$

AG is the median

$∴$ $ar.ΔAGB=ar.ΔAGC$

Similarly

BG is the median

$∴ar.ΔAGB=ar.ΔBGC$

So we can say that

$ar.ΔAGB=ar.ΔAGC=ar.ΔBGC$

Now,

$ΔAGB+ΔAGC+ΔBGC=ar.ΔABC$

$31 .ΔAGB+31 ΔAGC+31 ΔBGC=ar.ΔABC$

(they are in equal area)

$⇒31 (ΔAGB+ΔAGC+ΔBGC)=ar.ΔABC$

$⇒ΔAGB+ΔAGC+ΔBGC=31 .arΔABC$

Hence proved.

Answer verified by Toppr

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