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Question

If the medians of a triangle ABC intersect at G, prove that ar(ΔAGB)=ar(ΔBGC)=ar(CGA)=13ar(ΔABC).

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Solution

Let ABC be any triangle whose vertices A,B and C respectively.

We consider that D,E and F be mid point of ΔABC and G be the centroid ΔABC


Then, AE,BF&CD are medians

To prove- area ΔAGB= area ΔBGC area ΔAGB= area ΔBGC= area ΔCGA=13 area ΔABC

Now,

In ΔAGB and ΔAGC

AG is the median

Therefore ,

arΔAGB=arΔAGC …….. (1)

Similarly,

BG is the median

Therefore,

arΔAGB=arΔBGC …… (2)

By equation (1)&(2) to,

arΔAGB=arΔAGC=arΔBGC ……. (3)

Now, We know that

arΔAGB+arΔBGC+arΔCGA=arΔABC ......(4)


By equation (3) and (4) to, we get

arΔAGB+arΔAGB+arΔAGB=arΔABC,

arΔAGB=13arABC

Substituting in equation(1),

arΔAGB=arΔBGC=arΔCGA=13arΔABC

Hence proved.


996103_1048176_ans_54ece6e7360049939c5496e29c90192f.jpg

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