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Question 3
The median BE and CF of a triangle ABC intersect at G. Prove that the area of ΔGBC = area of the quadrilateral AFGE.

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Solution

BE is the median of ΔABC and we know that a median of a triangle divides it into two parts of equal area.

ar(ΔABE)=ar(ΔCBE)
ar(ΔABE)=12×ar(ΔABC) ...(i)

Similarly, ar(ΔACF)=ar(ΔBCF)
ar(ΔBCF)=12×ar(ΔABC) ...(ii)

From eqs. (i) and (ii),
ar(ΔABE)=ar(ΔBCF) ....(iii)
On subtracting ar(ΔGBF) from both sides of Eq. (iii), we get,
ar(ΔABE)ar(ΔGBF)=ar(ΔBCF)ar(ΔGBF)
ar(AFGE)=ar(GBC)


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