If the medians of a `triangle`ABC intersect at G, show that `ar(triang – InterviewSolution

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1.	If the medians of a `triangle`ABC intersect at G, show that `ar(triangleAGB)=ar(triangleAGC)=ar(triangleBGC)` `=(1)/(3)ar(triangleABC)`.
Answer» GIVEN A `triangle`ABC. Its medians AD, BE and CF intersect at G. TO PROVE `ar(triangleAGB)=ar(triangleAGC)=ar(triangleBGC)=(1)/(3)ar(triangleABC)` PROOF We know that a median of a triangle divides it into two triangle of equal area. In `triangle`ABC,AD is the median. `therefore ar(triangleABD)=ar(triangleACD)." "...(i)` In `triangle`GBC, GD is the median. `therefore ar(triangleGBD)=ar(triangleGCD)." "...(ii)` From (i) and (ii), we get `ar(triangleABD)-ar(triangleGBD)=ar(triangleACD)-ar(triangleGCD)` `rArr ar(triangleAGB)=ar(triangleAGC)`. Similarly, `ar(triangleAGB)=ar(triangleBGC)`. `rArr ar(triangleAGB)=ar(triangleAGC)`. But, `ar(triangleABC)=ar(triangleAGB)+ar(triangleAGC)+ar(triangleBGC)` `=3ar(triangleAGB)" "["using (iii)"]`. `therefore ar(triangleAGB)=(1)/(3)ar(triangleABC)`. Hence, `ar(triangleAGB)=ar(triangleAGC)=ar(triangleBGC)=(1)/(3)ar(triangleABC).`

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