If the medians of a ΔABC intersect at G. Prove that ar(ΔBGC) is equal to 13ar(ΔABC).
A C B D E F G


Answer:


Step by Step Explanation:
  1. We are given that AD, BE and CF are the medians of ΔABC intersecting at G.

    Now, we have to find the area of ΔBGC.
  2. We know that a median of a triangle divides it into two triangles of equal area.

    Now, in ΔABC, AD is the median. ar(ΔABD)=ar(ΔACD)(i) Similarly, in ΔGBC, GD is the median. ar(ΔGBD)=ar(ΔGCD)(ii)
  3. From (i) and (ii), we get: ar(ΔABD)ar(ΔGBD)=ar(ΔACD)ar(ΔGCD)ar(ΔAGB)=ar(ΔAGC) Similarly, ar(ΔAGB)=ar(ΔBGC) ar(ΔAGB)=ar(ΔAGC)=ar(ΔBGC)(iii)
  4. But,ar(ΔABC)=ar(ΔAGB)+ar(ΔAGC)+ar(ΔBGC)=3 ar(ΔBGC)[Using eq (iii)]ar(ΔBGC)=13ar(ΔABC)

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