If the medians of a ΔABC intersect at G. Prove that ar(ΔBGC) is equal to 13ar(ΔABC).
Answer:
- 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. - 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) - 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)
- But,ar(ΔABC)=ar(ΔAGB)+ar(ΔAGC)+ar(ΔBGC)=3 ar(ΔBGC)[Using eq (iii)]∴ar(ΔBGC)=13ar(ΔABC)