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Question
If the medians of a ΔABBC intersect at G, show that ar(ΔAGB) = ar(ΔAGC) = ar(ΔBGC) = `(1)/(3)"ar(ΔABC)"`.
Solution
The median of a triangle divides it into two triangles of equal areas.
In ΔABC, AD is the median
⇒ ar(ΔABD) = ar(ΔACD) .......(i)
In ΔGBC, GD is the medan
⇒ ar(ΔGBD) = ar(ΔGCD) .......(ii)
Subtracting (ii) and (i),
ar(ΔABD) - ar(ΔGBD) = ar(ΔACD) - ar(ΔGCD)
⇒ ar(AGB) = ar(ΔAGC) .......(iii)
Similarly, ar(ΔAGB) = ar(ΔBGC) .........(iv)
From (iii) and (iv),
ar(ΔAGB) = ar(ΔBGC) = ar(ΔAGC) ........(v)
But ar(ΔAGB) = ar(ΔBGC) = ar(ΔAGC) = ar(ΔABC)
Therefore, 3 ar(ΔAGB) = ar(ΔABC)
⇒ ar(ΔAGB) = `(1)/(3)"ar(ΔABC)"`
Hence, ar(ΔAGB) = ar(ΔAGC) = ar(ΔBGC) = `(1)/(3)"ar(ΔABC)"`.
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