Question

If the medians of a ΔABBC intersect at G, show that ar(ΔAGB) = ar(ΔAGC) = ar(ΔBGC) = `(1)/(3)"ar(ΔABC)"`.

Answer 1

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)"`.