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We know that, the median of a triangle divide it into two triangles of equal area.
In ΔABC, AD is the median
∴ ar (ΔABD) = ar (ΔACD) ...(1)
In ΔGBC, GD is the median.
∴ ar(ΔGBD) = ar(ΔGCD) ...(2)
Subtracting (2) from (1), we get
ar(ΔABD) – ar(ΔGBD) = ar(ΔACD) – ar(ΔGCD)
∴ ar(ΔAGB) = ar(ΔAGC) ...(3)
Similarly, ar(ΔAGB) = ar(ΔBGC) ...(4)
From (3) and (4), we get
ar(ΔAGB) = ar(ΔAGC) = ar(ΔBGC) ...(5)
Now, ar(ΔAGB) + ar(ΔAGC) + ar(ΔBGC) = ar(ΔABC)
⇒ ar(ΔAGB) + ar(ΔAGB) + ar(ΔAGB) = ar(ΔABC) (Using (5))
⇒ 3ar(ΔAGB) = ar(ΔABC)
⇒ ar(ΔAGB) .......(6)
From (5) and (6), we get
ar(ΔAGB) = ar (ΔAGC) = ar(ΔBGC)
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