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Question
If the medians of a ∆ABC intersect at G, show that ar (AGB) = ar (AGC) = ar (BGC) = `1/3` ar (ABC)
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Solution
Given: In ΔABC, AD, BE and CF are medians and intersect at G.
To prove: ar (ΔAGB) = ar (ΔAGC) = ar (ΔBGC) = `1/3` ar (ΔABC)
Proof: We know that, a median of a triangle divides it into two triangles of equal area.
In ΔABC, AD is a median
∴ ar (ΔABD) = ar (ΔACD) ...(i)
In ΔBGC, GD is a median
∴ ar (ΔGBD) = ar (ΔGCD) ...(ii)
On subtracting equation (ii) from equation (i), we get
ar (ΔABD) – ar (ΔGBD) = ar (ΔACD) – ar (ΔGCD)
⇒ ar (ΔAGB) = ar (ΔAGC) ...(iii)
Similarly, ar (ΔAGB) = ar (ΔBGC) ...(iv)
From equations (iii) and (iv),
ar (ΔAGB) = ar (ΔBGC) = ar (ΔAGC) ...(v)
Now, ar (ΔABC) = ar (ΔAGB) + ar (ΔBGC) + ar (ΔAGC)
⇒ ar (ΔABC) = ar (ΔAGB) + ar (ΔAGB) + ar (ΔAGB) ...[From equation (v)]
⇒ ar (ΔABC) = 3 ar (ΔAGB)
⇒ ar (ΔAGB) = `1/3` ar (ΔABC) ...(vi)
From eqattions (v) and (vi),
ar (ΔBGC) = `1/3` ar (ΔABC)
And ar (ΔAGC) = `1/3` ar (ΔABC)
Hence proved.
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