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Question
The medians of a triangle ABC intersect each other at point G. If one of its medians is AD,
prove that:
(i) Area ( ΔABD ) = 3 x Area ( ΔBGD )
(ii) Area ( ΔACD ) = 3 x Area ( ΔCGD )
(iii) Area ( ΔBGC ) = `1/3` x Area ( ΔABC ).
Solution
(i) The figure is shown below
(i) Medians intersect at centroid.
Given that C is the point of intersection of medians and hence G is the centroid of the triangle ABC.
Centroid divides the medians in the ratio 2: 1
That is AG: GD = 2: 1.
Since BG divides AD in the ratio 2: 1, we have,
`"A( ΔAGB )"/"A( ΔBGD)" = 2/1`
⇒ Area( ΔAGB ) = 2Area( ΔBGD )
From the figure, it is clear that,
Area( ΔABD ) = Area( ΔAGB ) + Area( ΔBGD )
⇒ Area( ΔABD ) = 2Area( ΔBGD ) + Area( ΔBGD )
⇒ Area( ΔABD ) = 3Area( ΔBGD ) ......(1)
(ii) Medians intersect at centroid.
Given that G is the point of intersection of medians and hence G is the centroid of the triangle ABC.
Centroid divides the medians in the ratio 2: 1
That is AG: GD = 2: 1.
Since CG divides AD in the ratio 2: 1, we have,
`"A( ΔAGC )"/"A( ΔCGD)" = 2/1`
⇒ Area( ΔAGC ) = 2Area( ΔCGD )
From the figure, it is clear that,
Area( ΔACD ) = Area( ΔAGC ) + Area( ΔCGD )
⇒ Area( ΔACD ) = 2Area( ΔCGD ) + Area( ΔCGD )
⇒ Area( ΔACD ) = 3Area( ΔCGD ) ......(2)
(iii) Adding equations (1) and (2), We have,
Area( ΔABD ) + Area( ΔACD ) = 3Area( ΔBGD ) + 3Area( ΔCGD )
⇒ Area( ΔABC ) = 3[ Area( ΔBGD ) + Area( ΔCGD ) ]
⇒ Area( ΔABC ) = 3[ Area( ΔBGC ) ]
⇒ `"Area( ΔABC )"/3 = [ Area( ΔBGC )]`
⇒ Area( ΔBGC ) = `1/3` Area( ΔABC )
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