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 ).

Answer 1

(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 )