In the adjoining figure, BD||CA, E is the midpoint of CA and `BD =(1)/(2) CA`. Prove that
`ar(triangleABC)=2ar(triangleDBC)`.

Explanation

`BD =(1)/(2)CA` and E is a midpoint of `CA rArr BD = CE`.
Now, BD||CE and BD = `CE rArr BCED` is a ||gm.
`ar(triangleDBC)=ar(triangleBCE)=(1)/(2)ar(triangleABC)" " [therefore BC " is a median of" triangleABC]`
`rArr ar(triangleABC)=2ar(triangleDBC)`.