D, E and F are respectively the mid-points of the sides BC, CA and AB of a`DeltaA B C`.Show that (i) BDEF is a parallelogram. (ii) `a r\ (D E F)=1/4a r\ (A B C)`(iii) `a r\ (B D E F)=1/2a r\ (A B C)`

Explanation

Given: A `∆ABC` in which `D`,`E`,`F` are the mid-point of the side `BC`, `CA` and `AB` respectively. To prove: (i) Quadrilateral `BDEF` is parallelogram.
(ii) `ar(║gm BDEF)=1/2ar(/_\ABC)`.
(iii) `ar(/_\DEF)=1/4ar(/_\ABC)`. Proof: (i) In `/_\ABC`,
`∴ F` is the mid-point of side `AB` and `E` is the midpoint of side `AC`.
`∴ EF║BD` [`∵`Line joining the mid-points of any two sides of a triangle is parallel to the third side.]
Similarly, `ED║FB`.
Hence, `BDEF` is a parallelogram.
Hence Proved.

(ii) Similarly, prove that `AFDE` and `FDCE` are parallelograms.
`∴ FD` is diagonals of parallelogram `BDEF`.
`∴ar(/_\FBD)=ar(/_\DEF)` ...(i)
Similarly, `ar(/_\FAE)=ar(/_\DEF)` . ..(ii)
And `ar(/_\DCE)=ar(/_\DEF)` ...(iii)
From equations (i), (ii) and (iii), we get:
`ar(/_\FBD)=ar(/_\FAE)=ar(/_\DCE)=ar(/_\DEF)`
And `ar(/_\FBD)+ar(/_\DCE)+ar(/_\DEF)+ar(/_\FAE)=ar(/_\ABC)`
`⇒ 2[ar(/_\FBD) + ar(/_\DEF)] = ar(/_\ABC)`
`⇒2[ar(║gmBDEF)]=ar(/_\ABC)`
`⇒ar(║gmBDEF)=1/2ar(/_\ABC)`

(iii) Since, `/_\ABC` is divided into four non-overlapping triangles `FBD`, `FAE`, `DCE` and `DEF`.
From (i), (ii) and (iii)
`∴ ar(/_\ABC)=ar(/_\FBD)+ar(/_\FAE)+ar(/_\DCE)+ar(/_\DEF)`
`⇒ar(/_\ABC)=4ar(/_\DEF)`
`⇒ar(/_\DEF)=1/2ar(/_\ABC)`.