Q: 5 D, E and F are respectively the mid-points of the sides BC, CA and AB of a . Show that
(ii)
We already proved that BDEF is a ||gm.
Similarly, DCEF and DEAF are also parallelograms.
Now, ||gm BDEF and ||gm DCEF is on the same base EF and between same parallels BC and EF
Ar (BDEF) = Ar (DCEF)
Ar(BDF) = Ar (DEF) .............(i)
It is known that diagonals of ||gm divides it into two triangles of equal area.
Ar(DCE) = Ar (DEF).......(ii)
and, Ar(AEF) = Ar (DEF)...........(iii)
From equation(i), (ii) and (iii), we get
Ar(BDF) = Ar(DCE) = Ar(AEF) = Ar (DEF)
Thus, Ar (ABC) = Ar(BDF) + Ar(DCE) + Ar(AEF) + Ar (DEF)
Ar (ABC) = 4 . Ar(DEF)
Hence proved.