D, E and F are respectively the mid-points of the sides BC, CA and AB of a △ABC. Show that
(i) BDEF is a parallelogram.(ii) ar (△DEF) = 14 ar (△ABC). [4 MARKS]
Statement of the theorem : 1 mark
Application of the theorem : 1 mark
Calculation : 2 marks
Given: D, E and F are respectively the mid-points of the sides BC, CA and AB of △ABC.
i.) In △ABC, F is the midpoint of AB and E is the midpoint of AC
Therefore, EF∥ BD---------(1) [Converse of mid-point theorem]
Similarly, BF ∥ DE----------(2)
In view of (1) and (2), BDEF is a parallelogram.
ii.) As in (i) , we can prove that AFDE and FDCE are parallelograms.
FD is the diagonal of the parallelogram BDEF.
So, ar(△FBD) = ar(△DEF)----(3)
Similarly, ar(△DEF) = ar(△FAE)-----(4)
And ar(△DEF) = ar(△DCE)----(5)
From (3), (4) and (5)
△ABC is divided into 4 triangles of equal areas.
Therefore, ar(△ABC) = ar(△FBD) + ar(△DEF) + ar(△FAE) + ar(△DCE)
= 4 × ar(△DEF)
⇒ ar(△DEF) = 14ar(△ABC)