Question

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) = $\frac{1}{4}$ ar ($△$ABC). [4 MARKS]

Answer 1

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$\parallel$ BD---------(1) [Converse of mid-point theorem]

Similarly, BF $\parallel$ 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)

$\Rightarrow$ ar($△$DEF) = $\frac{1}{4}$ar($△$ABC)