Given: In ∆ABC, D, E and F are mid-points
of sides BC, CA and AB respectively.
To prove:
(i) BDEF is a parallelogram
(ii) ar (∆DEF) = \(\frac { 1 }{ 4 }\) ar (∆ABC)
(iii) ar (BDEF) = \(\frac { 1 }{ 2 }\) ar (∆ABC)
Proof: (i) In ∆ABC, F is the mid-point of AB and E is the mid-point of AC.
Therefore from mid-point theorem
EF || BC and EF = \(\frac { 1 }{ 2 }\) AB = BD …(i)
Similarly, DE || AB and DE
= \(\frac { 1 }{ 2 }\) AB = BF … (ii)
EF || BD and EF = BD
⇒ BDEF is a parallelogram.
(ii) Here DF is diagonal of ||gm BDEF
ar (∆BDF) = ar (∆DEF) …(iii)
Similarly, ar (ADEF) = ar (∆AEF) …(iv)
and ar (∆CDE) = ar (∆DEF) …(v)
From (iii), (iv) and (v), we have
ar (∆BDF) = ar (∆AFE)
ar (∆DEF) = ar (∆CDE)
But ar (∆BDF) + ar (∆AFE) + ar (∆DEF) + ar
(∆CDE) = ar (∆ABC)
⇒ 4 ar (∆DEF) = ar (∆ABC)
⇒ ar (∆DEF) = \(\frac { 1 }{ 4 }\) ar (∆ABC) …(vi)
(iii) Now, area of parallelogram BDEF
= ar (∆BDF) + ar (∆DEF) [∵ ar (∆BDF) ar (∆DEF)]
⇒ ar (||gm BDEF) = 2 ar(∆DEF)
= 2x \(\frac { 1 }{ 4 }\) ar (∆ABC) [using relation (vi)]
= \(\frac { 1 }{ 2 }\) ar (∆ABC)
Hence, ar (||gm BDEF) = \(\frac { 1 }{ 2 }\) ar (∆ABC)
