Question

ABCD is a kite having AB=AD and BC=CD. Prove that the figure formed by joining the mid-points of the sides, in order, is a rectangle.

Answer 1

AB =AD and BC =CD (Given)

P,Q,R,S are mid points of side AB,BC,CD  and DA (Given)

In ∆ABC , P&Q are mid points of AB & BC

Therefore, PQ││AC  and PQ = 1/2AC -- 1

Similarly,

In ∆ADC , R & S are mid points of CD & AD

Therefore, RS││AC and RS = 1/2 AC -- 2

From 1 and 2 -

PQ││RS , PQ=RS

Since AB=AD

Thus, AB = 1/2 AD

AP=AS -- 3

= ∠1 = ∠2 -- 4

In ∆PBQ and ∆SDR

PB= SD

BQ = DR

PQ = SR

∆PBQ≅∆SOR ( By SSS Congruency)

∠3=∠4

∠3+SPQ+∠2 =180°

∠1+∠PSR+∠4 = 180°

Thus,

∠3+∠SPQ+∠2= ∠1+∠PSR+∠4

∠SPQ=∠PSR (∠2=∠1 and ∠3=∠4)

∠SPQ+∠PSR = 180°

=2∠SPQ = 180°

= ∠SPQ = 90°

Hence , PQRS is a parallelogram.