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

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Answer 2

Given,

ABCD is a kite , in which

AB=AD and BC =CD

P,Q,R,S are mid points of side AB,BC,CD &DA

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

∴ PQ││AC , PQ = ½AC

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

∴RS││AC and RS = 1/2 AC………….(ii)

From (i) and(ii) we have

PQ││RS , PQ=RS

Since AB=AD

AB=½ AD

AP=AS……………(iii)

=∠1=∠2…………….(iv)

Now in ∆PBQ and ∆SDR

PB= SD ∵ AD=AB

BQ = DR ∴ PB=SD

And PQ = SR ∵ PQRS is a parallelogram

So by sss congruency

∆PBQ≅∆SOR

∠3=∠4

Now, ∠3+SPQ+∠2 =180°

∠1+∠PSR+∠4 = 180°

∴ ∠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.