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Prove that the angle bisectors of a parallelogram forms a rectangle.


Given ABCD is a parallelogram 
To prove PQRS is a rectangle​

Now    ∠DAB +∠ADC = 180°(consecutive interior angles as AB∥DC and DA is a transverse)⇒∠DAB2+∠ADC2=90°⇒∠PAD +∠ADP =90°Now in ∆APD ,∠APD+∠PAD +∠ADP =180°(sum of interior angles of a triangle is 180°)⇒∠APD =180°-(∠PAD +∠ADP ) =180°-90°=90°So ∠SPQ =∠APD = 90°(vertically opposite angles)Similarly we can find ∠PQR =90°,∠QRS=90°and∠PSR =90°Thus quadrilateral PQRS having each angle =90°Hence PQRS is a rectangle

 

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Hi! To view the solution, go through Class IX NCERT book Chapter 8 Page 144 Example 5 If you face some problem, write the specific step (s) so that we can provide you meaningful help. Cheers!
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Hi! To view the solution, go through Class IX NCERT book Chapter 8 Page 144 Example 5 If you face some problem, write the specific step (s) so that we can provide you meaningful help. Cheers!
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Hi!

Here is the answer to your question.

Let PQRS be the given parallelogram and ABCD be the quadrilateral formed by its angle bisectors.

In parallelogram PQRS, we have

P = R and S = Q

In ∆PSL and ∆RQN

S = Q

SPL = QRN

and PS = RQ

∴ ∆PSL ≅ ∆RQN (by ASA criterion)

PLS = QNR (by CPCT)

Since the line segments PL and RN make equal angle on the parallel lines SR and PQ respectively, PL || RN.

This means that CD || AB.

Similarly, from ∆MPS and ∆ROQ

P = R

MSP = OQR

and PS = RQ

∴ ∆MPS ≅ ∆ROQ (by ASA criterion)

PMS = ROQ (by CPCT)

Since the line segments SM and QO make equal angle on the parallel lines PQ and RS respectively, SM || OQ.

This means that DA || BC.

we have already proved that CD || AB.

Thus, ABCD is a parallelogram.

Cheers!

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