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Question

Prove that the bisector of interior angles of a parallelogram form a rectangle.


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Solution

STEP 1 : Assumptions

Assume that ABCD is a parallelogram.

Let P,Q,R,S be the point of intersection of bisectors of A and B, B and C, C and D, D and A, respectively as shown in figure.

STEP 2 : Finding the value of DSA

From the triangle, ASD, we can observe that DS bisects D and AS bisects A.

Therefore,

ADS+DAS=12D+12A

ADS+DAS=12A+D

ADS+DAS=12180° [Since A and D are the interior angles on the same side of the transversal]

Hence,

ADS+DAS=90°

Using the angle sum property of a triangle, we can write:

DAS+ADS+DSA=180°

Now, substitute ADS+DAS=90° in the above equation, we get

90°+DSA=180°

DSA=180°-90°

DSA=90°

STEP 3 : Finding the value of APB,DRC,PSR and PQR

Since, PSR is being vertically opposite to DSA,

We can say PSR=90°

Likewise, it can be shown that APB=90° or APB=90°

Similarly, PQR=90° and SRQ=90°.

We have proved that PSR=PQR=90° and APB=CRD=90°.

As, both the pairs of opposite angles are equal to 90°, we can conclude that PQRS is a rectangle.

Hence, proved.


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