Question

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Please solve this question








SHOW THAT THE BISECTORS OF A PARALLELOGRAM FORM A RECTANGLE




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Answer 1

Let PQRS be a parallelogram and let the bisectors of the angles P, Q, R and S form a quadrilateral ABCD. To prove ABCD is a rectangle.

Because PQRS is a parallelogram, angle P + angle S = 180 degrees. Hence P/2 +S/2 = 90 degrees. i.e., angle APS + angle ASP = 90 degrees.

So angle PAS = 90 degrees. Therefore angle DAB = 90 degrees (vertically opposite angle). Similarly we can prove the other angles of the quadrilateral ABCD are 90 degrees each. Hence ABCD is a rectangle.

OR

Given: ABCD is a parallelogram.

To prove that the angle bisectors form a rectangle.

Construction: Draw angle bisectors of <A, <B, <C and <D.

Proof: Since adjacent angles of the parallelogram are supplementary, the angle bisectors will meet each other at 90 degrees. So you have a quadrilateral with all four angles 90 deg each, so the quadrilateral is a rectangle.

Proved.

Answer 2

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QUESTION➡

SHOW THAT THE BISECTORS OF A PARALLELOGRAM FORM A RECTANGLE.

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Let P, Q , R , S be the points of intersection of the bisectors of Angle A and Angle B, Angle B and Angle C, Angle C and Angle D, and Angle D and Angle A respectively of the parallelogram.

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When, DS bisects Angle D and AS bisects Angle A, therefore,

$Angle DAS + Angle ADS = \frac{1}{2} angle \: A + angle \: B \:$

$= \frac{1}{2} (angleA + angleB)$

$= \frac{1}{2} \times 180(int \: angles \: on \: the \: same \: transversal)$

$= 90$

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Also,

Angle DAS+ Angle ADS + Angle DSA= 180° ( Angle Sum Property of a triangle)

=> 90+ angle DSA= 180°

=> Angle DSA= 90°

=> Angle PSR = 90°

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Similarly,

Angle APB = 90° and Angle SPQ = 90°

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Similarly,

Angle PQR= 90° and Angle SRQ= 90°

So, PQRS is a quadrilateral in which all angles are right angles.

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CONCLUSION,

Angle PSR= Angle PQR = 90°

And,

Angle SPQ= Angle SRQ=90°. So both of opp angles are equal.



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Hence,



PQRS is a parallelogram in which one angle is 90° and so, PQRS is a rectangle.

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PROVED