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If bisectors of ∠A and ∠B of a quadrilateral ABCD intersect each other at P, of ∠B and ∠C at Q, of ∠C and ∠D at R and of ∠D and ∠A at S, then PQRS is a
  • a)
    Rectangle
  • b)
    Quadrilateral whose opposite angles are supplementary
  • c)
    Parallelogram
  • d)
    Rhombus
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
If bisectors of∠Aand∠Bof a quadrilateral ABCD intersect each o...
To show: ∠PSR + ∠PQR = 180°
∠SPQ + ∠SRQ = 180°
In △DSA,
∠DAS + ∠ADS + ∠DSA = 180° (angle sum property)
+ ∠ SA = 180° (since RD and AP are bisectors of ∠D and ∠A)
∠DSA = 180°
∠PSR = 180°−
(∵ ∠DSA = ∠PSR are vertically opposite angles)
Similarly,
∠PQR = 180°− 

Adding (i) and (ii), we get, ∠PSR + ∠PQR = 180°
=360° − 1/2 ​× (∠A + ∠B + ∠C + ∠D)
 
=360°− 1/2​ × 360° = 180° ∴ ∠PSR + ∠PQR = 180°
In quadrilateral PQRS,
∠SPQ + ∠SRQ + ∠PSR + ∠PQR = 360°
=> ∠SPQ + ∠SRQ + 180° = 360°
=> ∠SPQ + ∠SRQ = 180°
Hence, showed that opposite angles of PQRS are supplementary.
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Most Upvoted Answer
If bisectors of∠Aand∠Bof a quadrilateral ABCD intersect each o...
To show: ∠PSR + ∠PQR = 180°
∠SPQ + ∠SRQ = 180°
In △DSA,
∠DAS + ∠ADS + ∠DSA = 180° (angle sum property)
+ ∠ SA = 180° (since RD and AP are bisectors of ∠D and ∠A)
∠DSA = 180°
∠PSR = 180°−
(∵ ∠DSA = ∠PSR are vertically opposite angles)
Similarly,
∠PQR = 180°− 

Adding (i) and (ii), we get, ∠PSR + ∠PQR = 180°
=360° − 1/2 ​× (∠A + ∠B + ∠C + ∠D)
 
=360°− 1/2​ × 360° = 180° ∴ ∠PSR + ∠PQR = 180°
In quadrilateral PQRS,
∠SPQ + ∠SRQ + ∠PSR + ∠PQR = 360°
=> ∠SPQ + ∠SRQ + 180° = 360°
=> ∠SPQ + ∠SRQ = 180°
Hence, showed that opposite angles of PQRS are supplementary.
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If bisectors of∠Aand∠Bof a quadrilateral ABCD intersect each o...
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