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Prove that quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle.

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Given: ABCD is a parallelogram. 

Rays AS, BQ, CQ and DS bisect ∠A, ∠B, ∠C and ∠D respectively. 

To prove: PQRS is a rectangle. 

Proof: 

∠BAS = ∠DAS = x° …(i) [ray AS bisects ∠A] 

∠ABQ = ∠CBQ =y° ….(ii) [ray BQ bisects ∠B] 

∠BCQ = ∠DCQ = u° …..(iii) [ray CQ bisects ∠C] 

∠ADS = ∠CDS = v° ….(iv) [ray DS bisects ∠D] 

ABCD is a parallelogram. [Given] 

∴ ∠A + ∠B = 180° [Adjacent angles of a parallelogram are supplementary] 

∴ ∠BAS + ∠DAS + ∠ABQ + ∠CBQ = 180° [Angle addition property] 

∴ x° + x° + v° + v° = 180 [From (i) and (ii)] 

∴ 2x° + 2v° =180

∴ x° + v° = 90° …..(vi) [Dividing both sides by 2] 

In ∆ARB, 

∠RAB + ∠RBA + ∠ARB = 180° [Sum of the measures of the angles of a triangle is 180°] 

∴ x° + y° + ∠SRQ = 180° [A – S – R, B – Q – R] 

∴ 90° + ∠SRQ = 180° [From (v)] 

∴ ∠SRQ = 180°- 90° = 90° …..(vi) 

Similarly, we can prove 

∠SPQ = 90° …(viii) 

In ∆ASD, 

∠ASD + ∠SAD + ∠SDA = 180° [Sum of the measures of angles a triangle is 180°] 

∴ ∠ASD + x° + v° = 180° [From (vi)] 

∴ ∠ASD + 90° = 180° 

∴∠ ASD = 180°- 90° = 90° 

∴ ∠PSR = ∠ASD [Vertically opposite angles] 

∴ ∠PSR = 90° …..(ix) 

Similarly we can prove ∠PQR = 90° ..(x) 

∴ In PQRS, 

∠SRQ = ∠SPQ = ∠PSR = ∠PQR = 90° [From (vii), (viii), (ix), (x)] 

∴ PQRS is a rectangle. [Each angle is of measure 90°]

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