Question

Prove that a quadrilateral formed by the intersection of bisectors of interior angle made by a transersal on two parallel line is a rectangles

Answer 1

Rsjayant Ambitious

but angle LGH+ ANGLE LHG+ Angle GLH=180°

(ANGLE SUM PROPERTY)

90°+angle GLH=180°

(°.° angle LHG + angle LGH =90° )

Angle GLH = 90°

THUS in a parallelogram we have angle GLH =90°

hence , GMHL is a rectangle

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Answer 2

Given:
Line l || Line m and Line p is the transversal 

To prove:
PQRS is a rectangle 

Proof: 
RS, PS, PQ and RQ are bisectors of interior angles formed by the transversal with the parallel lines. 

∠RSP = ∠RPQ (Alternate angles) 
Hence Rs || PQ

Similarly, PS||RQ (∠RPS = ∠PRQ)

 Therefore quadrilateral PQRS is a parallelogram as both the pairs of opposite sides are parallel.

From the figure, we have
∠b + ∠b + ∠a + ∠a = 180° 
⇒ 2(∠b + ∠a) = 180° 
∴ ∠b + ∠a = 90°

That is PQRS is a parallelogram and one of the angle is a right angle. 
Hence PQRS is a rectangle

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