Question

if two parallel lines are intersected by a transversal then prove that the bisectors of the interior angles on both sides of transversal form a rectangle​

Answer 1

Step-by-step explanation:

f

AB and CD are two parallel line intersected by a transverse L

X and Y are the point of intersection of L with AB and CD respectively.

XP, XQ, YP and YQ are the angle bisector of ∠AXY,∠BXY,∠CYXand∠DYX

AB∥CD and L is transversal.

∴∠AXY=∠DYX (pair of alternate angle)

⇒

2

1

∠AXY=

2

1

∠DXY

⇒∠1=∠4(∠1=

2

1

∠AXYand∠4=

2

1

∠DXY)

⇒

YQ

PX

(If a transversal intersect two line in such a way that a pair of alternate

interior angle are equal, then the two line are parallel)

(1)

Also ∠BXY=∠CYX (pair of alternate angle)

⇒

2

1

∠BXY=

2

1

∠CYX

⇒∠2=∠3(∠2=

2

1

∠BXYand∠3=

2

1

∠CYX)

⇒

XQ

PY

(If a transversal intersect two line in such a way that a pair of alternate

interior angle are angle, then two line are parallel)

(2)

from (1) and (2), we get

PXQY is parallelogram ....(3)

∠CYD=180

0

⇒

2

1

∠CYD=

2

180

=90

0

⇒

2

1

(∠CYX+∠DYX)=90

0

⇒

2

1

∠CYX+

2

1

∠DYX=90

0

⇒∠3+∠4=90

0

⇒∠PYQ=90

0

.......(4)

So using (3) and (4) we conclude that PXQY is a rectangle.

Hence proved.