Question

If two parrallel lines are intersectedly
a transuersal, prove that bisectors of the
fairs of interior angles enclass o rectangle​

Answer 1

Answer:

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)

⇒21​∠AXY=21​∠DXY

⇒∠1=∠4(∠1=21​∠AXYand∠4=21​∠DXY)

⇒YQPX​

(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)

⇒21​∠BXY=21​∠CYX

⇒∠2=∠3(∠2=21​∠BXYand∠3=21​∠CYX)

⇒XQPY​

(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=1800

⇒21​∠CYD=2180​=900⇒21​(∠CYX+∠DYX)=900⇒21​∠CYX+21​∠DYX=900⇒∠3+∠4=900⇒∠PYQ=900.......(4)​

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

Hence proved.

solution

Step-by-step explanation: