Question

.Prove that the bisectors of alternate interior angles are parallel, if two parallel lines are intersected by a transversal.​

Answer 1

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)

⇒1/2∠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=180°

⇒1/2∠CYD=2180=90°⇒21(∠CYX+∠DYX)=900⇒21∠CYX+21∠DYX=90°⇒∠3+∠4=90°⇒∠PYQ=90°...(4)

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

Now, since alternate lines are parallel to each other in a rectangle

We can say that, the bisectors of alternate interior angles are parallel, if two parallel lines are intersected by a transversal.

Hence proved.

Hope this helps you

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Answer 2

Answer:

Given :- Two parallel lines AB and CD and transversal EF intersects them at G and H respectively. GM and HN are the bisectors of the alternate angles AGH and GHD.

To prove :- GM ll HN.

Proof :- Since AB ll CD and transversal EF cuts them at G and H respectively. Therefore,

AGH = GHD

1/2 ∠GHD = 1/2 ∠EGB  

⇒ MGH = GHN

Thus, two lines and HN are intersected by a transversal GH at G and H respectively such that a pair of alternate angles are equal.

Hence, GM ll HN.