Question

if two parallel lines are intersected by a transversal then prove that the bisectors of any two alternate angles are parallel​

Answer 1

Answer:

Given: Two parallel lines AB and CD and a transversal EF intersect them at G and H respectively. GM, HM, GL and HL are the bisectors of the two pairs of interior angles.

To Prove: GMHL is a rectangle.

Proof:

∵AB∥CD

∴∠AGH=∠DHG (Alternate interior angles)

⇒  

2

1

​  

∠AGH=  

2

1

​  

∠DHG

⇒∠1=∠2

(GM & HL are bisectors of ∠AGH and ∠DHG respectively)

⇒GM∥HL

(∠1 and ∠2 from a pair of alternate interior angles and are equal)

Similarly, GL∥MH

So, GMHL is a parallelogram.

∵AB∥CD

∴∠BGH+∠DHG=180  

o

 

(Sum of interior angles on the same side of the transversal =180  

o

)

⇒  

2

1

​  

∠BGH+  

2

1

​  

∠DHG=90  

o

 

⇒∠3+∠2=90  

o

 .....(3)

(GL & HL are bisectors of ∠BGH and ∠DHG respectively).

In ΔGLH,∠2+∠3+∠L=180  

o

 

⇒90  

o

+∠L=180  

o

 Using (3)

⇒∠L=180  

o

−90  

o

 

⇒∠L=90  

o

 

Thus, in parallelogram GMHL, ∠L=90  

o

 

Hence, GMHL is a rectangle.

Step-by-step explanation:

Answer 2

Answer:

Given: AB and CD are two parallel lines and transversal EF intersects then at G and H respectively. GM and HN are the bisectors of two corresponding angles ∠EGB and ∠GHD respectively.

To prove: GM∥HN

Proof:

∵AB∥CD

∴∠EGB=∠GHD (Corresponding angles)

⇒

2

1

∠EGB=

2

1

∠GHD

⇒∠1=∠2

(∠1 and ∠2 are the bisector of ∠EGB and ∠GHD respectively)

⇒GM∥HN

(∠1 & ∠2 are corresponding angles formed by transversal GH and GM and HN and are equal.)

Hence, proved.

solution