Verified answer

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.)

Concept Introduction:

Lines on a surface that are consistently spaced apart are known as parallel lines. Parallel lines don't cross each other.

Solution:

Given,

[tex]AB[/tex]║[tex]CD[/tex]

[tex]EP[/tex] and [tex]FQ[/tex] are the bisectors of ∠[tex]APG[/tex] and ∠[tex]CQP[/tex]

We have to prove [tex]EP[/tex]║[tex]FQ[/tex]

[tex]AB[/tex]║[tex]CD[/tex]

∴∠[tex]EGB[/tex]=∠[tex]GHD[/tex] (Corresponding angles)

the bisector of these two are [tex]GM[/tex] and [tex]HN[/tex]

So, these two are parallel

So, the corresponding angles formed by transversal [tex]GH[/tex] and [tex]GM[/tex] [tex]HN[/tex] are equal.

Final Answer:

The final answer is proved

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