Question

if two parallel lines are intersected by a transversal, then show that the bisectors of any pair of alternate angles, are also parallel​

Answer 1

Step-by-step explanation:

If two parallel lines are intersected by a transversal then prove that the bisectors of any pair of alternate interior angles are parallel. GIVEN AB∣∣CD are cut by a transversal t at E and F respectively, EG and FH are the bisectors of a pair of alt. int. ∠s,∠AEF and anl≥EFD respectively.

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Answer 2

Step-by-step explanation:

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.

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

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∥HNProof:

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∥HNProof:∵AB∥CD

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∥HNProof:∵AB∥CD∴∠EGB=∠GHD (Corresponding angles)

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∥HNProof:∵AB∥CD∴∠EGB=∠GHD (Corresponding angles)⇒21∠EGB=21∠GHD

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∥HNProof:∵AB∥CD∴∠EGB=∠GHD (Corresponding angles)⇒21∠EGB=21∠GHD⇒∠1=∠2

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∥HNProof:∵AB∥CD∴∠EGB=∠GHD (Corresponding angles)⇒21∠EGB=21∠GHD⇒∠1=∠2(∠1 and ∠2 are the bisector of ∠EGB and ∠GHD respectively)

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∥HNProof:∵AB∥CD∴∠EGB=∠GHD (Corresponding angles)⇒21∠EGB=21∠GHD⇒∠1=∠2(∠1 and ∠2 are the bisector of ∠EGB and ∠GHD respectively)⇒GM∥HN

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∥HNProof:∵AB∥CD∴∠EGB=∠GHD (Corresponding angles)⇒21∠EGB=21∠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.)

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∥HNProof:∵AB∥CD∴∠EGB=∠GHD (Corresponding angles)⇒21∠EGB=21∠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.