Question

If a transversal intersects two lines such that the
bisectors of a pair of corresponding angles are
equal, then prove that the two lines are parallel.​

Answer 1

In the given figure , a transversal AD intersects two lines PQ and RS at points B and C respectively. Ray BE is the bisector of angle ABQ and ray CG is the bisector of angle BCS and BE || CG

To prove that PQ || RS :-

It is given that ray BE is the bisector of angle ABQ

Therefore , ABE = $\frac{1}{2}ABQ$ (equation 1)

Similarly , Ray CG is the bisector of angle BCS

Therefore , BCG = $\frac{1}{2}BCS$ (equation 2)

But BE || CG and AD is the transversal

Therefore , ABE = BCG (Corresponding angles) (equation 3)

Substituting equation (1) and (2) in equation (3) , we get

$\frac{1}{2}ABQ$ = $\frac{1}{2}BCS$

That is , ABQ = BCS

But they are the corresponding angles formed by transversal AD with PQ and RS and are equal

Therefore PQ || RS

Note :- Refer to the attachment

Answer 2

Answer:

The transversal AD intersects the two lines PQ and RS at points B and C respectively. BE is the bisector of ∠ABQ and CF is the bisector of ∠BCS.

As, BE is the bisector of ∠ABQ, then,

∠ABE=  

2

1

​  

∠ABQ

In the same way,

∠BCF=  

2

1

​  

∠BCS

Since BE and CF are parallel and AD is the transversal, therefore, by corresponding angle axiom,

∠ABE=∠BCF

2

1

​  

∠ABQ=  

2

1

​  

∠BCS

∠ABQ=∠BCS

Therefore, by the converse of corresponding angle axiom,

PQ∥RS.

Step-by-step explanation:

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