Sum

A transversal EF of line AB and line CD intersects the lines at point P and Q respectively. Ray PR and ray QS are parallel and bisectors `angle` BPQ and `angle` PQC respectively.

Prove that line AB || line CD.

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#### Solution

Since ray PR bisects ∠BPQ and ray QS bisects ∠PQC, then

∠RPQ = ∠RPB = `1/2`∠BPQ and ∠SQP = ∠SQC = `1/2`∠PQC

∴ ∠BPQ = 2∠RPQ and ∠PQC = 2∠SQP ....(1)

Since PR || QS and PQ is a transversal intersecting them at P and Q, then

∠RPQ = ∠SQP (Alternate interior angles)

On multiplying both sides by '2', we get

2∠RPQ = 2∠SQP

Now, using (1), we get

∠BPQ = ∠PQC

But ∠BPQ and ∠PQC are alternate interior angles formed by a transversal EF of line AB and line CD.

∴ line AB || line CD (Alternate angles test)

Concept: Test for Parallel Line

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