Question

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 ∠BPQ and ∠PQC respectively. Prove that line AB || line CD.

Answer 1

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)