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.
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)