a transversal EF intersects two parallel lines AB and CD at P and Q respectively
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Since ray PR bisects ∠BPQ and ray QS bisects ∠PQC, then
∠RPQ=∠RPB=
2
1
∠BPQ and ∠SQP=∠SQC=
2,1
∴ line AB ∥ line CD (Alternate angles triangle
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