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 $\angle$BPQ and $\angle$PQC respectively. Prove that line AB || line CD.

Solution

## Since ray PR bisects ∠BPQ and ray QS bisects ∠PQC, then ∠RPQ = ∠RPB = $\frac{1}{2}$∠BPQ and ∠SQP = ∠SQC = $\frac{1}{2}$∠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)MathematicsMathematics Part - II (Solutions)Standard IX

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