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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.

 


Solution


Since ray PR bisects ∠BPQ and ray QS bisects ∠PQC, then
∠RPQ = ∠RPB = 12∠BPQ and ∠SQP = ∠SQC = 12∠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)

Mathematics
Mathematics Part - II (Solutions)
Standard IX

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