Maharashtra State BoardSSC (English Medium) 9th Standard
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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. - Geometry

Sum

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.

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Solution

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)

  Is there an error in this question or solution?
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APPEARS IN

Balbharati Mathematics 2 Geometry 9th Standard Maharashtra State Board
Chapter 2 Parallel Lines
Practice Set 2.2 | Q 6 | Page 22
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