Question

prove that ray AB is parallel to ray EF​

Answer 1

Answer:

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)