**Given**: A circle with centre P, AB is the diameter.

QA || RP, where QR is tangent to the circle.

**To prove**: RB is tangent to the circle.

We will prove ∠RBP = 90°.

**Construction**: Join BQ and PQ.

**Proof**: RQ is tangent to circle and PQ is the radius at point of contact Q.

∴ PQ ⊥ RQ (radius ⊥ tangent at point of contact)

⇒ ∠PQR = 90° ...(1)

QA || RP and PQ is the transversal.

∴ ∠2 = ∠3 ...(2) (alternate interior angles)

Also, PQ = PA (radii of same circle)

∴ In ∆PQA, ∠3 = ∠4 ...(3) (angles opposite to equal sides are equal)

(2) and (3) ⇒ ∠2 = ∠3 = ∠4

∠BPQ and ∠BAQ are the angles made by same arc BQ at the centre P and on the remaining part of the circle respectively.

∴ ∠BPQ = 2 ∠BAQ

i.e., ∠1 + ∠2 = 2∠4

⇒ ∠1 + ∠2 = ∠4 + ∠4

⇒ ∠1 = ∠4 (as ∠2 = ∠4)

So, we have ∠1 = ∠2 = ∠3 = ∠4

⇒ ∠1 = ∠2

Consider ∆BPR and ∆RPQ,

BP = PQ (radii of same circle)

∠1 = ∠2 (proved above)

RP = RP (common)

∴ ∆BPR ≅ ∆QPR (By SAS congruency)

⇒ ∠PBR = ∠PQR (c.p.c.t)

⇒ ∠PBR = 90° (using (1))

i.e., PB ⊥ BR

Thus, RB is tangent to circle at a point B.