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.