Question

In a circle with center P and AB as the diameter. Prove that RB is tangent to the circle.

Answer 1

Given: A circle with centre P, AB is the diameter.
$QA\parallel RP,$ where RQ is the tangent to the circle.
To prove: RB is tangent to the circle ie. $\angle RBP={90}^{\circ }$
Construction: Join BQ and PQ
Proof: RQ is tangent to the circle and PQ is the radius at point of contact $Q.\therefore PQ\perp RQ$ (radius of a circle is perpendicular to the tangent at point of contact)
$\angle PQR={90}^{\circ }\dots \dots \left(1\right)$
$\Rightarrow QA\parallel RPandPQ$ is the transversal
$\Rightarrow \angle 2=\angle 3\dots \dots$ (2) (Alternate interior angles)
But, PQ = PA (radii of the circle)
$\therefore In\Delta PQA,\angle 3=\angle 4\dots \dots \left(3\right)$ (Angles opposite to equal sides are equal)
From (2) and (3) $\Rightarrow \angle 2=\angle 3=\angle 4$
$\angle BPQand\angle BAQ$ are the angles made by the arc BQ at the centre P and on the remaining part of the circle respectively.
$\therefore \angle BPQ=2\angle BAQi.e.,\angle 1+\angle 2=2\angle 4\Rightarrow \angle 1=\angle 4(as\angle 2=\angle 4)So,\angle 1=\angle 2=\angle 3=\angle 4\Rightarrow \angle 1=\angle 2$
In $\Delta BPRand\Delta RPQ$,
BP = PQ (radii of the circle)
$\angle 1=\angle 2$ (proved)
RP = RP (common)
$\therefore \Delta BPR\cong \Delta QPR$ (SAS congruency)
$\Rightarrow \angle PBR=\angle PQR$ (corresponding angles)
But $\angle PBR={90}^{\circ }$ from equation (1)
i.e., $PB\perp BR$
Thereofre, RB is a tangent to the circle at point B.