Question

In the given figure, $QR$ is a tangent at $Q$, $P$ is centre of the circle and $PR||AQ$, where $AQ$ is a chord through $A$, an end point of the diameter $AB$. Prove that $BR$ is tangent at $B$.

Answer 1

R.E.F image

$\angle AQB=90\because$ diameter

Subtends ${90}^{\circ }$

Drop $\perp$ from P to Q then $\angle PQR={90}^{\circ }$

$\because Q$ is a tangent

also Let $\angle PAQ=a$

then $\because AQ$ is $\parallel PR$

$\angle BPR=a\Rightarrow \angle ABQ=90-a=b$

$\because$ Sum of $\angle$ of $\Delta ={180}^{\circ }$

also $PB=PQ=r$ (radius)

$\Rightarrow \angle PQB=\angle PBQ=90-a=b$

as equal sides $\Rightarrow$ equal angles.

$\therefore \angle QPR=180-2b-a=a$

then by $SAS\Delta PBR$ is congruent to $\Delta PQR$

(as $PQ=PB,\angle BPR=\angle QPR,PR=PR$)

$\therefore \because \angle PQR=90$

$\angle PBR=90\therefore BR$ is also

a tangent.