Question

Two chords, PQ and PR of a circle are equal. Prove that the bisector of $\angle RPQ$ passes through the centre of the circle.

Answer 1

Given, chords $RP=RQ$

In $△PSQ$ and $△PSR$

$PQ=PR$ (given)

$\angle RPS=\angle QPS$ (given)

$PS=PS$ (common)

$△PSQ\cong △PSR$ (by SAS)

$\Rightarrow RS=QS\angle PSR=\angle PSQ$

But,

$\angle PSR+\angle PSQ={180}^{o}2\angle PSR={180}^o\angle PSQ=\angle PSR={90}^o$

then, $RS=QS$ and $\angle PSR={90}^o$

$PS$ is the perpendicular bisector of chord $RQ$

$PS$ passes through center of circle.