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Question

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

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Solution

## $\mathrm{Let}\mathrm{PQ}\mathrm{and}\mathrm{PR}\text{be}\mathrm{two}\mathrm{equal}\mathrm{chords}\mathrm{and}\mathrm{PL}\mathrm{the}\mathrm{bisector}\mathrm{of}\angle \mathrm{RPQ}.\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{in}△\mathrm{PMQ}\mathrm{and}△\mathrm{PMR},\phantom{\rule{0ex}{0ex}}\mathrm{PQ}=\mathrm{PR}\left(\mathrm{given}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{QPM}=\angle \mathrm{RPM}\left(\mathrm{given}\right)\phantom{\rule{0ex}{0ex}}\mathrm{PM}=\mathrm{PM}\left(\mathrm{common}\mathrm{sides}\right)\phantom{\rule{0ex}{0ex}}△\mathrm{PMQ}\cong △\mathrm{PMR}\left(\text{by}\mathrm{SAS}\mathrm{congruency}\right)\phantom{\rule{0ex}{0ex}}\therefore \mathrm{RM}=\mathrm{QM}----\left(1\right)\phantom{\rule{0ex}{0ex}}\left(\text{by}\mathrm{CPCT}\right)\phantom{\rule{0ex}{0ex}}\mathrm{And}\phantom{\rule{0ex}{0ex}}\angle \mathrm{PMR}=\angle \mathrm{PMQ}----\left(2\right)\phantom{\rule{0ex}{0ex}}\left(\text{by}\mathrm{CPCT}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{PMR}+\angle \mathrm{PMQ}=180°\left(\mathrm{angle}\mathrm{of}\mathrm{the}\mathrm{straight}\mathrm{line}\right)\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{PMQ}+\angle \mathrm{PMQ}=180°\left[\mathrm{using}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}⇒\angle \mathrm{PMQ}=\angle \mathrm{PMR}=90°----\left(3\right)\phantom{\rule{0ex}{0ex}}\mathrm{From}\left(1\right)\mathrm{and}\left(3\right),\phantom{\rule{0ex}{0ex}}\mathrm{PM}\mathrm{is}\mathrm{the}\mathrm{perpendicular}\mathrm{bisector}\mathrm{of}\mathrm{QR}.\phantom{\rule{0ex}{0ex}}\mathrm{And}\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{perpendicular}\mathrm{from}\mathrm{the}\mathrm{centre}\mathrm{bisect}\text{s}\mathrm{the}\mathrm{chord}.\phantom{\rule{0ex}{0ex}}\mathrm{Therefore},\mathrm{either}\mathrm{PM} \mathrm{or}\mathrm{extended}\mathrm{PM} \mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{centre}.\phantom{\rule{0ex}{0ex}}\text{That is},\mathrm{bisector}\mathrm{of}\angle \mathrm{RPQ}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{centre}.$

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