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.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{PQ}\mathrm{and}\mathrm{PR}\text{be}\mathrm{two}\mathrm{equal}\mathrm{chords}\mathrm{and}\mathrm{PL}\mathrm{the}\mathrm{bisector}\mathrm{of}\angle \mathrm{RPQ}. \\ \mathrm{Now},\mathrm{in}△\mathrm{PMQ}\mathrm{and}△\mathrm{PMR}, \\ \mathrm{PQ}=\mathrm{PR}\left(\mathrm{given}\right) \\ \angle \mathrm{QPM}=\angle \mathrm{RPM}\left(\mathrm{given}\right) \\ \mathrm{PM}=\mathrm{PM}\left(\mathrm{common}\mathrm{sides}\right) \\ △\mathrm{PMQ}\cong △\mathrm{PMR}(\text{by}\mathrm{SAS}\mathrm{congruency}) \\ \therefore \mathrm{RM}=\mathrm{QM}----\left(1\right) \\ \left(\text{by}\mathrm{CPCT}\right) \\ \mathrm{And} \\ \angle \mathrm{PMR}=\angle \mathrm{PMQ}----\left(2\right) \\ \left(\text{by}\mathrm{CPCT}\right) \\ \angle \mathrm{PMR}+\angle \mathrm{PMQ}=180°\left(\mathrm{angle}\mathrm{of}\mathrm{the}\mathrm{straight}\mathrm{line}\right) \\ \Rightarrow \angle \mathrm{PMQ}+\angle \mathrm{PMQ}=180°\left[\mathrm{using}\left(2\right)\right] \\ \Rightarrow \angle \mathrm{PMQ}=\angle \mathrm{PMR}=90°----\left(3\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(3\right), \\ \mathrm{PM}\mathrm{is}\mathrm{the}\mathrm{perpendicular}\mathrm{bisector}\mathrm{of}\mathrm{QR}. \\ \mathrm{And}\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{perpendicular}\mathrm{from}\mathrm{the}\mathrm{centre}\mathrm{bisect}\text{s}\mathrm{the}\mathrm{chord}. \\ \mathrm{Therefore},\mathrm{either}\mathrm{PM}\hspace{0.17em}\mathrm{or}\mathrm{extended}\mathrm{PM}\hspace{0.17em}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{centre}. \\ \text{That is},\mathrm{bisector}\mathrm{of}\angle \mathrm{RPQ}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{centre}. \end{gathered}$$