Question

In fig. 4.34 seg OP $\perp$ seg AB and seg OQ $\perp$ CD, OP = OQ.
Then prove that chord AB $\cong$ chord CD.

Answer 1

Draw BO and DO.
$$\begin{gathered} \mathrm{In}\mathrm{right}△\mathrm{OPB}\mathrm{and}△\mathrm{OQD}, \\ \mathrm{OP}\hspace{0.17em}=\mathrm{OQ}=\left(\mathrm{given}\right) \\ \angle \mathrm{OPB}=\angle \mathrm{OQD}\left(b\mathrm{oth}\mathrm{equal}\mathrm{to}90°\right) \\ \mathrm{OB}=\mathrm{OD}\left(\mathrm{radii}\mathrm{of}\mathrm{same}\mathrm{circle}\right) \\ \therefore △\mathrm{OPB}\cong △\mathrm{OQD}(by\mathrm{RHS}\mathrm{congruency}) \\ \mathrm{PB}=\mathrm{QD}\left(by\mathrm{CPCT}\right) \\ \Rightarrow 2\mathrm{PB}=2\mathrm{QD} \\ \Rightarrow \mathrm{AB}=\mathrm{CD}\left(\mathrm{perpendicular}\mathrm{from}\mathrm{the}\mathrm{centre}\mathrm{bisects}\mathrm{the}\mathrm{respective}\mathrm{chord}\right) \\ \mathrm{That}\mathrm{is},\mathrm{chord}\mathrm{AB}\cong \mathrm{chord}\mathrm{CD}. \end{gathered}$$