Question

In a circle with centre O, diameter AD and diameter BC are given then prove that chord AB$\cong$chord CD.

Answer 1

$$\begin{gathered} \mathrm{In}△\mathrm{AOB}\mathrm{and}△\mathrm{COD}, \\ \mathrm{AO}=\mathrm{CO}\left(\mathrm{radii}\mathrm{of}\mathrm{same}\mathrm{circle}\right) \\ \angle \mathrm{AOB}=\angle \mathrm{COD}\left(\mathrm{vertically}\mathrm{opposite}\mathrm{angles}\right) \\ \mathrm{BO}=\mathrm{DO}\left(\mathrm{radii}\mathrm{of}\mathrm{same}\mathrm{circle}\right) \\ \therefore △\mathrm{AOB}\cong △\mathrm{COD}\left(\text{by}\mathrm{SAS}\mathrm{congruency}\mathrm{rule}\right) \\ \mathrm{Now},\mathrm{their}\mathrm{corresponding}\mathrm{sides}\mathrm{will}\mathrm{be}\mathrm{equal}. \\ \therefore \text{C}\mathrm{hord}\mathrm{AB}\cong \text{C}\mathrm{hord}\mathrm{CD}(\mathrm{by}\mathrm{c}.\mathrm{p}.\mathrm{c}.\mathrm{t}) \end{gathered}$$

Hence proved.