Question

In the fig. 3.10, seg AO $\cong$ seg OD, seg BO $\cong$ seg OC Then show that $\angle$ABC $\cong$ $\angle$DCB

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{seg}\mathrm{AO}\cong \mathrm{seg}\mathrm{OD},\mathrm{seg}\mathrm{BO}\cong \mathrm{seg}\mathrm{OC} \\ \mathrm{i}.\mathrm{e}.\mathrm{AO}=\mathrm{OD},\mathrm{BO}=\mathrm{OC} \\ \mathrm{In}△\mathrm{OBC}, \\ \angle \mathrm{OBC}=\angle \mathrm{OCB}(\because \mathrm{OB}=\mathrm{OC})....\left(\mathrm{i}\right) \\ \mathrm{In}△\mathrm{ABC}\mathrm{and}△\mathrm{DCB}, \\ \mathrm{AO}+\mathrm{OC}=\mathrm{DO}+\mathrm{OB}\left(\because \mathrm{AO}=\mathrm{OD},\mathrm{BO}=\mathrm{OC}\right) \\ \Rightarrow \mathrm{AC}=\mathrm{DB}, \\ \angle \mathrm{OBC}=\angle \mathrm{OCB},\left\{\mathrm{from}\left(\mathrm{i}\right)\right\} \\ \mathrm{BC}=\mathrm{BC}\left(\mathrm{common}\right) \\ \therefore △\mathrm{ABC}\cong △\mathrm{DCB}(by\mathrm{SAS}\mathrm{congruency}\mathrm{test}) \\ \mathrm{Thus},\angle \mathrm{ABC}\cong \angle \mathrm{DCB}(\mathrm{by}\mathrm{c}.\mathrm{a}.\mathrm{c}.\mathrm{t}). \end{gathered}$$