Question

Ray BD is angle bisector of $\angle$ABC of $∆$ABC and seg AB $\cong$ seg BC, then show that seg. AD $\cong$ seg DC i.e. seg BD is median of $∆$ABC.

Answer 1

$$\begin{gathered} \mathrm{Given}:\angle \mathrm{ABD}=\angle \mathrm{DBC}, \\ \mathrm{seg}\mathrm{AB}\cong \mathrm{seg}\mathrm{BC} \\ \mathrm{In}△\mathrm{ABD}\mathrm{and}△\mathrm{CBD}, \\ \mathrm{seg}\mathrm{AB}\cong \mathrm{seg}\mathrm{BC}\left(\mathrm{given}\right) \\ \angle \mathrm{ABD}=\angle \mathrm{CBD}\left(\mathrm{given}\right) \\ \mathrm{seg}\mathrm{BD}\cong \mathrm{seg}\mathrm{BD}\left(\text{c}\mathrm{ommon}\right) \\ \therefore △\mathrm{ABD}\cong △\mathrm{CBD}(\text{by}\mathrm{SAS}\mathrm{congruence}\mathrm{test}) \\ \Rightarrow \mathrm{seg}\mathrm{AD}\cong \mathrm{seg}\mathrm{CD}(\text{by}\mathrm{cpct}) \\ \mathrm{Thus},\mathrm{seg}\mathrm{BD}\hspace{0.17em}\mathrm{is}\text{the}\mathrm{median}\mathrm{of}△\mathrm{ABC}. \end{gathered}$$