Question

In the figure 3.51$\square$ PQRS is given and Side QS is angle bisector of $\angle$PQR and $\angle$PSR.

Prove that : $∆$PQS $\cong$ $∆$RQS and also $\angle$P $\cong$ $\angle$R

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{QS}\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{bisector}\mathrm{of}\angle \mathrm{PQR}\mathrm{and}\angle \mathrm{PSR}. \\ \Rightarrow \angle \mathrm{PQS}=\angle RQS...\left(1\right) \\ \Rightarrow \angle \mathrm{PSQ}=\angle \mathrm{RSQ}...\left(2\right) \\ \mathrm{In}△\mathrm{PQS}\mathrm{and}△\mathrm{RQS}, \\ \angle \mathrm{PQS}=\angle \mathrm{RQS}\left(\text{f}\mathrm{rom}1\right) \\ \mathrm{QS}=\mathrm{QS}\left(\mathrm{common}\right) \\ \angle \mathrm{PSQ}=\angle \mathrm{RSQ}\left(\text{f}\mathrm{rom}2\right) \\ \therefore △\mathrm{PQS}\cong △\mathrm{RQS}(\text{by}\mathrm{ASA}\mathrm{congruency}\mathrm{test}) \\ \mathrm{Therefore},\angle \mathrm{P}\cong \angle \mathrm{R}(\text{by}\mathrm{c}.\mathrm{a}.\mathrm{c}.\mathrm{t}) \end{gathered}$$