Question

In fig. 3.52 consider $\square$MNPQ, diagonal MP is perpendicular bisector of diagonal NQ, Then show seg MN $\cong$ seg MQ, and seg PN $\cong$ seg PQ. Find the pairs of congruent triangle and justify your answer.

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{MP}\mathrm{is}\mathrm{perpendicular}\mathrm{bisector}\mathrm{of}\mathrm{NQ}. \\ \Rightarrow \angle \mathrm{MTN}=\angle \mathrm{MTQ}=\angle \mathrm{NTP}=\angle \mathrm{PTQ}=90° \\ \mathrm{MT}=\mathrm{PT},\mathrm{NT}=\mathrm{QT} \\ \mathrm{In}△\mathrm{MTN}\mathrm{and}△\mathrm{MTQ}, \\ \mathrm{NT}=\mathrm{QT}, \\ \angle \mathrm{MTN}=\angle \mathrm{MTQ}=90° \\ \mathrm{MT}=\mathrm{MT}, \\ \therefore △\mathrm{MTN}\cong △\mathrm{MTQ}(by\mathrm{SAS}\mathrm{congruency}\mathrm{test}) \\ \mathrm{Therefore},\mathrm{MN}=\mathrm{MQ}(by\mathrm{c}.\mathrm{s}.\mathrm{c}.\mathrm{t})...\left(\mathrm{i}\right) \\ Or,\mathrm{seg}\mathrm{MN}\cong \mathrm{seg}\mathrm{MQ} \\ \mathrm{In}△\mathrm{NTP}\mathrm{and}△\mathrm{QTP}, \\ \mathrm{NT}=\mathrm{TQ}, \\ \angle \mathrm{NTP}=\angle \mathrm{QTP}=90° \\ \mathrm{TP}=\mathrm{TP}\left(\mathrm{common}\right) \\ \therefore △\mathrm{NTP}\cong △\mathrm{QTP}(by\mathrm{SAS}\mathrm{congruency}\mathrm{test}) \\ \mathrm{Therefore},\mathrm{PN}=\mathrm{PQ}(by\mathrm{c}.\mathrm{s}.\mathrm{c}.\mathrm{t}).......\left(\mathrm{ii}\right) \\ Or,\mathrm{seg}\mathrm{PN}\cong \mathrm{seg}\mathrm{PQ} \\ \mathrm{Also},\mathrm{in}△\mathrm{MNP}\mathrm{and}△\mathrm{MQP}, \\ \mathrm{MN}=\mathrm{MQ}\left\{\text{f}\mathrm{rom}\right(\mathrm{i}\left)\right\} \\ \mathrm{PN}=\mathrm{PQ}\left\{\text{f}\mathrm{rom}\right(\mathrm{ii}\left)\right\} \\ \mathrm{MP}=\mathrm{PM}\left(\text{c}\mathrm{ommon}\right) \\ \therefore △\mathrm{MNP}\cong △\mathrm{MQP}(\text{b}y\mathrm{SSS}\mathrm{congruency}\mathrm{test}) \end{gathered}$$