Question

As per shown in the figure 3.77 point S is not equidistant from sides of $\angle$ABC. Show that ray BS is not its the angle bisector.

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{PS}\ne \hspace{0.17em}\mathrm{SQ} \\ \mathrm{Let}\mathrm{BS}\mathrm{be}\mathrm{the}\mathrm{angle}\mathrm{bisector}\mathrm{of}\angle \mathrm{ABC}. \\ \mathrm{Then},\mathrm{in}△\mathrm{PBS}\mathrm{and}△\mathrm{QBS}, \\ \angle \mathrm{BPS}=\mathrm{BQS}=90°\left(\mathrm{given}\right) \\ \angle \mathrm{PBS}=\angle \mathrm{QBS}(\mathrm{by}\mathrm{assumption}) \\ \mathrm{BS}=\mathrm{BS}\left(\mathrm{common}\right) \\ \mathrm{Then},△\mathrm{PBS}\cong △\mathrm{QBS}(\mathrm{AAS}\mathrm{congruency}\mathrm{test}) \\ \mathrm{Thus},\mathrm{SP}=\mathrm{SQ}\hspace{0.17em}\left(\mathrm{cpct}\right),\mathrm{which}\mathrm{is}\mathrm{a}\mathrm{contradiction}. \\ \mathrm{Hence},\mathrm{our}\mathrm{assumption}\mathrm{is}\mathrm{wrong}\mathrm{and}\mathrm{BS}\mathrm{is}\mathrm{not}\mathrm{the}\mathrm{angle}\mathrm{bisector}\text{of}\angle \mathrm{ABC}. \end{gathered}$$