In the given figure, the bisectors of ∠B and ∠C of ∆ABC meet at I. If IP ⊥ BC, IQ ⊥ CA and IR ⊥ AB, prove that (i) IP = IQ = IR, (ii) IA bisects ∠A.

$$\begin{gathered} \mathrm{In}△\mathrm{IPC}\mathrm{and}△\mathrm{IQC},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{IPC}=\angle \mathrm{IQC}...(90°\mathrm{each}) \end{gathered}$$

$$\begin{gathered} \angle \mathrm{PCI}=\angle \mathrm{QCI}(\mathrm{CI}\mathrm{is}\mathrm{the}\mathrm{bisector}\mathrm{of}\mathrm{angle}\mathrm{C}) \\ \mathrm{IC}=\mathrm{IC}(\mathrm{Common}\mathrm{side}) \\ \mathrm{Thus},∆\mathrm{IPC}\cong △\mathrm{IQC}\hspace{0.17em}(\mathrm{AAS}\hspace{0.17em}\mathrm{criterion}) \end{gathered}$$

$\Rightarrow \mathrm{IP}=\mathrm{IQ}(\mathrm{Corresponding}\mathrm{parts}\mathrm{of}\mathrm{congruent}\mathrm{triangles})......\left(1\right)$

$$\begin{gathered} \mathrm{Similarly},\mathrm{in}△\mathrm{IPB}\mathrm{and}△\mathrm{IRB},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{IPB}=\angle \mathrm{IRB}(90°\mathrm{each}) \end{gathered}$$

$$\begin{gathered} \angle \mathrm{PBI}=\angle \mathrm{RBI}(\mathrm{BI}\mathrm{is}\mathrm{the}\mathrm{bisector}\mathrm{of}\mathrm{angle}\mathrm{B}) \\ \mathrm{IB}=\mathrm{IB}(\mathrm{Common}\mathrm{side}) \\ \mathrm{Thus},∆\mathrm{IPB}\cong △\mathrm{IRB}(\mathrm{AAS}\hspace{0.17em}\mathrm{criterion}) \end{gathered}$$

$\Rightarrow \mathrm{IP}=\mathrm{IR}(\mathrm{Corresponding}\mathrm{parts}\mathrm{of}\mathrm{congruent}\mathrm{triangles}).....\left(2\right)$

From (1) and (2), we have,

IP = IQ = IR

Now consider the right triangles, ARI and AQI.

$$\begin{gathered} \mathrm{IR}=\mathrm{IQ}(\mathrm{Proved}\mathrm{above}) \\ \mathrm{AI}=\mathrm{AI}(\mathrm{Common}\mathrm{side}) \\ \angle \mathrm{ARI}=\angle \mathrm{AQI}(90°\mathrm{each}) \\ ∆\mathrm{ARI}\cong ∆\mathrm{AQI}(\mathrm{RHS}\mathrm{criterion}) \\ \Rightarrow \angle \mathrm{RAI}=\angle \mathrm{QAI}(\mathrm{Corresponding}\mathrm{parts}\mathrm{of}\mathrm{congruent}\mathrm{triangles}) \end{gathered}$$

Thus, IA bisects angle A.