Question

The bisectors of the acute angles of a right triangle meet at O. Find the angle at O between the two bisectors.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180°. \\ \mathrm{Hence},\mathrm{for}△\mathrm{ABC},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180° \\ \Rightarrow \angle \mathrm{A}+90°+\angle \mathrm{C}=180° \\ \angle \mathrm{A}+\angle \mathrm{C}=180°-90° \\ \angle \mathrm{A}+\angle \mathrm{C}=90° \\ \mathrm{For}△\mathrm{OAC}: \\ \angle \mathrm{OAC}=\frac{\angle \mathrm{A}}{2}(\mathrm{OA}\mathrm{bisects}\angle \mathrm{A}.) \\ \angle \mathrm{OCA}=\frac{\angle \mathrm{C}}{2}(\mathrm{OC}\mathrm{bisects}\angle \mathrm{C}.) \\ \mathrm{On}\mathrm{applying}\mathrm{the}\mathrm{above}\mathrm{logic}\mathrm{to}△\mathrm{OAC},\mathrm{we}\mathrm{get}: \\ \angle \mathrm{AOC}+\angle \mathrm{OAC}+\angle \mathrm{OCA}=180°(\mathrm{Sum}\mathrm{of}\mathrm{angles}\mathrm{of}△\mathrm{AOC}) \\ \Rightarrow \angle \mathrm{AOC}+\frac{\angle \mathrm{A}}{2}+\frac{\angle \mathrm{C}}{2}=180° \\ \angle \mathrm{AOC}+\frac{\angle \mathrm{A}+\angle \mathrm{C}}{2}=180° \\ \angle \mathrm{AOC}+\frac{90°}{2}=180°(\because \angle \mathrm{A}+\angle \mathrm{C}=90°) \\ \angle \mathrm{AOC}=180°-45° \\ \angle AOC=135° \end{gathered}$$