Question

In ∆ABC, ∠A = 50° and BC is produced to a point D. The bisectors of ∠ABC and ∠ACD meet at E. Find ∠E.

Answer 1

$$\begin{gathered} \mathrm{In}\mathrm{the}\mathrm{given}\mathrm{triangle},\angle \mathrm{ACD}=\angle \mathrm{A}+\angle \mathrm{B}.(\mathrm{Exterior}\mathrm{angle}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{two}\mathrm{opposite}\mathrm{interior}\mathrm{angles}.) \\ \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{Therefore},\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{triangle},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{ABC}+\angle \mathrm{BCA}+\angle \mathrm{CAB}=180°.(\mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{angles}\mathrm{of}△\mathrm{ABC}) \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{BCA}=180° \\ \angle \mathrm{BCA}=180°-(\angle \mathrm{A}+\angle \mathrm{B}) \\ \angle \mathrm{ECA}=\frac{\angle \mathrm{ACD}}{2}(\because \mathrm{EC}\mathrm{bisects}\angle \mathrm{ACD}) \\ \angle \mathrm{ECA}=\frac{\angle \mathrm{A}+\angle \mathrm{B}}{2}(\because \angle \mathrm{ACD}=\angle \mathrm{A}+\angle \mathrm{B}) \\ \angle \mathrm{EBC}=\frac{\angle \mathrm{ABC}}{2}=\frac{\angle \mathrm{B}}{2}(\because \mathrm{EB}\mathrm{bisects}\angle \mathrm{ABC}) \\ \angle \mathrm{ECB}=\angle \mathrm{ECA}+\angle \mathrm{BCA} \\ \Rightarrow \angle \mathrm{ECB}=\frac{\angle \mathrm{A}+\angle \mathrm{B}}{2}+180°-(\angle \mathrm{A}+\angle \mathrm{B}) \\ \mathrm{If}\mathrm{we}\mathrm{use}\mathrm{the}\mathrm{same}\mathrm{logic}\mathrm{for}△\mathrm{EBC},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{EBC}+\angle \mathrm{ECB}+\angle \mathrm{BEC}=180°(\mathrm{Sum}\mathrm{of}\mathrm{all}\mathrm{angles}\mathrm{of}△\mathrm{EBC}) \\ \frac{\angle \mathrm{B}}{2}+\frac{\angle \mathrm{A}+\angle \mathrm{B}}{2}+180°-(\angle \mathrm{A}+\angle \mathrm{B})+\angle \mathrm{BEC}=180° \\ \angle \mathrm{BEC}=\angle \mathrm{A}+\angle \mathrm{B}-\left(\frac{\angle \mathrm{A}+\angle \mathrm{B}}{2}\right)-\frac{\angle \mathrm{B}}{2} \\ \angle \mathrm{BEC}=\frac{\angle \mathrm{A}}{2} \\ \Rightarrow \angle \mathrm{BEC}=\frac{50°}{2}=25° \end{gathered}$$