Question

Given : In the fig 3.11 seg AB $\cong$ seg AC Ray AF || Side BC show that Ray AF is angle bisector of $\angle$ EAC.

Answer 1

$$\begin{gathered} \mathrm{In}△\mathrm{ABC}, \\ \mathrm{seg}\mathrm{AB}\cong \mathrm{seg}\mathrm{AC} \\ \mathrm{i}.\mathrm{e}.\mathrm{AB}=\mathrm{AC} \\ \Rightarrow \angle \mathrm{B}=\angle \mathrm{C}....\left(\mathrm{i}\right) \\ \text{A}\mathrm{lso},r\mathrm{ay}\mathrm{AF}\parallel r\mathrm{ay}\mathrm{BC} \\ \Rightarrow \angle \mathrm{FAC}=\angle \mathrm{ACB}.....\left(\mathrm{ii}\right) \\ \mathrm{Let}\angle \mathrm{B}=\angle \mathrm{C}=\mathit{}x \\ \mathrm{From}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right), \\ \angle \mathrm{B}=\angle \mathrm{ACB}=\angle \mathrm{FAC}=\mathit{}x\mathit{}\mathit{}...\left(\mathrm{iii}\right) \\ \mathrm{Also},\angle \mathrm{EAC}=\angle \mathrm{ABC}+\angle \mathrm{ACB}(\because \mathrm{exterior}\mathrm{angle}\mathrm{theorem}) \\ \Rightarrow \angle \mathrm{EAF}+x=x+x \\ \Rightarrow \angle \mathrm{EAF}=x\mathit{}.....\left(\mathrm{iv}\right) \\ \mathrm{From}\left(\mathrm{iii}\right)\mathrm{and}\left(\mathrm{iv}\right),\mathrm{we}\mathrm{get}: \\ \angle \mathrm{EAF}=\mathit{}x\mathit{}=\angle \mathrm{FAC} \\ \mathrm{Thus},\mathrm{ray}\mathrm{AF}\mathrm{bisects}\angle \mathrm{EAC}. \end{gathered}$$