Question

In the fig 2.60 side BC of $∆$ABC is produced to D. The bisectors BE and CE of $\angle$B and $\angle$ACD respectively meet at the point E. Then prove $\angle$E = $\frac{1}{2}\angle A$

Answer 1

$$\begin{gathered} \mathrm{Let}\angle \mathrm{ABE}=\angle \mathrm{EBC}=x \\ \mathrm{and}\angle \mathrm{ACE}=\angle \mathrm{ECD}=y \\ \mathrm{Now},\angle \mathrm{ACB}+\angle \mathrm{ACD}=180° \\ \Rightarrow \angle \mathrm{ACB}+2y=180° \\ \Rightarrow \angle \mathrm{ACB}=180°-2y...\left(\mathrm{i}\right) \\ \mathrm{In}△\mathrm{ABC},\mathrm{we}\mathrm{have}, \\ \angle \mathrm{BAC}+\angle \mathrm{ABC}+\angle \mathrm{ACB}=180° \\ \Rightarrow \angle \mathrm{BAC}+2x+180°-2y=180°[\mathrm{From}(\mathrm{i}\left)\right] \\ \Rightarrow \angle \mathrm{BAC}=180°-(2x+180°-2y) \\ \Rightarrow \angle \mathrm{BAC}=2y-2x=2\left(y\mathit{-}x\right)...\left(\mathrm{ii}\right) \\ \mathrm{In}△\mathrm{EBC},\mathrm{we}\mathrm{have}, \\ \angle \mathrm{BEC}+\angle \mathrm{EBC}+\angle \mathrm{ECB}=180° \\ \Rightarrow \angle \mathrm{BEC}+x+y+180°-2y=180° \\ \Rightarrow \angle \mathrm{BEC}=y-x....\left(\mathrm{iii}\right) \\ \mathrm{From}\left(\mathrm{ii}\right)\mathrm{and}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{get}, \\ \angle \mathrm{BEC}\hspace{0.17em}=\frac{1}{2}\angle \mathrm{BAC} \end{gathered}$$