Question

In Fig., show that AB || EF.

Answer 1

Extend line CE to E'.


$$\begin{gathered} \angle \mathrm{BAC}=57°=22°+35°=\angle \mathrm{ACE}+\angle \mathrm{ECD} \\ \therefore \mathrm{AB}\left|\right|\mathrm{CD} \\ \mathrm{Here},\angle \mathrm{E}\text{'}\mathrm{EF}+\angle \mathrm{FEC}=180°\left(\mathrm{Linear}\mathrm{pair}\right) \\ \Rightarrow \angle \mathrm{E}\text{'}\mathrm{EF}=180°-\angle \mathrm{FEC}=180°-145°=35°=\angle \mathrm{ECD} \\ \therefore \mathrm{EF}\left|\right|\mathrm{CD} \\ \mathrm{Thus},\mathrm{AB}\left|\right|\mathrm{CD}\left|\right|\mathrm{EF} \end{gathered}$$