In the given figure, \(BA || DE\). Prove that \(\angle ABC + \angle BCD = 180^{\circ} + \angle CDE\).
Proof:
Draw \(CF\) parallel to \(AB\) and \(DE\).
\(\angle\) \(+ \angle\) = 180^{\circ}\) ---- (\(1\)) [Co-interior angles]
\(\angle FCD = \angle\) ---- (\(2\)) [Alternate interior angles]
Adding equations (\(1\)) and (\(2\)), we get:
\(\angle\) \(+ \angle\) \(+ \angle FCD = 180^{\circ} + \angle\)
\(\angle ABC + \angle BCD = 180^{\circ} + \angle CDE\)
Hence, proved.