l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{CDA}\).

From figure, we have,

\(\angle{1}\) = \(\angle{2}\) (Vertically opposite angles) ...(i)

\(\angle{1}\) = \(\angle{6}\) (Corresponding angles) ...(ii)
\(\angle{6}\) = \(\angle{4}\) (Corresponding angles) ...(iii)

From Equations, (i), (ii) and (iii), we have

\(\angle{1}\) = \(\angle{4}\) and \(\angle{2}\) = \(\angle{6}\) ...(iv)

In \(\triangle{ABC}\) and \(\triangle{CDA}\), we have

\(\angle{4}\) = \(\angle{2}\) ...(from (iii) and (iv))
\(\angle{5}\) = \(\angle{3}\) ...(Alternate angles)

and AC is a common side.

Therefore, we get,
\(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{CDA}\) ...(By SAS congruency test)
Hence, proved.