ABCD is a trapezium in which AB || CD and AD = BC (see figure).

Show that
i) \(\angle{A}\) = \(\angle{B}\)
ii) \(\angle{C}\) = \(\angle{D}\)
iii) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\)
iv) Diagonal AC = Diagonal BD
[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E].

Given: ABCD is trapezium.
AB || CD and AD = BC
Now, extend AB and draw a line through C parallel to DA intersecting AB produced at E.

Now, ADCE is a parallelogram.
\(\therefore \) AD || CE and AD = CE
But AD = BC
\(\therefore \) AD = BC = CE

i) We know that, \(\angle{A}\) + \(\angle{E}\) = \(180^\circ\)
(Since, interior angles on the same side of the transversal )
\(\Rightarrow \) \(\angle{E}\) = \(180^\circ\) - \(\angle{A}\)
\(\therefore \) \(\angle{E}\) = \(180^\circ\) - \(\angle{A}\)
(Since, BC = EC)

Also, \(\angle{ABC}\) = \(180^\circ\) - \(\angle{CBE}\)
(Since, ABE is straight line)
\(\angle{ABC}\) = \(180^\circ\) - \(180^\circ\) + \(\angle{A}\)
\(\angle{B}\) = \(\angle{A}\)
Hence, it is proved.

ii)Now, \(\angle{A}\) + \(\angle{D}\) = \(180^\circ\)
(Since, interior angles on the same side of the transversal)
\(\Rightarrow \) \(\angle{D}\) = \(180^\circ\) - \(\angle{A}\)
\(\Rightarrow \) \(\angle{D}\) = \(180^\circ\) - \(\angle{B}\)
(from eq.(i)) ...(ii)

Also, \(\angle{C}\) + \(\angle{B}\) = \(180^\circ\) ...180 (Since, interior angles on the same side of the transversal BC)
\(\Rightarrow \) \(\angle{C}\) = \(180^\circ\) - \(\angle{B}\) ...(iii)
from Eq. (ii) and (iii), we get,
\(\angle{C}\) = \(\angle{D}\)
Hence, proved.

iii)Now, in \(\triangle{ABC}\) and \(\triangle{BAD}\), we have,
AD = BC ...(Given)
\(\angle{A}\) = \(\angle{B}\) ...(From Eq.(i))
AB = BA ...(Common side)
\(\therefore \) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\) ...(By SAS rule)
Hence, proved.

iv)
\(\because \) \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{BAD}\)
\(\therefore \) AC = BD
Hence, it is proved.