ABCD is a quadrilateral in which AD = BC and \(\angle{DAB}\) = \(\angle{CBA}\) (see figure). Prove that
(i) \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{BAC}\)
(ii) BD = AC
(iii) \(\angle{ABD}\) = \(\angle{BAC}\)

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

AD = BC ...(Given)
Also,
\(\angle{DAB}\) = \(\angle{CBA}\) ...(Given)

And AB is a common side.

Therefore, \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{BAC}\) ...(SAS congruency test)

Hence, BD = AC ...(By CPCT)

and \(\angle{ABD}\) = \(\angle{BAC}\) ...(By CPCT)
Hence, proved.