In quadrilateral ACBD, AC = AD and AB bisects \(\angle{A}\) (see figure). Show that \(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{ABD}\). What can you say about BC and BD ?

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

AC = AD ....(Given)

As, AB bisects \(\angle{A}\),

Therefore, \(\angle{CAB}\) = \(\angle{DAB}\)

Also, AB is a common side,

So, we can say,
\(\triangle{ABC}\) \(\displaystyle \cong \) \(\triangle{ABD}\) ...(by SAS test of congruency)

Also, BC = BD ...(By CPCT)
Hence, proved.