In the \(\triangle{ABC}\), AD is the perpendicular bisector of BC (see figure).
Show that \(\triangle{ABC}\) is an isosceles triangle in which AB = AC.

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

DB = DC ...(given)
\(\angle{ADB}\) = \(\angle{ADC}\) ...(Since, AD is the perpendicular bisector of BC)

and AD is the Common side.

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

Therefore, AB = AC ...(By CPCT)

Hence, \(\triangle{ABC}\) is an isosceles triangle.
Hence, proved.