AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects \(\angle{A}\)

In \(\triangle{ABD}\) and \(\triangle{ACD}\), we have,
AB = AC ...(Given)

\(\angle{ABD}\) = \(\angle{ADC}\) = \(90^\circ\) ...(Since AD is an altitude)
and AD is Common side.
Therefore, \(\triangle{ABD}\) \(\displaystyle \cong \) \(\triangle{ACD}\) ...(By RHS congruency axiom)
Therefore, BD = DC ...(By CPCT)
Thus, we can say that, AD bisects BC.

Also, \(\angle{BAD}\) = \(\angle{CAD}\) ...(By CPCT)
Thus, this shows that AD bisects \(\angle{A}\).
Hence, proved.