ABC is an isosceles triangle with AB = AC. Draw AP perpendicular to BC to show that \(\angle{B}\) = \(\angle{C}\)

In \(\triangle{ABP}\) and \(\triangle{ACP}\), we have,
AB = AC ...(Given)
AP is the Common side.
\(\angle{ABP}\) = \(\angle{ACP}\) = \(90^\circ\) ...(Since, AP perpendicular to BC)
Therefore, \(\triangle{ABP}\) \(\displaystyle \cong \) \(\triangle{ACP}\) ...(By RHS congruency test)
Thus, \(\angle{B}\) = \(\angle{C}\) ...(By CPCT)
Hence, proved.