ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure).
Show that these altitudes are equal.

In \(\triangle{ABE}\) and \(\triangle{ACF}\) , we have,

\(\angle{AEB}\) = \(\angle{AFC}\) ...(BE and CF are perpendiculars drawn to sides AC and AB respectively)

Also, \(\angle{A}\) is common angle.
and AB = AC ...(Given)

Therefore, \(\triangle{ABE}\) \(\displaystyle \cong \) \(\triangle{ACF}\) ...(By AAS congruency test)

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

Hence, BE = CF ...(By CPCT)
Hence, proved.