ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that
(i) \(\triangle{ABE}\) \(\displaystyle \cong \) \(\triangle{ACF}\)
(ii) AB = AC i.e., ABC is an isosceles triangle.

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

\(\angle{AEB}\) = \(\angle{AFC}\)(\(\because BE and CF are perpendicular to sides AC and AB)

\(\angle{BAE}\) = \(\angle{CAF}\) ...(\(\because \(\angle{A}\) is the Common angle)

and BE = CF ...(Given)

\(\therefore \) \(\triangle{ABE}\) \(\displaystyle \cong \) \(\triangle{ACF}\) ...(By AAS Congruency test)

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

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