ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.

Given: ΔABC is an isosceles triangle

To prove: BE = CF

Proof: In ΔAEB and ΔAFC,

∠AEB = ∠AFC (Each 90° as BE and CF are altitudes)

∠A = ∠A (Common angle)

AB = AC (Given ΔABC is an isosceles triangle)

∴ ΔAEB ≅ ΔAFC (By AAS congruence rule)

∴ BE = CF (By CPCT)

Hence, proved.