3. 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.
Solution: Given: (i) BE and CF are altitudes. (ii) AC = AB To prove: BE = CF Proof: Triangles ΔAEB and ΔAFC are similar by AAS congruency since ∠A = ∠A (It is the common arm) ∠AEB = ∠AFC (They are right angles) AB = AC (Given in the question) ∴ ΔAEB ≅ ΔAFC and so, BE = CF (by CPCT).
