Given :
BE = CF, where BE and CF are altitudes
So, ∠AEB = 90° and ∠AFC = 90°
(i) In Δ ABE and Δ ACF,
⇒ ∠AEB = ∠AFC (Each 90°)
⇒ ∠A = ∠A (Common angle)
⇒ BE = CF (Given)
∴ Δ ABE ≅ Δ ACF (By A.A.S. congruence rule)
Hence, proved that Δ ABE ≅ Δ ACF.
(ii) As,
Δ ABE ≅ Δ ACF
We know that,
Corresponding parts of congruent triangles are equal.
∴ AB = AC (By C.P.C.T.)
Hence, proved that ABC is an isosceles triangle with AB = AC.
