ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that

(i) Δ ABE ≅ Δ ACF

(ii) AB = AC, i.e., ABC is an isosceles triangle

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.