In triangle ABC, AB = AC; BE ⊥ AC and CF ⊥ AB. Prove that :

(i) BE = CF

(ii) AF = AE

(i) In △ ABC,

⇒ AB = AC (Given)

∴ ∠B = ∠C (Angles opposite to equal sides are equal)

In △ BCF and △ CBE,

⇒ ∠B = ∠C (Proved above)

⇒ BC = BC (Common side)

⇒ ∠F = ∠E (Both equal to 90°)

∴ △ BCF ≅ △ CBE (By A.A.S. axiom)

We know that,

Corresponding sides of congruent triangle are equal.

∴ BE = CF.

Hence, proved that BE = CF.

(ii) Since, △ BCF ≅ △ CBE

∴ BF = CE = y (let) [By C.P.C.T.C.]

⇒ AB = AC = x (let)

From figure,

⇒ AF = AB - BF = x - y

⇒ AE = AC - AE = x - y

∴ AF = AE.

Hence, proved that AF = AE.