In △ ABC,
AB = BC = CA,
AD ⊥ BC, BE ⊥ AC,
Proof: In △ ADC and △ BEC
∠ADC = ∠BEC (each 90°)
∠ACD = ∠BCE (common)
and AC = BC (Sides of an equilateral triangle)
∴ △ ADC ≅ △ BEC (A.A.S. Axiom)
Hence, (i) AD = BE (c.p.c.t)
And (ii) BD = CE (c.p.c.t)
Hence proved.