Answer 1

In Δ ABC,

AB = BC = CA,

AD ⊥ BC, BE ⊥ AC.

Proof: In Δ ADC and Δ BEC

∠ADC = ∠BEC ...........(each 90°)

∠ACD = ∠BCE ..............(common)

and AC = BC ............(sides of an equilateral triangle)

∴ Δ ADC ≅ Δ BEC ...........(A.A.S. Axiom)

Hence (i) AD = BE ...........(c.p.c.t.)

and (ii) BD = CE ...........(c.p.c.t.)

Hence proved.