Proof: As shown in the figure, EH‖AB and AC intersect at point H and point E. 。

Then ∠ BAC+∠ AHE = 180, ∠BAC=∠CHE,

∵AB=BC=AC,∴∠BAC=∠ACB=60，

∴∠CHE=∠ACB=∠B=60，

∴EH=EC.

∵ad‖bc,∴∠fce= 180-∠b = 120，

∠ ahe = 180-∠ BAC = 120，

∴∠AHE=∠FCE，

∵∠aoe=∠cof,∠aef=∠acf,∴∠eac=∠efc，

∴△AEH≌△FEC，

∴ae=ef；

(2) Guess: The conclusion in (1) is that there is no change.

Proof: As shown in the figure, if point E is EH‖AB and AC is at point H, then ∠ BAC+∠ AHE = 180, ∠BAC=∠CHE.

∵AB=BC∴∠BAC=∠ACB

∴∠CHE=∠ACB∴EH=EC

∵AD‖BC∴∠D+∠DCB= 180。

∵∠BAC=∠D∴∠AHE=∠DCB=∠ECF

∠∠AOE =∠COF∠AEF =∠ACF，

∴∠EAC=∠EFC，

∴△AEH≌△FEC，

∴ae=ef；

(3) Conjecture: The conclusion in (1) has changed.

It is proved that the intersection e is EH‖AB and AC is at H.

From (2) ≈EAC =≈EFC,

∠AHE=∠DCB=∠ECF，

∴△AEH∽△FEC，

∴AE:EF=EH:EC，

∫EH‖AB，

∴△ABC∽△HEC，

∴EH:EC=AB:BC=k，

∴AE:EF=k，

∴AE=kEF.

Hope to adopt ~