Equilateral Delta ADC and Delta BCE

∴AC=CD,BC=CE,∠DCA=∠ECB=60

∴∠DCA+∠DCE=∠ECB+∠DCE

∴∠ACE=∠DCB

In △ACE and △DCB, AC=DC, BC=EC, ∠ACE=∠BCD.

∴△ACE≌△DCB

∴AE=BD

(2) Proof: ∫△ACE?△DCB

∴∠DBC=∠AEC

∠∠DCE = 180 ﹣∠acd﹣∠bce=60 =∠BCE

In △EMC and △BNC, ∠ECB=∠ECM, ∠AEC=∠CBD, EC=BC.

∴△EMC≌△BNC

∴CM=CN

∠∠MCN = 60

∴△CMN is an equilateral triangle.

(3) The conclusion (1) holds for the following reasons:

No matter how many degrees of rotation, AC=CD, BC=CE, ∠ DCA = ∠ ECB = 60, ∠ACE=∠BCD are derived.

∴△ACE≌△DCB

∴AE=BD