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Junior high school moving point math problems
Let DE=x and AE = 3 √ 3-X. According to the meaning of the question, we can get 0≤x≤3√3.

∵AD is the symmetry axis of equilateral triangle ABC.

∴AD⊥BC? BD=DC

∴CD=3 (length of equilateral triangle BC = 6 = 6)

∴CE? =x? +9

∵CF is obtained by CE rotation.

∴CE=CF

∫△ABC is an equilateral triangle

∴∠ACB=60

Rotation angle ∠ BCF = 60.

∴∠ACE=∠DCF

∴cos∠ACE=cos∠DCF

∵cos∠ACE=(AC? +CE? -AE? )/(2xACxCE) cos∠DCF=(CD? +CF? -DF? )/(2xCDxCF)

∴(AC? +CE? -AE? )/2=CD? +CF? -DF?

36+CE? -AE? = 18+2xDF?

DF? =x? -3√3x+9

Formula, get DF? =(x-3√3/2)? +9/4

∫0≤x≤3√3

When x = 3 ∴ 3/2, DF? There is a minimum value of 9/4.

∴ When x=3√3/2, DF has a minimum value of 3/2.