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(1) From the meaning of the question, Tana =BCAC=33=3.

∴∠A=60。

∫DE∨AB,

∴∠CDE=∠A=60。

As shown in the answer sheet 1, the EH⊥AC passing through point E is point H.

EH=DE? sin∠CDE=a? 32=32a。

The distance from point E to AC is a constant.

(2) If AD= 14, when a=2, as shown in Figure 2.

Let AB intersect DF and EF at M and N points respectively.

∫△def is an equilateral triangle, ∴∠ MDE = 60.

According to (1), CDE = 60,

∴∠ADM= 180 -∠MDE-∠CDE=60,

∫∠A = 60,

△ ADM is an equilateral triangle,

∴DM=AD= 14.

If the intersection m is MG∨AC and the intersection DE is g, then ∠ dmg = ∠ ADM = 60,

∴△DMG is an equilateral triangle,

∴DG=MG=DM= 14.

∴GE=DE-DG=2- 14=74.

∠∠mgd =∠e = 60 ,∴mg∥ne,

And ∵DE∨AB,

A quadrilateral is a parallelogram.

∴NE=MG= 14,MN=GE=74.

∴t=de+dm+mn+ne=2+ 14+74+ 14= 174.

(3) If point D moves to the midpoint of AC, the sub-case is discussed as follows:

① If 0 < a ≤ 32, △DEF is within △ABC, as shown in Figure 3:

∴t=3a;

② If 32 < a ≤ 3, point E is within △ABC and point F is outside △ABC, as shown in Figure 4:

Let AB intersect DF and EF at points M and N respectively, and let M be MG∑AC and DE intersect at point G. 。

As in (2), it can be seen that △ADM and △DMG are equilateral triangles, and quadrilateral MGEN is a parallelogram.

∴dm=dg=ne=ad=32,mn=ge=de-dg=a-32,

∴t=de+dm+mn+ne=a+32+(a-32)+32=2a+32;

③ If 3 < a < 3, then points E and F are outside △ABC, as shown in Figure 5:

Let AB intersect DF and EF at points M and N respectively, and BC intersect DE and EF at points P and Q respectively.

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