Analysis: Since BC and DE are perpendicular to the beam AC, BC∑DE can be obtained, and D is the midpoint of AB, so AB=BD, and AE: Ce = AD: BD can be obtained by using the parallel line segment ratio theorem, so it can be proved that AE=CE is the center line of △ABC, and DE= 1/2 BC can be obtained at RT.

Solution:

Columns BC and DE are perpendicular to beam AC,

∴BC∥DE，

∫D is the midpoint of AB,

∴AD=BD，

∴AE:CE=AD:BD，

∴AE=CE，

∴DE is the center line of △ABC,

∴DE= 1/2 years BC,

In Rt△ABC, BC= 1 /2 AB=4,

∴DE=2.

Comments: This question examines the proportion theorem of parallel lines and the midline theorem of triangles. The opposite side of a right triangle with an angle of 30 is equal to half of the hypotenuse. The key to solve the problem is to prove that DE is the center line of △ABC.

I think it is to find the length.