People's education printing plate eighth grade (on) special training of mathematics synchronization class 2.4 equilateral triangle page 19 20 questions. Problem solving process ~ ~ urgent! ! ! ! ! ! ! ! ! ! △ BCD in this question is an equilateral triangle, which can be proved without giving it.

△ABC rotates 60 clockwise in the plane, so there are:

∠ BDC = 60, and △ABD and △ECD are identical,

So AD=ED, BD=CD, AB=EC, ∠ABD=∠ECD, ∠E=∠BAD.

∠BAC = 120∠BDC = 60, so ∠ ACD+∠ Abd = ∠ ACD+∠ ECD = 180, so A, C and D are in a straight line.

AD = DE = & gt△ADE is an isosceles triangle = & gt∠DAE=∠E=∠BAD.

And ∠ DAE+∠ bad = 120, so ∠ bad = ∠ e = 60, so △ADE is an equilateral triangle.

So AD=AE=5.