Please enter a square ABCD with a side length of 3, point E is on the ray BC, be = 2ce, and connect the intersecting ray DC of AE to point F. If the triangle ABE is folded along the straight line AE, point B falls on point B', and find the sine value of angle DAB'.

Let ∠ DAB' = α, ∠ BAE = ∠ B 'AE = β.

Then α+2β = 90.

So sin ∠ dab' = sinα = sin (90-2β) = cos2β.

As long as an acute angle in a right triangle is equal to 2β, the problem will be solved.

Extend AB to m, make BM = AB, connect EM, make MN⊥ ray AF, and let the vertical foot be n.

Then ∠ men = 2β.

Obviously, AB = BM = 3. Grade three mathematics, BE = 2, AE = ME = √ 13.

According to s △ AEM = am * be/2 = AE * Mn/2 (or by similarity)

Calculate Mn = 12/√ 13.

So en = 5/√ 13.

So sin ∠ dab' = sinα = sin (90-2β) = cos2β.

= cos∠MEN = EN/EM =(5/√ 13)√ 13 = 5/ 13