Because A and B are the two roots of X 2+4 (C+2) = (C+4) X, that is, the two roots of X 2-(C+4) X+4 (C+2).

So Vieta's theorem gives a*b=4(c+4) a+b=c+4.

So (a+b) 2 = (c+4) 2, that is, A 2+b 2+2ab = c 2+8c+16 is replaced by A 2+b 2 = c 2, because ab=4c+8.

It conforms to Pythagorean theorem, so triangle ABC is right triangle AC vertical BC.

Let AE be x, because DE=BD tanA=3/4, there is AE = X DE = BD = 3/4 A = 3/4 B.

Because a 2+b 2 = c 2a = 3/4b, there is a radical sign 7 b with c=2.

Judging from the above steps, triangle ABC is similar to triangle ADE.

Therefore, from C = radical 7b, AD = radical 7ae = radical 7 x.

Because AE 2+de 2 = AD, there is x 2+(3/4x) 2 = (2x root) 2.

Solving this equation x is the length of AE.