a = BD+DC = 10；

s△ABC = 1/2bc * sinA =(√2/4)* BC； - ( 1)

s△ABC = 1/2a * h = 1/2 * 10 * h = 5h； - (2)

From (2)( 1):

(√2/4)*bc=5h

BC/h = 10√2-(3)；

(3) The squares on both sides of the equation are:

b^2c^2/h^2=200； - (4)

c^2=h^2+bd^2=h^2+36； ; - (5)

b^2=h^2+cd^2=h^2+ 16； - (6)

(5) Substituting (6) into (4) gives:

(h^2+36)*(h^2+ 16)/h^2=200；

Solve the equation about H 2: H 2 =144; h^2=4；

h = 12； H=2 (give up);

When h = 2: tan ∠ bad = BD/h = 3/2 >1;

∠BAD & gt； 45, and give up the contradiction!

Substitute h= 12 into-(5);

Get:

c^2=h^2+36= 144+36= 180；

AB=c=6√5