(1) From the domain under the root sign, we can get, a? =4, similarly, a>0, therefore, a=2,

The original equation is simplified to 2√(ab)=a+b, then b=2,

(2) The coordinates of point B are (2,2), point A is (2,0) and point C is (0,2).

AF=AB=AO Therefore, A can be regarded as the center of the circle, and B, F, 0 F and 0 are all in the circle: (x-2)? +y？ =4 or more.

The linear equation of CD is 2x+y-2=0,

Simultaneous, the coordinates of point F can be solved: (2/5, 6/5) The abscissa of point F is less than 1, so k(BE)= 1/2.

Then, the equation of BE is x-2y+2=0.

(3): Two methods for your reference,

< 1 > the simple way is to take special circumstances. When BAF = 0, then at this time, GA=GC=2, GO=2√2.

So √2|GO|=|GA|+|GC|.

< 2 > as usual, let ∠BAF=a, then ∠FOA= 1/2*(90? -a)=(45？ -a/2)

Then the equation of the straight line FO is, y=x*tan(45? -a/2)， 1

Similarly, ∠GAX=90? -a/2

Then the equation of the straight line GA is, y=(x-2)*tan(90? -a/2)2

Produced by 1 and 2 at the same time,

The coordinates of point G are 2 {1+tan (a/2)}/{1+tan? (a/2)}，2 { 1-tan(a/2)}/{ 1+tan？ (a/2)}

In fact, the coordinates of point G can be simplified as cosa+sina+ 1 and cosa-sina+ 1.

So |GO|=√[(cosa+sina+ 1)? +(cosa-sina+ 1)？ ]=2√2[cos(a/2)]，

|GA|=√[(cosa+sina- 1)？ +(cosa-sina+ 1)？ ]=2[(cos(a/2)-sin(a/2)]，

|GC|=√[(cosa+sina+ 1)？ +(cosa-sina- 1)？ ]=2[(cos(a/2)+sin(a/2)]，

So √2|GO|=|GA|+|GC|.

In addition, there is another relationship: GA? +GC？ =4