Solution: (1) As shown in figure 1, extend BD to E, make BE=AB, and connect AE, CE,

∫∠ABD = 60，

∴△ABE is an equilateral triangle,

∴AE=AB,∠AEB=60，

AB = AC，

∴AC=AE，

∴∠ACE=∠AEC，

∫∠ACD = 60，

∴∠ACE-∠ACD=∠AEC-∠AEB，

That is ∠DCE=∠DEC,

∴DE=CD，

∴BE=BD+DE=BD+CD，

∴ab=bd+cd；

So, the answer is: AB = BD+CD;

2) guess: AB=

2

2

(BD+CD)。

The reasons are as follows: As shown in Figure 2, point A is AE⊥AB, and the extension line of BD is at point E, connecting CE.

∫∠ABD = 45，

∴△ABE is an isosceles right triangle,

∴AE=AB,∠AEB=45，

AB = AC，

∴AC=AE，

∴∠ACE=∠AEC，

∫∠ACD = 45，

∴∠ACE-∠ACD=∠AEC-∠AEB，

That is ∠DCE=∠DEC,

∴DE=CD，

∴BE=BD+DE=BD+CD，

In Rt△ABE, AB=BE? cos∠ABD=(BD+CD)？ cos45 =

2

2

(BD+CD)，

That is AB= 2 (BD+CD)