(2) Guess: the ratio of ∠DBC to ∠ABC is the same as the conclusion in (1).

Proof: As shown in Figure 2, as ∠KCA=∠BAC.

Make BK‖AC pass through CK at point K and connect DK.

∵ ∠BAC≠90，

The quadrilateral ABKC is an isosceles trapezoid.

∴ CK=AB。

dc=da,∴DCA = DAC。

* ∠kca=∠bac,∴∠kcd=∠3.

∴△kcd?△ bad. ∴∠2=∠4,KD=BD.

∴KD=BD=BA=KC.

∫bk‖ac,∴∠ACB =∠6。

∫∠kca=2∠acb,∴∠5 =∠ACB。 ∴ ∠5=∠6.

∴ KC=KB ∴KD=BD=KB.∴ ∠KBD=60。

* ∠acb=∠6=60-∠ 1,∴∠bac=2∠acb= 120-2∠ 1,.

∵ ∠ 1+(60-∠ 1)+( 120-2∠2)+∠2= 180,

∴ ∠2=2∠ 1.

The ratio of ∴ ∠DBC to ∠ABC degrees.