∠ABD=∠ABC+∠DBC=∠BAC+45

=∠ 1+∠BAD+45；

∠ADB=∠BDC-∠ 1

=45 -∠ 1;

∠ABD+∠ADB+∠BAD = 180； So we can get: 90+2 ∠ bad = 180.

So: bad = 45.

(2)

① CE divides AD vertically, so AE equals DE, and AED is an isosceles right triangle starting from (1).

② According to ①, let DE /DC = AE/BC = DA/DB = K and let DC = BC = 1.

Then db = √ 2, de = AE = k, da = (√ 2) k, df = Fe = (√ 2) k/2.

Calculate cf = √ (1-k? /2), so ce = cf+Fe = √ (1-k? /2)+(√2)k/2

Similarly be = √ (2-k? )。 According to the above, it can be proved that (√ 2) Ce = DE+BE.