1)∠ABC=∠ACB=∠EAC/2=∠CAD, so: AD//BC, correct;

2)BD bisects ∠ABC, AD//BC, ∠ADB=∠CBD=∠ABC/2=∠ACB/2, ∠ACB=2∠ADB, correct;

3)∠ADC= 180-∠CAD-∠ACD=

180-(∠ACB+∠ACD)= 180-[2∠ABD+( 180-2∠ABD)/2]

=180-2 ∠ Abd-90+∠ Abd = 90-∠ Abd, correct;

4)∠ADB= 180-∠BAD-∠ABD

∠CDB = 180-∠BCD-∠CBD = 180-∠BCD-∠ABD

So:180-∠ Abd = ∠ ADB+∠ Bad = ∠ CDB+∠ BCD.

∠BAD =∠BAC+∠CAD =∠BAC+∠ACB =∠BAC+∠BCD-∠ACD

Because: ∠BAC-∠ACD≠0 (because CD is not necessarily parallel to AB)

So: bad BCD

So: ∠ ADB ADB

So: BD does not necessarily divide ∠ADC equally, which is incorrect;

5)∠BDC = 180-∠BCD-∠DBC = 180-[∠ABC+(∠BAC+∠ABC)/2]-∠ABC/2

= 180-3∠ABC/2-∠BAC/2-∠ABC/2

= 180-2∠ABC-∠BAC/2

=∠BAC-∠BAC/2

=∠BAC/2

correct

To sum up, the correct number is four, namely 1, 2, 3, 5, and C.