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.