(1) diameter AB⊥CD,

∴CH=DH (vertical diameter theorem), ∴A is correct.

(2) If AD is connected with DF, it can be proved that CF=DF (vertical diameter theorem).

∠AFD=∠B+∠BDF=∠C+∠BDF=∠CDF+∠BDF

=∠BDC=∠A

∴DA=DF

∴AH=FH, ∴B correct.

(3)DFE =∠C+∠CDF = 2∠C = 2∠B =∠E

∴DF=DE，

And CF=DF.

∴ cf = de ∴ d is correct.

In some places, the process is very short, so you need to reconsider.