When OA and OC are connected, OA=OC, so ∠ACO=∠CAO=∠A/2.

So ∠DCO=∠ACB/2-∠A/2.

And because < OEF = 90-< EFD = 90-(< FEC+< ACD).

=90 -(∠A +∠ACB/2)

=(∠A/2+∠ACB/2)-(∠A +∠ACB/2)

=∠ACB/2-∠A/2

So ∠DCO=∠OCF=∠OEF.

Therefore, CEOF is at four o'clock.