I found a something interesting. If $CO\parallel{DB}$, through the diagram below, we see $AC\parallel{DB}$. $CO$ and $AC$ both contain $C$, and so $A$ and $C$ must be collinear.

Attachments:

Let $CE$ be a diameter. Then $CA = ED$ since $CAED$ is a rectangle and $ED = CB$ since $EDBC$ is an isosceles trapezoid. Hence, $CA = CB$ and $\triangle ABC$ is isosceles. [asy][asy] pair A, B, C, D, E, O; A = dir(180); D = dir(0); C = dir(130); E = dir(310); B = dir(80); O = (0,0); draw(circumcircle(A,B,C), dashed+lightblue); draw(A--C--B, blue); draw(E--D, blue); draw(A--D); draw(C--E); draw(B--D); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$O$", O, -dir(B)); [/asy][/asy]

We have: $CO$ $\parallel$ $BD$, $BD$ $\perp$ $AB$ then: $CO$ $\perp$ $AB$ But: $OA = OB$ so: $CO$ is perpendicular bisector of $AB$ or $AC = BC$ Hence: $\triangle ABC$ is $C$ - isosceles