Given all four angles of a quadrilateral $ABCD$, the fact that $AC$ bisects angle $A$, and that angles $A$ and $B$ are equal, how do we find $ACD$, $ADB$, $ABD$, $CBD$? Better yet, is there a formula for each of those angles in terms of angles $A, B, C, D$?

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We can certainly compute all the angles mentioned. In principle, the computation procedure described below yields a formula. It is not a formula I would care to write down, but it can be done explicitly for use, say, in a computer program. There may be nice formulas for the various angles. This answer does not address that.

The fact that the angles at $A$ and $B$ are equal does not play a role in the calculation. It may be important in seeking nice formulas.

Without loss of generality, we can assume that $AC=1$. We know $\angle DAC$ (it is half of the angle at $A$). We also know the angle at $D$. So $\angle ACD$ is known, since the angles of $\triangle ACD$ have sum $180^\circ$.

In $\triangle ACD$, we know all the angles and one side $AC$. So by the Sine Law, we know the sides $AD$ and $CD$.

Similarly, we can find the sides $AB$ and $BC$. And by the Cosine Law we can find $BD$.

Now for example for $\triangle ABD$, we know all the sides, so we can find the two unknown angles by using the Cosine Law. Similarly, we can find any other of the angles mentioned in the OP.

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EDIT: I added an extra condition. I think that the information is now sufficient. – Ilya Dec 8 '12 at 0:42