If $D$ was on the line $AB$, then either $D = A$, $B = C$, or $A, B, D, C$ would on a line (in the order). $D = A$ is impossible because that would imply $C = A$ and $||CD|| = 0$, and $B = C$ is impossible because that would imply $||AD|| = 2$but $||AC|| = 1$. The last case is impossible, because $||AC|| \neq ||AD||$.
Therefore, $D$ is not colinear with $A, B, C$. Additionally, $C \neq A$.
(edit: When $C = A$, $||CB|| = ||BD|| = ||DC|| = 1$, so an ᅟequilateral triangle $\triangle CBD$ can be formed.)
If $C$ was inside $AB$, since $D$ is on a circle with center $A$ and radius $||AC||$, projection of $D$ on $AB$ should be between $A$ and $C$. However, since $||BD|| = ||CD||$, projection of $D$ should be located at midpoint of $BC$. Therefore, it is impossible.
Therefore, either $C$ is located on  ray $AB$ - $AB$, or  ray $BA$ - $AB$.
The first case
$\angle BAD = \angle BDA$ since $||AB|| = ||BD||$. $\angle DBC = \angle DCB$ since $||BD|| = ||CD||$. Since $\angle DBC = 2 \angle DAB$, the sum of inner angles of triangle $ADC$ can be calculated by
$$\angle DAC + \angle ADC + \angle ACD = 5 \angle DAC = 180°$$
Therefore, $\angle DAC = 36°$. Now, since $\triangle ACD \sim \triangle DBC$,
$$||AC|| : ||CD|| \equiv ||DB|| : ||BC||$$
$$||AC|| : 1 \equiv 1 : ||AC|| - 1$$
And now $||AC||$ can be calculated by solving a quadratic equation and the fact that $||AC|| > 1$.
The second case
Do the same thing. But now $||BC|| = ||AC||+1$.