1. Given: Four points $A, C, B, D$ lie on a circle $\omega_1$ in that order. We need to prove that $CD^2 = AC \cdot BC + AD \cdot BD$ if and only if at least one of $C$ and $D$ is the midpoint of arc $AB$.

2. **Construct a point $B'$**: Create a point $B'$ on the arc $CD$ which does not contain $A$ such that $B'D = CB$ and $B'C = BD$.

3. Apply Ptolemy's Theorem: According to Ptolemy's Theorem, for a cyclic quadrilateral $ABCD$, the sum of the products of its two pairs of opposite sides is equal to the product of its diagonals:
$$AC \cdot BD + AD \cdot BC = AB \cdot CD$$
Here, we need to show that $AB' = CD$.

4. Rearrange the given condition: The given condition is $CD^2 = AC \cdot BC + AD \cdot BD$. We can rewrite this as:
$$CD^2 = AC \cdot BC + AD \cdot BD$$
By Ptolemy's Theorem, we have:
$$AC \cdot BD + AD \cdot BC = AB \cdot CD$$
Therefore, if $AB' = CD$, then:
$$CD^2 = AB' \cdot CD$$
Simplifying, we get:
$$CD = AB'$$

5. Consider configurations: We need to consider the configurations where $C$ or $D$ is the midpoint of arc $AB$. If $C$ is the midpoint of arc $AB$, then $AC = BC$ and $AD = BD$. Similarly, if $D$ is the midpoint of arc $AB$, then $AD = BD$ and $AC = BC$.

6. Verify the condition: If $C$ is the midpoint of arc $AB$, then:
$$CD^2 = AC \cdot BC + AD \cdot BD$$
Since $AC = BC$ and $AD = BD$, we have:
$$CD^2 = AC^2 + AD^2$$
Similarly, if $D$ is the midpoint of arc $AB$, then:
$$CD^2 = AC \cdot BC + AD \cdot BD$$
Since $AC = BC$ and $AD = BD$, we have:
$$CD^2 = AC^2 + AD^2$$

7. Conclusion: Therefore, the given condition $CD^2 = AC \cdot BC + AD \cdot BD$ holds if and only if at least one of $C$ and $D$ is the midpoint of arc $AB$.

$\blacksquare$