1. Identify the points of tangency and their properties:
Let $M, N, P, Q$ be the points where the sides $AB, BC, CD, DA$ touch the circle $\Gamma$. Since $\Gamma$ is inscribed in the quadrilateral $ABCD$, the tangents from a point to a circle are equal. Therefore, we have:
$$AQ = AM = MB = BN = x$$
and
$$DQ = DP = y, \quad CP = CN = z$$

2. Use the given angles to find relationships between the segments:
Given $\angle A = \angle B = 120^\circ$ and $\angle D = 90^\circ$, we can use the properties of tangents and the given angles to find the lengths of the segments.

3. **Express $AD$ in terms of $r$:**
Since $\angle A = 120^\circ$, the tangents from $A$ to the circle are equal, and similarly for $B$. Therefore:
$$AQ = AM = MB = BN = \frac{r\sqrt{3}}{3}$$
Thus, the length $AD$ can be expressed as:
$$AD = AQ + DQ = \frac{r\sqrt{3}}{3} + r$$

4. **Find the length of $BC$:**
Given $BC = 1$, we can use the tangents from $B$ and $C$:
$$BC = BN + NC = \frac{r\sqrt{3}}{3} + r(2 + \sqrt{3})$$
Simplifying, we get:
$$1 = \frac{r\sqrt{3}}{3} + r(2 + \sqrt{3})$$

5. **Solve for $r$:**
Combine the terms involving $r$:
$$1 = \frac{r\sqrt{3}}{3} + 2r + r\sqrt{3}$$
$$1 = r \left( \frac{\sqrt{3}}{3} + 2 + \sqrt{3} \right)$$
$$1 = r \left( 2 + \frac{4\sqrt{3}}{3} \right)$$
$$1 = r \left( \frac{6 + 4\sqrt{3}}{3} \right)$$
$$r = \frac{3}{6 + 4\sqrt{3}}$$

6. **Simplify $r$:**
Rationalize the denominator:
$$r = \frac{3}{6 + 4\sqrt{3}} \cdot \frac{6 - 4\sqrt{3}}{6 - 4\sqrt{3}}$$
$$r = \frac{3(6 - 4\sqrt{3})}{(6 + 4\sqrt{3})(6 - 4\sqrt{3})}$$
$$r = \frac{18 - 12\sqrt{3}}{36 - 48}$$
$$r = \frac{18 - 12\sqrt{3}}{-12}$$
$$r = \frac{18 - 12\sqrt{3}}{-12} = \frac{18}{-12} + \frac{-12\sqrt{3}}{-12}$$
$$r = -\frac{3}{2} + \sqrt{3}$$

7. **Find $AD$:**
Substitute $r$ back into the expression for $AD$:
$$AD = \frac{r\sqrt{3}}{3} + r$$
$$AD = \frac{(-\frac{3}{2} + \sqrt{3})\sqrt{3}}{3} + (-\frac{3}{2} + \sqrt{3})$$
Simplify:
$$AD = \frac{-3\sqrt{3}/2 + 3}{3} + (-\frac{3}{2} + \sqrt{3})$$
$$AD = -\frac{\sqrt{3}}{2} + 1 + (-\frac{3}{2} + \sqrt{3})$$
$$AD = -\frac{\sqrt{3}}{2} + 1 - \frac{3}{2} + \sqrt{3}$$
$$AD = \frac{\sqrt{3}}{2} - \frac{3}{2} + 1$$
$$AD = \frac{\sqrt{3} - 3 + 2}{2}$$
$$AD = \frac{\sqrt{3} - 1}{2}$$

The final answer is $\boxed{\frac{\sqrt{3} - 1}{2}}$.