1. Let the length of $AB = x$.
2. Let the lengths of $AD$ and $DC$ be $a$ and $b$ respectively.
3. Given that $\angle B = \angle D = 90^\circ$, quadrilateral $ABCD$ can be divided into two right triangles $\triangle ABD$ and $\triangle BCD$.
4. The area of quadrilateral $ABCD$ is the sum of the areas of $\triangle ABD$ and $\triangle BCD$.

The area of $\triangle ABD$ is:
$$\text{Area of } \triangle ABD = \frac{1}{2} \times AB \times AD = \frac{1}{2} \times x \times a$$

The area of $\triangle BCD$ is:
$$\text{Area of } \triangle BCD = \frac{1}{2} \times BC \times DC = \frac{1}{2} \times x \times b$$

5. Therefore, the total area of quadrilateral $ABCD$ is:
$$\text{Area of } ABCD = \frac{1}{2} \times x \times a + \frac{1}{2} \times x \times b = \frac{1}{2} x (a + b)$$

6. Given that $|AD| + |DC| = 1$, we have:
$$a + b = 1$$

7. Substituting $a + b = 1$ into the area formula, we get:
$$\text{Area of } ABCD = \frac{1}{2} x \times 1 = \frac{1}{2} x$$

8. We also know that $|AB| = |BC| = x$, and since $\angle B = \angle D = 90^\circ$, we can use the Pythagorean theorem in $\triangle ABD$ and $\triangle BCD$:
$$AB^2 + AD^2 = BD^2 \quad \text{and} \quad BC^2 + DC^2 = BD^2$$

Since $AB = BC = x$, we have:
$$x^2 + a^2 = BD^2 \quad \text{and} \quad x^2 + b^2 = BD^2$$

Therefore:
$$x^2 + a^2 = x^2 + b^2$$

This implies:
$$a^2 = b^2$$

9. Since $a$ and $b$ are positive lengths, we have $a = b$. Given $a + b = 1$, we get:
$$2a = 1 \implies a = \frac{1}{2} \quad \text{and} \quad b = \frac{1}{2}$$

10. Substituting $a = \frac{1}{2}$ and $b = \frac{1}{2}$ into the area formula, we get:
$$\text{Area of } ABCD = \frac{1}{2} x \times 1 = \frac{1}{2} x$$

11. To find $x$, we use the fact that $2x^2 = a^2 + b^2$:
$$2x^2 = \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$$

Therefore:
$$x^2 = \frac{1}{4} \implies x = \frac{1}{2}$$

12. Substituting $x = \frac{1}{2}$ into the area formula, we get:
$$\text{Area of } ABCD = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$$

Thus, the area of quadrilateral $ABCD$ is always $\frac{1}{4}$.

The final answer is $\boxed{\frac{1}{4}}$.