Let $ ABCD$ a cuadrilateral inscribed in a circumference. Suppose that there is a semicircle with its center on $ AB$, that
is tangent to the other three sides of the cuadrilateral. 

(i) Show that $ AB \equal{} AD \plus{} BC$. 

(ii) Calculate, in term of $ x \equal{} AB$ and $ y \equal{} CD$, the maximal area that can be reached for such quadrilateral.
### Part (i)

1. **Define the center and points of tangency:**
   Let \( O \) be the center of the semicircle \(\Gamma\) that is tangent to \( BC \), \( CD \), and \( DA \). Let \( P \), \( Q \), and \( R \) be the points where \(\Gamma\) touches \( BC \), \( CD \), and \( DA \) respectively.

2. **Angles and triangles:**
   Let \( 2\alpha = \angle DAB \) and \( 2\beta = \angle CBA \). The triangles \( OPC \) and \( OQC \) are right triangles at \( P \) and \( Q \) respectively, so \( \angle POC = \angle QOC = \alpha \). Similarly, \( \angle QOD = \angle ROD = \beta \).

3. **Tangents and trigonometric relations:**
   Let \( r \) be the radius of \(\Gamma\). Then:
   \[
   AR = \frac{r}{\tan(2\alpha)}
   \]
   and
   \[
   CP = r \tan(\alpha)
   \]

4. **Sum of tangents:**
   Using the tangent addition formula:
   \[
   \frac{1}{\tan(2\alpha)} + \tan(\alpha) = \frac{\cos(2\alpha) + 2\sin^2(\alpha)}{\sin(2\alpha)} = \frac{1}{\sin(2\alpha)} = \frac{OA}{r}
   \]
   Therefore:
   \[
   AR + CP = OA
   \]

5. **Similar relations for other sides:**
   Similarly, we have:
   \[
   BP + DR = OB
   \]

6. **Conclusion:**
   Since \( OA = OB \) (both are radii of the semicircle), we get:
   \[
   AD + BC = AB
   \]
   Hence, \( AB = AD + BC \).

\(\blacksquare\)

### Part (ii)

1. **Parallel sides:**
   If \( BC \) and \( AD \) are parallel, then \( ABCD \) is a rectangle because it is cyclic. The area of \( ABCD \) is:
   \[
   \text{Area} = \frac{AB \cdot (BC + DA)}{2} = \frac{x^2}{2}
   \]

2. **Non-parallel sides:**
   If \( BC \) and \( AD \) are not parallel, let \( E \) be the point where they meet. Since \( ABCD \) is cyclic, triangles \( ABE \) and \( DCE \) are similar with a proportionality constant \( \frac{AB}{CD} = \frac{x}{y} \).

3. **Area relation:**
   The area of \( ABCD \) is:
   \[
   \text{Area} = \left(1 - \frac{y^2}{x^2}\right) \times \text{Area of } ABE
   \]

4. **Perimeter of \( ABE \):**
   The perimeter of \( ABE \) is determined by \( x \) and \( y \):
   \[
   DE + CE = \frac{y}{x}(AE + BE) = AE + BE - (AD + BC) = AE + BE - x
   \]
   Therefore:
   \[
   AE + BE + AB = \frac{x^2}{x - y} + x
   \]

5. **Maximal area condition:**
   The area reaches a maximum when \( ABE \) is isosceles at \( E \), i.e., when \( ABCD \) is an isosceles trapezoid with \( AB \parallel CD \) and \( BC = AD = \frac{x}{2} \).

6. **Altitude calculation:**
   The altitude from \( C \) or \( D \) onto \( AB \) is:
   \[
   \sqrt{AD^2 - \left(\frac{x - y}{2}\right)^2} = \frac{1}{2} \sqrt{2xy - y^2}
   \]

7. **Maximal area:**
   Multiply the altitude by \( \frac{x + y}{2} \):
   \[
   \text{Maximal Area} = \frac{1}{2} \sqrt{2xy - y^2} \cdot \frac{x + y}{2} = \frac{(x + y) \sqrt{2xy - y^2}}{4}
   \]

The final answer is \( \boxed{ \frac{(x + y) \sqrt{2xy - y^2}}{4} } \)