Let $AB$ be a diameter of a circle $(O)$ and let $P$ be any point on the tangent drawn at $B$ to $(O).$ Define $AP\cap (O)=C\neq A,$ and let $D$ be the point diametrically opposite to $C.$ If $DP$ meets $(O)$ second time in $E,$ then,

[b](i)[/b] Prove that $AE, BC, PO$ concur at $M.$

[b](ii)[/b] If $R$ is the radius of $(O),$ find  $P$ such that the area of $\triangle AMB$ is maximum, and calculate the area in terms of $R.$
### Part (i)

1. **Define Points and Intersections:**
   - Let \( M \equiv AE \cap BC \).
   - Let \( G \equiv AE \cap BP \).

2. **Power of a Point:**
   - By the Power of a Point theorem, we have:
     \[
     AB^2 = \overline{AC} \cdot \overline{AP} = \overline{AE} \cdot \overline{AG}
     \]
   - This implies that \( C, E, G, P \) lie on a circle.

3. **Angles and Parallel Lines:**
   - Since \( C, E, G, P \) lie on a circle, we have:
     \[
     \widehat{CGP} = \widehat{CEP} = 180^\circ - \widehat{CED} = 180^\circ - 90^\circ = 90^\circ
     \]
   - Therefore, \( CG \parallel AB \).

4. **Collinearity:**
   - Since \( O \) is the midpoint of \( AB \), it is well-known that \( P, O, M \) are collinear.

5. **Conclusion:**
   - Hence, \( AE, PO, BC \) concur at \( M \).

\(\blacksquare\)

### Part (ii)

1. **Area of Triangle \( \triangle AMB \):**
   - Denote \( S_{XYZ} \) as the area of \( \triangle XYZ \).
   - We have:
     \[
     S_{AMB} = \frac{MB \cdot S_{ABC}}{BC}
     \]
   - Since \( M \) lies on \( PO \), we can express \( MB \) in terms of \( PA \):
     \[
     S_{AMB} = \frac{PA \cdot S_{ABC}}{PA + PC}
     \]
   - Using the fact that \( S_{ABC} = \frac{1}{2} \cdot AC \cdot BC \), we get:
     \[
     S_{AMB} = \frac{PA \cdot AC \cdot BC}{2(2PC + AC)}
     \]

2. **Maximizing the Area:**
   - To maximize \( S_{AMB} \), we need to find the maximum value of:
     \[
     \frac{PA \cdot AC \cdot BC}{2(2PC + AC)}
     \]
   - Given that \( AB = 2R \) (diameter of the circle), we have:
     \[
     S_{AMB} \leq \frac{AB^2}{4\sqrt{2}} = \frac{(2R)^2}{4\sqrt{2}} = \frac{4R^2}{4\sqrt{2}} = \frac{R^2}{\sqrt{2}}
     \]

3. **Equality Condition:**
   - Equality holds when \( AP = R\sqrt{2} \).

The final answer is \( \boxed{ \frac{R^2}{\sqrt{2}} } \)