1. Define the Circle and Points:
Let $(O)$ be the given circle with center $O$ and radius $R$. Let $C$ be a point not lying on this circle. Consider all triangles $\triangle ABC$ such that points $A$ and $B$ lie on the given circle.

2. Midpoint and Perpendicular:
Let $M$ be the midpoint of the chord $AB$, and let $F$ be the foot of the perpendicular from $C$ to the line $AB$.

3. Area of Triangle:
The area of triangle $\triangle ABC$ can be expressed as:
$$[ABC] = \frac{1}{2} \times AB \times CF$$
where $CF$ is the perpendicular distance from $C$ to $AB$.

4. **Expression for $CF$:**
The length $CF$ can be expressed in terms of $OM$ and $CO$. Since $M$ is the midpoint of $AB$, $OM$ is the perpendicular distance from $O$ to $AB$. We can write:
$$CF = OM + CO \cdot \cos \angle OCF$$
However, since $F$ is the foot of the perpendicular from $C$ to $AB$, we have:
$$CF = OM - CO \cdot \cos \angle MOC$$

5. **Maximizing $CF$:**
To maximize the area $[ABC]$, we need to maximize $CF$. The term $CO \cdot \cos \angle MOC$ reaches its minimum value when $\cos \angle MOC = -1$, which occurs when $\angle MOC = \pi$. This implies that $M$ and $C$ are collinear with $O$, and $F \equiv M$.

6. Conclusion:
When $\angle MOC = \pi$, the points $A$ and $B$ are symmetric with respect to $O$, making $\triangle ABC$ isosceles with $AB$ as the base and $C$ as the apex. Therefore, the triangle of maximal area is isosceles.

$\blacksquare$