In a circle of center $O$ and radius $r$, a triangle $ABC$ of orthocenter $H$ is inscribed. It is considered a triangle $A'B'C'$ whose sides have by length the measurements of the segments $AB, CH$ and $2r$. Determine the triangle $ABC$ so that the area of the triangle $A'B'C'$ is maximum.
1. Let \( O \) be the center of the circle with radius \( r \), and let \( H \) be the orthocenter of the triangle \( ABC \). We need to determine the triangle \( ABC \) such that the area of the triangle \( A'B'C' \) is maximized, where the sides of \( A'B'C' \) are \( AB \), \( CH \), and \( 2r \).

2. Recall that the points \( H \), \( G \) (the centroid), and \( O \) are collinear, and the line passing through these points is called the Euler line. It is also known that \( HG = 2GO \).

3. Let \( M_c \) be the midpoint of the side \( AB \). Since \( O \) is the center of the circle, \( OM_c \) is perpendicular to \( AB \), making \( M_c \) the midpoint of \( AB \).

4. We need to express the perimeter of \( \triangle A'B'C' \) in terms of \( r \) and the angles of \( \triangle ABC \). We have:
   \[
   P_{\triangle A'B'C'} = AB + CH + 2r
   \]

5. Since \( M_c \) is the midpoint of \( AB \), we have:
   \[
   AB = 2AM_c
   \]

6. The length \( CH \) can be expressed using the fact that \( H \) is the orthocenter. We know that \( CH = 2OM_c \) because \( H \) is twice the distance from \( O \) to the midpoint of \( AB \).

7. Therefore, we can write:
   \[
   P_{\triangle A'B'C'} = 2AM_c + 2OM_c + 2r
   \]

8. Using trigonometric identities, we have:
   \[
   AM_c = r \sin \angle C \quad \text{and} \quad OM_c = r \cos \angle C
   \]

9. Substituting these into the perimeter expression, we get:
   \[
   P_{\triangle A'B'C'} = 2r \sin \angle C + 2r \cos \angle C + 2r
   \]

10. We can factor out \( 2r \):
    \[
    P_{\triangle A'B'C'} = 2r (\sin \angle C + \cos \angle C + 1)
    \]

11. To maximize \( \sin \angle C + \cos \angle C \), we use the identity:
    \[
    \sin \angle C + \cos \angle C = \sqrt{2} \sin \left( \angle C + \frac{\pi}{4} \right)
    \]

12. Therefore, we have:
    \[
    P_{\triangle A'B'C'} = 2r \left( \sqrt{2} \sin \left( \angle C + \frac{\pi}{4} \right) + 1 \right)
    \]

13. The maximum value of \( \sin \left( \angle C + \frac{\pi}{4} \right) \) is 1, which occurs when:
    \[
    \angle C + \frac{\pi}{4} = \frac{\pi}{2} \quad \text{or} \quad \angle C = \frac{\pi}{4}
    \]

14. Substituting this maximum value, we get:
    \[
    P_{\triangle A'B'C'} = 2r (\sqrt{2} \cdot 1 + 1) = 2r (\sqrt{2} + 1)
    \]

15. Therefore, the maximum perimeter is:
    \[
    P_{\triangle A'B'C'} = (2\sqrt{2} + 2)r
    \]

16. The condition for this maximum is \( \angle C = \frac{\pi}{4} \).

The final answer is \( \boxed{ (2\sqrt{2} + 2)r } \).