Let $ABCD$ be a convex quadrilateral whose vertices do not lie on a circle. Let $A'B'C'D'$ be a quadrangle such that $A',B', C',D'$ are the centers of the circumcircles of triangles $BCD,ACD,ABD$, and $ABC$. We write $T (ABCD) = A'B'C'D'$. Let us define $A''B''C''D'' = T (A'B'C'D') = T (T (ABCD)).$

[b](a)[/b] Prove that $ABCD$ and $A''B''C''D''$ are similar.

[b](b) [/b]The ratio of similitude depends on the size of the angles of $ABCD$. Determine this ratio.
### Part (a): Prove that \(ABCD\) and \(A''B''C''D''\) are similar.

1. **Define the circumcenters and circumradii:**
   Let \(A', B', C', D'\) be the circumcenters of triangles \(BCD, ACD, ABD,\) and \(ABC\) respectively. Let \(a, b, c, d\) be the circumradii of these triangles.

2. **Calculate the lengths \(A'C'\) and \(B'D'\):**
   \[
   A'C' = \left| a \cos \alpha + c \cos \gamma \right| = \left| \frac{BD}{2 \sin \alpha} \cos \alpha + \frac{BD}{2 \sin \gamma} \cos \gamma \right| = BD \cdot \frac{\left| \sin (\alpha + \gamma) \right|}{2 \sin \alpha \sin \gamma}
   \]
   \[
   B'D' = \left| b \cos \beta + d \cos \delta \right| = \left| \frac{AC}{2 \sin \beta} \cos \beta + \frac{AC}{2 \sin \delta} \cos \delta \right| = AC \cdot \frac{\left| \sin (\beta + \delta) \right|}{2 \sin \beta \sin \delta}
   \]

3. **Calculate the area of \(A'B'C'D'\):**
   \[
   [A'B'C'D'] = \frac{1}{2} A'C' \cdot B'D' \cdot \sin \widehat{(A'C', B'D')}
   \]
   Since \(\sin \widehat{(A'C', B'D')} = \sin \widehat{(AC, BD)}\), we have:
   \[
   [A'B'C'D'] = \frac{1}{2} AC \cdot BD \cdot \sin \widehat{(AC, BD)} \cdot \frac{\left| \sin (\alpha + \gamma) \sin (\beta + \delta) \right|}{4 \sin \alpha \sin \beta \sin \gamma \sin \delta}
   \]
   \[
   [A'B'C'D'] = [ABCD] \cdot k
   \]
   where \(k = \frac{\left| \sin (\alpha + \gamma) \sin (\beta + \delta) \right|}{4 \sin \alpha \sin \beta \sin \gamma \sin \delta}\).

4. **Relate the angles of \(A'B'C'D'\) to those of \(ABCD\):**
   Let \(\alpha', \beta', \gamma', \delta'\) be the angles of \(A'B'C'D'\) at \(A', B', C', D'\) respectively. We have:
   \[
   \sin \alpha' = \sin \gamma, \quad \sin \beta' = \sin \delta, \quad \sin \gamma' = \sin \alpha, \quad \sin \delta' = \sin \beta
   \]
   and
   \[
   \left| \sin (\alpha' + \gamma') \right| = \left| \sin (\beta' + \delta') \right| = \left| \sin (\alpha + \gamma) \right| = \left| \sin (\beta + \delta) \right|
   \]

5. **Calculate the area of \(A''B''C''D''\):**
   \[
   [A''B''C''D''] = [T(T(ABCD))] = [T(A'B'C'D')] = [A'B'C'D'] \cdot k = [ABCD] \cdot k^2
   \]

6. **Conclude similarity:**
   Since the area of \(A''B''C''D''\) is a scaled version of \(ABCD\) by \(k^2\), and the angles are preserved, \(A''B''C''D''\) is similar to \(ABCD\).

\(\blacksquare\)

### Part (b): Determine the ratio of similitude.

1. **Recall the similarity coefficient \(k\):**
   \[
   k = \frac{\left| \sin (\alpha + \gamma) \sin (\beta + \delta) \right|}{4 \sin \alpha \sin \beta \sin \gamma \sin \delta}
   \]

2. **Calculate the ratio of similitude:**
   The ratio of similitude between \(ABCD\) and \(A''B''C''D''\) is \(k^2\):
   \[
   k^2 = \left( \frac{\left| \sin (\alpha + \gamma) \sin (\beta + \delta) \right|}{4 \sin \alpha \sin \beta \sin \gamma \sin \delta} \right)^2
   \]

The final answer is \( \boxed{ \left( \frac{\left| \sin (\alpha + \gamma) \sin (\beta + \delta) \right|}{4 \sin \alpha \sin \beta \sin \gamma \sin \delta} \right)^2 } \)