Let $ABCD$ be a quadrilateral satisfying $\angle BCD=\angle CDA$. Suppose rays $AD$ and $BC$ meet at $E$, and let $\Gamma$ be the circumcircle of $ABE$. Let $\Gamma_1$ be a circle tangent to ray $CD$ past $D$ at $W$, segment $AD$ at $X$, and internally tangent to $\Gamma$. Similarly, let $\Gamma_2$ be a circle tangent to ray $DC$ past $C$ at $Y$, segment $BC$ at $Z$, and internally tangent to $\Gamma$. Let $P$ be the intersection of $WX$ and $YZ$, and suppose $P$ lies on $\Gamma$. If $F$ is the $E$-excenter of triangle $ABE$, and $AB=544$, $AE=2197$, $BE=2299$, then find $m+n$, where $FP=\tfrac{m}{n}$ with $m,n$ relatively prime positive integers.

[i]Proposed by Michael Kural[/i]
1. **Claim**: \(FP\) is the \(F\)-median of the excentral triangle of \(\triangle ABE\).

2. **Proof of Claim**:
   - Consider \(\triangle ABE\) as the reference triangle.
   - We need to determine the position of \(C\) on \(BE\) such that \(WX\) and \(YZ\) concur on \(\Gamma\).
   - Let \(CD\) intersect \(\Gamma\) at points \(R\) and \(S\).

3. **Application of Sawayama-Thebault Lemma**:
   - According to the Sawayama-Thebault Lemma, \(WX\) passes through the incenter of \(\triangle ARS\) and \(YZ\) passes through the incenter of \(\triangle BRS\).
   - Denote these incenters as \(I_a\) and \(I_b\) respectively.
   - Let \(T\) be the midpoint of arc \(RS\) on \(\Gamma\).

4. **Properties of \(T\)**:
   - By the midpoint of arc lemma, \(TI_a = TI_b = TR = TS\).
   - \(I_a\) lies on \(WX\) and \(I_b\) lies on \(YZ\).

5. **Angle Chasing**:
   - We need to determine the angles that \(TA\) and \(TB\) make relative to the sides of \(\triangle ABE\).
   - Note that \(OT\) (where \(O\) is the circumcenter of \(\triangle ABE\)) is parallel to the \(E\)-angle bisector.

6. **Similarity of Quadrilaterals**:
   - After some angle chasing, we find that quadrilateral \(TI_aPI_b\) is similar to \(MAFB\), where \(M\) is the midpoint of arc \(AEB\).
   - This similarity implies \(\frac{\sin \angle I_aTP}{\sin \angle PTI_b} = \frac{\sin \angle AMF}{\sin \angle FMB}\).

7. **Conclusion from Similarity**:
   - Since \(M\) is the midpoint of the \(F\)-side of the excentral triangle of \(\triangle ABE\), \(FP\) is indeed the \(F\)-median of the excentral triangle of \(\triangle ABE\).

8. **Calculation of \(FP\)**:
   - The power of \(F\) with respect to \(\Gamma\) is given by \(FE \cdot FM\), where \(M\) is the midpoint of arc \(AB\).
   - Given \(AB = 544\), \(AE = 2197\), and \(BE = 2299\), we can compute the necessary lengths and use them to find \(FP\).

9. **Final Calculation**:
   - Eventually, we find \(FP = \frac{2431}{9}\).

The final answer is \(2431 + 9 = \boxed{2440}\).