Problem 1555

1. Identify the given conditions and setup the problem:
- Circle $\Gamma$ is centered at $O$.
- Chord $AB$ with tangents at $A$ and $B$ meeting at $C$.
- Secant from $C$ intersects $AB$ at $D$ and $\Gamma$ at $E$ such that $D$ lies on segment $CE$.
- Given: $\angle BOD + \angle EAD = 180^\circ$, $AE = 1$, and $BE = 2$.

2. Analyze the given angle condition:
- $\angle BOD + \angle EAD = 180^\circ$ implies $\angle BOD + \angle EAB = 180^\circ$.
- This means $\angle BOD = 180^\circ - \angle EAB$.
- Since $\angle EAB$ is an inscribed angle, $\angle EAB = \frac{\overarc{EAB}}{2}$.
- Therefore, $\angle BOD = 180^\circ - \frac{\overarc{EAB}}{2} = \frac{\angle BOE}{2}$.
- This implies that line $OD$ bisects $\angle BOE$.

3. Use the perpendicular bisector property:
- Since $O$ is the center of $\Gamma$, $\overline{OD}$ is the perpendicular bisector of $BE$.
- By the perpendicularity lemma, $BD = ED$.

4. Establish congruency and harmonic division:
- By ASA congruency, $\triangle BFD \cong \triangle EAD$.
- Thus, $BF = EA = 1$.
- The tangents at $A$ and $B$ intersect at $C$ on the extension of $\overline{EF}$.
- By harmonic division lemma, cyclic quadrilateral $AEBF$ is harmonic, i.e., $(AB;EF) = -1$.
- Therefore, $\frac{EA}{EB} \div \frac{FA}{FB} = -1$.
- Given $EA = 1$ and $EB = 2$, we have $\frac{1}{2} \div \frac{FA}{1} = -1$, leading to $FA = \frac{1}{2}$.

5. Apply Ptolemy's theorem:
- Since $AEBF$ is an isosceles trapezoid, by Ptolemy's theorem:
$$AF \cdot BE + AE \cdot BF = EF \cdot AB$$
- Substituting the known values:
$$\left(\frac{1}{2}\right) \cdot 2 + 1 \cdot 1 = EF^2$$
- Simplifying, we get:
$$1 + 1 = EF^2 \implies EF^2 = 2 \implies EF = \sqrt{2}$$

6. Use the Law of Cosines:
- Let $CF = x$. By the Law of Cosines on $\triangle ACF$:
$$AC^2 = x^2 + \left(\frac{1}{2}\right)^2 - 2 \left(\frac{1}{2}\right) x \cos(\angle AFC)$$
- We know $\cos(\angle AFC) = -\cos(\angle AFE)$.
- By the Law of Cosines on $\triangle AFE$:
$$1 = \left(\frac{1}{2}\right)^2 + 2 - \sqrt{2} \cos(\angle AFE)$$
- Solving for $\cos(\angle AFE)$:
$$1 = \frac{1}{4} + 2 - \sqrt{2} \cos(\angle AFE) \implies \cos(\angle AFE) = \frac{5\sqrt{2}}{8}$$
- Thus, $\cos(\angle AFC) = -\frac{5\sqrt{2}}{8}$.
- Plugging back in:
$$AC^2 = x^2 + \frac{1}{4} + \frac{5\sqrt{2}x}{8}$$

7. Apply the Power of a Point theorem:
- By the Power of a Point theorem:
$$AC^2 = CF \cdot CE \implies x^2 + \frac{1}{4} + \frac{5\sqrt{2}x}{8} = x(x + \sqrt{2})$$
- Simplifying:
$$x^2 + \frac{1}{4} + \frac{5\sqrt{2}x}{8} = x^2 + x\sqrt{2}$$
- Solving for $x$:
$$\frac{3\sqrt{2}x}{8} = \frac{1}{4} \implies x = \frac{\sqrt{2}}{3} = CF$$
- Therefore, $CE = CF + FE = \frac{\sqrt{2}}{3} + \sqrt{2} = \frac{4\sqrt{2}}{3}$.

The final answer is $\boxed{\frac{4\sqrt{2}}{3}}$.