Example 3 As shown in Figure 1.4.4, in quadrilateral $A B C D$, $\angle B=\angle C=60^{\circ}, B C=1$, a circle is drawn with $C D$ as its diameter intersecting $A B$ at point $M$ and tangent to $A B$ at point $M$, and intersects $B C$ at point $E$. Find the length of segment $B E$.
Solution 1 (Cutting out an isosceles triangle) Connect $O M$, then $O M \perp A B$, so by the given conditions, it is easy to know $\angle C O M=150^{\circ}$. Connect $O E$, it is easy to know $\triangle C O E$ is an equilateral triangle, thus $\triangle E O M$ is an isosceles right triangle, i.e., $O M=O E=C E, \angle B M E=\angle E M O=45^{\circ}$, hence
$$
M E=\sqrt{2} O E=\sqrt{2} C E=\sqrt{2}(1-B E)
$$

From $E M: \sin 60^{\circ}=B E: \sin 45^{\circ}$ we know $M E=\frac{\sqrt{6}}{2} B E$.
From $\sqrt{2}(1-B E)=\frac{\sqrt{6}}{2} B E$ we find $B E=4-2 \sqrt{3}$.
Solution 2 (Cutting out a parallelogram) Connect $O E$, it is easy to know $O E=C E$ and $O E \parallel A B$. Connect $O M$, then $O M \perp A B$, draw $O F \parallel B C$ intersecting $A B$ at $F$, we get $\square B E O F$. Thus
$$
M F=\frac{1}{2} O F=\frac{1}{2} B E, F B=O E=C E=1-B E
$$

And
$$
B M^{2}=B E \cdot 1=B E
$$

So
$$
\left(\frac{1}{2} B E+1-B E\right)^{2}=B E
$$

That is $B E^{2}-8 B E+4=0$. Hence $B E=4-\sqrt{3}$.
Solution 3 (Cutting out a square) Connect $O M, O E$, draw $E N \perp B M$ at $N$, then $E O M N$ is a square. Thus
$$
B N=\frac{1}{2} B E, M N=O E=C E=1-B E
$$

So
$$
B M=\frac{1}{2} B E+1-B E
$$

Thus
$$
\left(\frac{1}{2} B E+1-B E\right)^{2}=B M^{2}=B E \cdot 1=B E
$$

That is $B E=4-2 \sqrt{3}$.
Solution 4 (Completing a regular triangle) Extend $B A$ and $C D$ to intersect at $P$, then $\triangle B C P$ is a regular triangle, thus $C P=B P=B C=1$, by the secant-tangent theorem we know
$$
B M^{2}=B E \cdot B C=B E,(1-B M)^{2}=P M^{2}=P D \cdot P C=1-C D
$$

Connect $O E$, it is easy to know $O C=E C$, so
$$
(1-B M)^{2}=1-2 C E=1-2(1-B E)=2 B E-1
$$

Thus
$$
(1-\sqrt{B E})^{2}=2 B E-1
$$

That is $B E=4-2 \sqrt{3}$.
Solution 5 (Completing a right triangle) Connect $M O$ and extend it to intersect the extension of $B C$ at $Q$, then $M Q \perp A B$, and $\angle Q=30^{\circ}$, thus $B M=\frac{1}{2}(1+C P)$. Connect $O E$, it is easy to know $O C$ is the median of the hypotenuse of $\mathrm{Rt} \triangle E Q O$, i.e., $C P=C E=1-B E$, substituting into the above equation gives $B M=1-\frac{1}{2} B E$, and $B M^{2}=B E$, so $\sqrt{B E}=1-\frac{1}{2} B E$, thus $B E=4-2 \sqrt{3}$.