3. As shown in the figure, in quadrilateral $A B C D$, $B C>$
$C D>D A, O$ is the midpoint of $A B$, and connect $O C, O D$.
If $\angle A O D=\angle D O C$
$=\angle C O B=60^{\circ}$, then the relationship between $B C$ and $C D+D A$ is ( ).
(A) $B C>C D+D A$
(B) $B C=C D+D A$
(C) $B C<C D+D A$
(D) Cannot be determined
3. C.

Solution 1: As shown in the figure, draw a line $C D$ through $O M \therefore A B$ at: $M$. Then $A, B$ are symmetric about $C M$. Take the symmetric point $E$ of $D$ about $O M$. Since $O D$ and $O C$ are symmetric about $O M$, $E$ lies on $O C$. Connect $M E, E B$, then we have
$$
M D=M E, A D=B E \text {. }
$$

Thus,
$$
\begin{array}{l}
C D+D A=(C M+M D)+B E \\
=(C M+M E)+B E>C E+B E \\
>B C .
\end{array}
$$

Solution 2: As shown in the figure, take a point $E$ on $O C$ such that $O E=O D$, and connect $D E, E B$. Then $\triangle O A D \cong \triangle O B E$. We have $B E=A D$ and $\triangle O D E$ is an equilateral triangle, thus $\angle C E D=120^{\circ}$. In $\triangle C D E$, $C D$ is the longest side, so
$$
C D+D A=C D+B E>C E+B E>B C \text {. }
$$

Solution 3: Fold $\triangle A O D$ and $\triangle B O C$ along $O D$ and $O C$ respectively, so that $O A$ coincides with $O B$, forming a tetrahedron where $\triangle A(B) C D$ is the base triangle (as shown in the figure). By the triangle inequality, the sum of the lengths of any two sides of a triangle is greater than the length of the third side, so the correct choice is (C).