6. Let the diagonals of convex quadrilateral $A B C D$ intersect at $O$, and $A D / / B C$. The following are four statements about it:
(1) Given $A B+B C=A D+D C$. Then $A B C D$ is a parallelogram,
(2) Given $D C+D O=A O+A B$. Then $A B C D$ is a parallelogram;
(3) Given $B C+B O+A O=A D+D O+C O$. Then $A B C D$ is not a parallelogram:
(4) Given $A D+C O=B C+A O$. Then $A B C D$ is a parallelogram.
The number of true statements is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
6. (B).
(1) As shown in the figure, extend $CB$ to point $E$ such that $BE=AB$, and extend $AD$ to point $F$ such that $DF=DC$. Then $EC \parallel AF$, so $AECF$ is a parallelogram, $\angle E=\angle F$, $\angle ABE=\angle CDF$, $\angle ABC=\angle ADC$.
Since $AD \parallel BC$, $\angle ABC + \angle DAB = 180^{\circ}$.
$$
\therefore \angle ADC + \angle DAB = 180^{\circ}, AB \parallel DC.
$$

Given $AD \parallel BC$, then $ABCD$ is a parallelogram.
(2) and (4) are false propositions. For example, in the isosceles trapezoid $ABCD$ with diagonals intersecting at point $O$, $\angle AOD=60^{\circ}$. Clearly, this is a counterexample to (2) and (4).
(3) is a true proposition. Given $\triangle AOD \sim \triangle COB$. Let $\frac{AD}{BC}=\frac{AO}{CO}=\frac{DO}{BO}=t$.

Then $AD = BC \cdot t$.
$$
AO = CO \cdot t, DO = BO \cdot t.
$$

Substitute these into the given equation and simplify to get
$$
\begin{aligned}
& (BC + BO - CO) \cdot (1 \\
- t) = & 0,
\end{aligned}
$$

Since $BC + BO > CO$,
$$
\therefore t = 1.
$$
$\triangle AOD \cong \triangle COB$,
$\therefore ABCD$ is a parallelogram.