We define $A'$, $B'$ and $D'$ as follows: $A'$ is the reflection of $A$ wrt $C$. $B'$ is the reflection of $B$ wrt the midpoint of $CD$. $D'$ is the reflection of $D$ wrt the midpoint of $BC$. Then $C$ is the midpoint of segment $AA'$. Furthermore, $CD'=BD=B'C$, so $C$ is also the midpoint of segment $B'D'$. This implies that $AB'A'D'$ is a parallelogram. From the triangle inequality, it follows that $\sqrt2\cdot AC=AB+CD=AB+BD'\geq AD'$ and analogously $\sqrt2\cdot BD=BC+DA\geq B'A$. First squaring and then adding these inequalties gives: $2\cdot AC^2+2\cdot BD^2 \geq (AD')^2+(AB')^2 \ (1)$. But in parallelogram $AB'A'D'$ the diagonals have length $AA'=2\cdot AC$ and $B'D'=2\cdot CD'=2\cdot BD$, so the parallelogram law gives that: $(2\cdot AC)^2+(2\cdot BD)^2=2\cdot(AB')^2+2\cdot(AD')^2$, i.e., in $(1)$ equality must hold and so equality must hold in both triangle inequalities. So $B$ is on segment $AD'$ en $D$ is on segment $AB'$. Because $BD' \parallel CD$, we find that $AB \parallel CD$ and analogously $AD \parallel BC$, hence $ABCD$ is a parallelogram.