Let the reflections of $A$ across $BD$ be $A'$, and similar for $B',C',D'$ so that we get our parallelograms. Finally, let $P=AC\cap BD$. We first show that there cannot be more than one point outside $ABCD$, on the assumption that there is at least $1$ point outside. Assume WLOG that $B'$ is outside $ABCD$. Then $\angle B'AC<\angle DAC$ and $\angle B'CA<\angle DCA$, so \begin{align*} \angle BCA&<\angle DAC\\ \angle BAC&<\angle DCA. \end{align*}Clearly $D$ cannot satisfy its respective angle inequalities. Assume for contradiction that $A$ satisfies its angle inequalities. That is, \begin{align*} \angle ABD&<\angle BDC\\ \angle ADB&<\angle DBC. \end{align*}Then $\angle PAB<\angle PCD$ and $\angle PBD<\angle PDC$. However, this would contradict the fact that $\angle APB=\angle DPC$. So we cannot have more than $1$ of $A',B',C',D'$ inside $ABCD$. Now we show that at least of one $A',B',C',D'$ is inside $ABCD$. WLOG we can say that this implies \begin{align} \angle BAC&<\angle ACD\\ \angle CAD&<\angle BCA\\ \angle ABD&<\angle BDC\\ \angle CBD&<\angle BDA. \end{align}However, taking (1) and (3) gives the same contradiction again. Therefore, exactly one of $A',B',C',D'$ lies inside $ABCD$.