Is it possible to construct a convex quadrilateral $ABCD$ such that $\overline{AB} \cong \overline{DC}$ and $\overline{AC}$ bisects $\overline{DB}$ but not vice verse?

I'm having a difficult time constructing such a quadrilateral.


This is possible as long as $\overline{BD}$ is more than twice as long than $\overline{AB}$.

Concrete example: $A=(0,-10)$, $B=(4,-13)$, $C=(0,16)$, $D=(-4,13)$, where $\overline{AB}$ and $\overline{CD}$ are both $5$ units long and the diagonals interxect in the origin.

We make use of the fact that two sides and an angle do not determine a triangle if the given angle is opposite the shorter side.


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