Choose point $D$ on some suitable side (if isosceles choose a lateral side), say $BC$. Now draw the lines through $D$ parallel to $AC$ and $AB$, intersecting $AB$ and $AC$ at $F$ and $E$ respectively. The point $D$ we want satisfies the property that $B,F,E,C$ are concyclic, making $\triangle AEF,\triangle DFE,\triangle FBD,\triangle EDC$ the four triangles we want. Such a point must exist because when $D$ varies continuously from $B\to C$, $F$ moves continuously from $B\to A$ and $E$ goes $A\to C$, so there must be a point where $EF$ is antiparallel to $BC$. Since $D$ is clearly not the midpoint of $BC$, this construction does indeed work. In fact, $D$ is the intersection of $AK$ and $BC$, where $K$ is symmedian point.