Irska

Let $F,G,H,I$ and $J$ be the projections of $M$ on the diagonal $AC$, the side $BC$, the side $CD$, the side $AD$ and the diagonal $BD$ respectively. The points $G,H,J$ are collinear, because they are on Simons's line for $△BCD$. The points $F,H,I$ are on Simons's line for $△ACD$.

Because $\angle MID=\angle MJD=\angle MHD=90^{\circ }$, the points $H,I,J$ lie on the circle with diameter $MD$. This implies $\angle JIH=\angle JDH=\angle BDC$ and $\angle IHJ=180^{\circ }-\angle IDJ=\angle ADB=\angle ACB$. Hence, the triangles $HIJ$ and $CAB$ are similar. The same is true for the triangles $FGH$ and $ABD$. Because $FI$ intersects $GJ$ at $H$, the points $F,G,I$ and $J$ are on a circle if and only if the triangles $HIJ$ and $HGF$ are similar. Hence, $F,G,I$ and $J$ are on a circle iff the triangles $CAB$ and $DBA$ are similar. This is equivalent to $ABCD$ being an isosceles trapezoid with $AB$ parallel to $CD$.