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 $\triangle BCD$. The points $F, H, I$ are on Simons's line for $\triangle ACD$.

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 $CD$.