Calculating area of quadrilateral when distance of vertices from an arbitrary point is known

By AM-GM

$$\sum_{cyc}MA^2=\frac{1}{2}\sum_{cyc}\left(MA^2+MB^2\right)\geq\sum_{cyc}MA\cdot MB\geq$$ $$\geq2\sum_{cyc}S_{\Delta AMB}=2S_{ABCD}=r\sum_{cyc}AB,$$ where $r$ is a radius of our circle.

In another hand, by AM-GM again $$\left(\sum_{cyc}AB\right)^2\geq4(AB+CD)(AD+BC)=4\sum_{cyc}AB\cdot AD=$$ $$=2\sum_{cyc}(AB\cdot AD+BC\cdot CD)\geq2\cdot4\cdot2S_{ABCD}=16S_{ABCD},$$ which gives $$\sum_{cyc}AB\geq4\sqrt{S_{ABCD}}=4\sqrt{\frac{1}{2}r\sum_{cyc}AB}$$ or $$\sum_{cyc}AB\geq8r.$$ Thus, $$\sum_{cyc}MA^2\geq r\sum_{cyc}AB\geq8r^2=2.$$ The equality occurs for $$\measuredangle DAB=\measuredangle ABC=\measuredangle BCD=\measuredangle CDA=90^{\circ}$$ and $$\measuredangle AMB=\measuredangle BMC=\measuredangle CMD=\measuredangle DMA=90^{\circ},$$ which says that $ABCD$ is a square.

Id est, $$S_{ABCD}=1\cdot1=1.$$ Done!