$EF=AD$, $ED=AC$ $\Rightarrow$ $\triangle FDC$ is isosceles and $\angle FDC=\pi-\angle EDA$ $\angle KHI=\angle HIG$ $\Rightarrow$ $KH\parallel EG$. $KM\parallel HI$, $KM=HI$ $\Rightarrow$ $KMIH$ is a parallelogram. The last 2 observations imply that $M,G,E$ are collinear. $EM=MI-EI=EG-EI=IG=EF$ thus $\triangle MEF$ is isosceles and $\angle MEF=\pi-\angle GEF=\pi-\angle EDA$ Hence $\triangle MEF\sim \triangle FDC$ $\Rightarrow$ $\angle EFM=\angle CFD$ and as $E,F,D$ are collinear, we get that $M,F,C$ are collinear too