This problem is from Tournament of towns spring 2011(Junior-A level)

First Part: $\measuredangle EKD=\measuredangle EHD=\dfrac {\pi} {2}$.So,$E,K,H,D$ are concyclic. Again,$\measuredangle AEB=\measuredangle ADB=\dfrac {\pi} {2}$.So,$A,E,D,B$ are also concyclic.Then,by Reim's Theorem,$HK \parallel AB$ or $HK \parallel FG$. Second Part: $E,H,F,B$ are cyclic.So,$\angle HFE=\angle FBE,\angle EHK=\angle EDK=\angle DEB(DK \parallel BE)=\angle DAB$. So,$\angle FHK=\angle EBA+\angle DAB$. Similarly,$\angle HKG=\angle EBA + \angle DAB$.So,$HKFG$ is an isosceles trapezoid.Therefore,$FK=GH$.