Intensive One

Let $ABC$ be a triangle with $\angle BAC = 78 ^ \circ$ , $\angle CBA = 81 ^ \circ$ and circumradius $2017$ . Let $M$ be the midpoint of $BC$ and $H$ be the orthocenter of $\Delta$ $ABC$ . Again, let $P$ be the intersection of the angle bisector of $\angle$ $BAC$ with the circumcircle of $\Delta$ $ABC$ . The line through $M$ and perpendicular to $AP$ cuts $PH$ at $Q$ .

Now if you've found the length of $MQ$ , let the value be $x$ . Consider another arbitrary triangle $DEF$ . The circle with diameter $DF$ cuts the altitude $EG$ at $K$ . The circle with diameter $DE$ cuts the altitude $FS$ at $Z$ and the extension of $FS$ at $T$ . If known that $\angle$ $FTK$ = $\frac {x}{50}$ , find $\angle$ $DZK$ !