Let $T$ in line $BA$ and $R$ in line $BC$ such that $BT, BR>BA$, $AT=CE$ and $CR=AD$. So, the following Lemma is valid: Lemma The points $E, X$ and $T$ are collinear just like the points $D, Y$ and $R$. Let $U=MX\cap BC$ and $V=MY\cap AB$. As $MX//AB$ and $M$ is the midpoint of the segment $DE$, by midline, we must have $U$ being the midpoint of the segment $BE$. Likewise, from $MY//BC$ and $M$ being the midpoint of the segment $DE$, $V$ will have to be the midpoint of the segment $BD$. Therefore, by the Intercept Theorem(Tales Theorem) in $MX$ and in $MY$ we get: $\frac{AX}{XC}=\frac{BU}{UC}=\frac{\frac{BC-EC}{2}}{\frac{BC-EC}{2}+EC}=\frac{BC-EC}{BC+EC}=\frac{BE}{TB}$ and $\frac{YC}{YA}=\frac{BV}{VA}=\frac{\frac{AB-DA}{2}}{\frac{AB-DA}{2}+DA}=\frac{AB-DA}{AB+DA}=\frac{BD}{RB}$ Now, notice that, by Menelaus, in the points $E, X$ and $T$ and $D, Y$ and $R$ $\frac{TA}{TB}\cdot\frac{BE}{EC}\cdot\frac{CX}{XA}=\frac{BE}{TB}\cdot\frac{CX}{XA}=1$, thus $E, X$ and $T$ are collinear. $\frac{RC}{RB}\cdot\frac{BD}{DA}\cdot\frac{AY}{YC}=\frac{BD}{RB}\cdot\frac{AY}{YC}=1$, thus $D, Y$ and $R$ are collinear, proving our lemma. Therefore, as $MX//DT$ and, $X$ must be the midpoint of $ET$. Likewise, as $MY//ER$, $Y$ must be the midpoint $DR$. Thus, since $DT=DE$, $DF$ bissects $\angle TDE$ and, as $ER=ED$, $EY$ bissects the $\angle RED$ and the $F$ is the exincenter relative to $B$ in the triangle $BED$. But, since $DX$ is the perpendicular bissector of $PA$ and $EY$ is the perpendicular bissector of $PC$, $FA=FP=FC$, then the points $A$, $P$ and $C$ are the contact points of the exincircle in the lines $BA$, $BC$ and $DE$, and then $FP\bot DE$.

Attachments:

Clearly, $P$ is the point where the $B$-excircle of $DBE$ is tangent to $DE$, so $A$ and $C$ are also tangency points. Define $F'$ to be the $B$-excentre of $DBE$ and $X'$ and $Y'$ to be the feet of the altitudes from $E$ and $D$, respectively, on $DEF'$. Then if $\angle EDB=2\alpha$ we get $\angle X'DM=90^\circ -\alpha$, so $\angle DMX'=2\alpha$, and therefore $MX'\parallel AB$. On the other hand, since $A$ and $C$ are the reflections of $P$ on $DF'$ and $EF'$, respectively, and $PX'Y'$ is the orthic triangle of $DEF'$ by definition then $A,X',Y',C$ are collinear. Hence $X'=X$ and $Y'=Y$, so $F'=F$, and therefore $FP$ is perpendicular to $DE$, as desired.

Let $T$$=$$MY$$\cap$$AB$,$S$$=$$MX$$\cap$$BC$ and the lenths of $TB$, $SB$ and $AD$ be $a$,$b$ and $x$, respectively. As $YT$$\parallel$$BC$, and $M$ is the midpoint of $DE$, then $T$ is the midpoint of $BD$; likewisely, $S$ is the midpoint of $BE$. Notice that $TB=TD=a$, and $SB=SE=b$, so $CE=2a+x-2b$. Then, as quadrilateral $BSMT$ is a parallelogram, $TB=SM=a$, so $MX=a+x-b$, because $SX=SC$. Now, observe that $MD=ME=\frac{PD+PE}{2}=\frac{(x)+(2a+x-2b)}{2}=a+x-b$. Then $MX=MD=ME$, so $\angle DXE=90^{\circ}$. Analogously, $\angle DYE=90^{\circ}$. Then if $\angle BDE=\alpha$, $\angle DMX=\alpha$ and $\angle MXE = \angle MEX =\frac{\alpha}{2}$, so $\angle XDE=90^\circ-\frac{\alpha}{2}$, and $\angle XDA=90^\circ-\frac{\alpha}{2}$. Hence, $DX$ is the angle bisector of $\angle ADE$, which is isosceles, meaning $DX$ is the penpendicular bisector of $PA$. Analogously, $EY$ is the perpendicular bisector of $PC$. Therefore, F is the circumcenter of $\bigtriangleup APC$, then F is the $B$-excentre of $\bigtriangleup BDE$, which has contact points A, P and C, so that means $FP$$\bot$$DE$$\Box$