Let $P=XC \cap BY$. Now note that $\triangle{AXC}$, $\triangle{AYB}$ are congruent and isosceles; therefore we have that $\angle{CXA} = \angle{ACX} = 45^{\circ}-\frac{A}{2}$. Using the fact that $XA \perp AB$, through some basic angle chasing we have that $XP \perp BP$, and similarly we get $CP \perp YP$ so $XC \perp YB$. Since from symmetry $BP=CP$ we have that $BPC$ is a right isosceles triangle. Then we have that the quadrilateral $KFPE$ is a rectangle. Let $PB=CP=a$. Then the distance from $D$ to $PB$ and $CP$ is $\frac{a}{2}$. Note that if the distance from $K$ to $PB$ is $b$, then it is clear that the distance from $K$ to $CP$ is $a-b$. So then if we consider $D'$ to be the projection from $D$ to $PB$, then the right triangle $DD'E$ has legs of length $\frac{a}{2}, \left|b-\frac{a}{2}\right|$. By similar calculations we get that if $D''$ is the projection of $D$ onto $CP$ then right triangle $DD''F$ has the same legs; therefore from the Pythagorean theorem we have that $DE=DF$ as desired. To find the locus of the midpoint of $EF$, if we consider $\triangle{BPC}$ to be on the Cartesian plane such that $P = (0, 0)$, $B = (a, 0)$, $C=(0, a)$, then for some $x \in [0, a]$ we have $E=(x, 0)$, $F=(0, a-x)$. Then the midpoint is $\left(\frac{x}{2}, -\frac{x}{2}+\frac{a}{2}\right)$. So therefore all the midpoints lie on the line $f(x) = -x+\frac{a}{2}$. Therefore the locus of all such points is the line parallel to $BC$ that passes through the midpoints of $BP$ and $CP$.

Solution needs not be complicated. a) EM/KD = BE/BK = BM/BD = CN/CD = FN/KD and EM = FN implying that triangles EMD and FND (M and N are midpoints of PB and PC, respectively); therefore, DE = DF. b) Because KEPF is a rectangle, the midpoint of EF is also midpoint of PK; therefore, altitude from midpoint of EF to BC equals half that from P to BC and locus is the segment MN (not a line).

Oops... triangles EMD and FND are congruent.. I forgot to put in the congruent part.

Well, the solution is not very complicated; I just tend to use more words than necessary. The question didn't take more than a couple minutes.

I totally agree with you.

As already proved, $BY\bot CX$ and, by symmetry, $A-P-D$ are collinear. $PEDKF$ is cyclic, $PD$ angle bisector of $\angle EPF$, hence $DE=DF$. Alternately, see that $\frac{BE}{BP}=\frac{BK}{BC}=\frac{PF}{PC}$, i.e. a $90^\circ$ rotation about $D$ will map $B$ to $P$, $P$ to $C$ and $E$ tp $F$, done. Best regards, sunken rock

Some atomic bomb solution with transformation geometry. a) Rotation around $A$ with $90^{\circ}$ sends $CX$ to $BY$ so $CX\bot BY$. Let $CX\cap BY = \{S\}$. Then $BCS$ is half of a square so $KESF$ is rectangle. Since rotation around $D$ with $90^{\circ}$ sends $CS$ to $SB$ and $SE = CF$ it sends $F$ to $E$. So $DE=DF$. b) Finally, we can get $E$ from $F$ with translation for $\overrightarrow{SB}$ followed with reflection across $BC$. Now this isometry is indirect so midpoint $M$ for $EF$ describes line by Hjelmslev's theorem. Since it is bounded with midpoint on $BS$ and midpoint on $CS$, $M$ describes only this part of whole line.