We'll show that $X$ and $Y$ both lie on the perpendicular bisector of $\overline{EF}.$ Since $ME = MF$, this will imply that $M, X, Y$ are collinear. Let $D'$ be the reflection of $D$ in $AB$ and let $\ell$ be the perpendicular bisector of $\overline{EF}.$ Redefine $X \equiv AB \cap \ell$; it's enough to prove that $D, E, Q, X$ are concyclic. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.254866175442268, xmax = 23.88491752377898, ymin = -5, ymax = 4; /* image dimensions */ pen ffevev = rgb(1.,0.8980392156862745,0.8980392156862745); pen evfuev = rgb(0.8980392156862745,0.9568627450980393,0.8980392156862745); pen qqzzqq = rgb(0.,0.6,0.); filldraw((3.9994994744969294,3.3520993026467303)--(1.103927415974651,-4.435650072051066)--(12.28494755303328,-4.436442947650338)--cycle, ffevev, red); filldraw((3.499224083084014,2.0065899221476933)--(2.461641184232423,-0.7840281002725)--(3.9989472101105914,-4.435855365532881)--cycle, evfuev, qqzzqq); filldraw((3.499224083084014,2.0065899221476933)--(6.348725581974177,1.1437640840619447)--(9.898373346167874,2.9042052056718477)--cycle, evfuev, qqzzqq); /* draw figures */ draw((3.9994994744969294,3.3520993026467303)--(1.103927415974651,-4.435650072051066), red); draw((1.103927415974651,-4.435650072051066)--(12.28494755303328,-4.436442947650338), red); draw((12.28494755303328,-4.436442947650338)--(3.9994994744969294,3.3520993026467303), red); draw((3.9989472101105914,-4.435855365532881)--(2.461641184232423,-0.7840281002725), blue); draw((3.9989472101105914,-4.435855365532881)--(6.348725581974177,1.1437640840619447)); draw((2.461641184232423,-0.7840281002725)--(6.348725581974177,1.1437640840619447)); draw((6.348725581974177,1.1437640840619447)--(9.898373346167874,2.9042052056718477), blue); draw((xmin, -2.016339950607145*xmin + 9.062215236996648)--(xmax, -2.016339950607145*xmax + 9.062215236996648), linetype("2 2") + blue); /* line */ draw((-1.0880065799612757,-2.544469221882402)--(2.461641184232423,-0.7840281002725)); draw((3.499224083084014,2.0065899221476933)--(2.461641184232423,-0.7840281002725), qqzzqq); draw((2.461641184232423,-0.7840281002725)--(3.9989472101105914,-4.435855365532881), qqzzqq); draw((3.9989472101105914,-4.435855365532881)--(3.499224083084014,2.0065899221476933), qqzzqq); draw((3.499224083084014,2.0065899221476933)--(6.348725581974177,1.1437640840619447), qqzzqq); draw((6.348725581974177,1.1437640840619447)--(9.898373346167874,2.9042052056718477), qqzzqq); draw((9.898373346167874,2.9042052056718477)--(3.499224083084014,2.0065899221476933), qqzzqq); /* dots and labels */ dot((3.9994994744969294,3.3520993026467303),linewidth(3.pt) + dotstyle); label("$A$", (4.0157832944158045,3.5903056265094357), NE * labelscalefactor); dot((1.103927415974651,-4.435650072051066),linewidth(3.pt) + dotstyle); label("$B$", (0.9081723989365879,-4.876098860386704), N * labelscalefactor); dot((12.28494755303328,-4.436442947650338),linewidth(3.pt) + dotstyle); label("$C$", (12.408779649922664,-4.8026907289974305), E * labelscalefactor); dot((3.9989472101105914,-4.435855365532881),linewidth(3.pt) + dotstyle); label("$D$", (4.0157832944158045,-4.876098860386704), N * labelscalefactor); dot((6.348725581974177,1.1437640840619447),linewidth(3.pt) + dotstyle); label("$F$", (6.413782253132051,1.339122930571734), NE * labelscalefactor); dot((2.461641184232423,-0.7840281002725),linewidth(3.pt) + dotstyle); label("$E$", (2.0582331240351954,-0.5939578626791187), NE * labelscalefactor); dot((9.898373346167874,2.9042052056718477),linewidth(3.pt) + dotstyle); label("$Q$", (9.98631131407666,3.051979329654768), NE * labelscalefactor); dot((3.499224083084014,2.0065899221476933),linewidth(3.pt) + dotstyle); label("$X$", (3.13488571774453,1.9753267359454323), NE * labelscalefactor); dot((6.694437484503966,-4.436046509850702),linewidth(3.pt) + dotstyle); label("$M$", (6.511659761651081,-4.973976368905734), NE * labelscalefactor); dot((-1.0880065799612757,-2.544469221882402),linewidth(3.pt) + dotstyle); label("$D'$", (-1.2940715427415967,-2.282344884632395), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Since $AB$ bisects $\angle DEF$, it follows that $D' \in EF.$ Moreover, $ED' = ED = FQ$, implying that $D'$ and $Q$ are symmetric in $\ell.$ Therefore, the composition $\sigma$ of reflection in $AB$ followed by reflection in $\ell$ sends $D \mapsto Q.$ Moreover, $\sigma: E \mapsto F.$ Since the composition of two reflections is a rotation about the intersection point of the two lines, it follows that $\sigma$ is a rotation centered at $X \equiv AB \cap \ell$ that sends $\overline{DE} \mapsto \overline{FQ}.$ Hence, $\triangle XDE \sim \triangle XQF$, so $\angle XDE = \angle XQF = \angle XQE$, so $D, E, Q, X$ are concyclic.

Since $EX$ is a external bisector of $\measuredangle DEQ$ and $XD=XQ$ we get $DEXQ$ is cyclic $\Longrightarrow$ $\measuredangle XQF$ $=$ $\measuredangle XDE$ $...(1)$. Combining $(1)$, $XQ$ $=$ $XD$ and $FQ=DE$ we get $\triangle XQF\cong \triangle XDE$ $\Longrightarrow$ $\measuredangle XFE$ $=$ $180^{\circ}-\measuredangle XFQ$ $=$ $180^{\circ}-\measuredangle XED$ $=$ $\measuredangle BED$ $=$ $\measuredangle XEF$ $\Longrightarrow$ $XE$ $=$ $XF$ similarly $YE$ $=$ $YF$, but $ME$ $=$ $MF$ $\Longrightarrow$ $M$, $X$ and $Y$ belongs in the perpendicular bisector of $EF$ $\Longrightarrow$ $M$, $X$ and $Y$ are collinear.

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%209.pdf p. 31-32. Sincerely Jean-Louis

Dukejukem wrote: Let $D'$ be the reflection of $D$ in $AB$ and let $\ell$ be the perpendicular bisector of $\overline{EF}.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.254866175442268, xmax = 23.88491752377898, ymin = -5, ymax = 4; /* image dimensions */ pen ffevev = rgb(1.,0.8980392156862745,0.8980392156862745); pen evfuev = rgb(0.8980392156862745,0.9568627450980393,0.8980392156862745); pen qqzzqq = rgb(0.,0.6,0.); filldraw((3.9994994744969294,3.3520993026467303)--(1.103927415974651,-4.435650072051066)--(12.28494755303328,-4.436442947650338)--cycle, ffevev, red); filldraw((3.499224083084014,2.0065899221476933)--(2.461641184232423,-0.7840281002725)--(3.9989472101105914,-4.435855365532881)--cycle, evfuev, qqzzqq); filldraw((3.499224083084014,2.0065899221476933)--(6.348725581974177,1.1437640840619447)--(9.898373346167874,2.9042052056718477)--cycle, evfuev, qqzzqq); /* draw figures */ draw((3.9994994744969294,3.3520993026467303)--(1.103927415974651,-4.435650072051066), red); draw((1.103927415974651,-4.435650072051066)--(12.28494755303328,-4.436442947650338), red); draw((12.28494755303328,-4.436442947650338)--(3.9994994744969294,3.3520993026467303), red); draw((3.9989472101105914,-4.435855365532881)--(2.461641184232423,-0.7840281002725), blue); draw((3.9989472101105914,-4.435855365532881)--(6.348725581974177,1.1437640840619447)); draw((2.461641184232423,-0.7840281002725)--(6.348725581974177,1.1437640840619447)); draw((6.348725581974177,1.1437640840619447)--(9.898373346167874,2.9042052056718477), blue); draw((xmin, -2.016339950607145*xmin + 9.062215236996648)--(xmax, -2.016339950607145*xmax + 9.062215236996648), linetype("2 2") + blue); /* line */ draw((-1.0880065799612757,-2.544469221882402)--(2.461641184232423,-0.7840281002725)); draw((3.499224083084014,2.0065899221476933)--(2.461641184232423,-0.7840281002725), qqzzqq); draw((2.461641184232423,-0.7840281002725)--(3.9989472101105914,-4.435855365532881), qqzzqq); draw((3.9989472101105914,-4.435855365532881)--(3.499224083084014,2.0065899221476933), qqzzqq); draw((3.499224083084014,2.0065899221476933)--(6.348725581974177,1.1437640840619447), qqzzqq); draw((6.348725581974177,1.1437640840619447)--(9.898373346167874,2.9042052056718477), qqzzqq); draw((9.898373346167874,2.9042052056718477)--(3.499224083084014,2.0065899221476933), qqzzqq); /* dots and labels */ dot((3.9994994744969294,3.3520993026467303),linewidth(3.pt) + dotstyle); label("$A$", (4.0157832944158045,3.5903056265094357), NE * labelscalefactor); dot((1.103927415974651,-4.435650072051066),linewidth(3.pt) + dotstyle); label("$B$", (0.9081723989365879,-4.876098860386704), N * labelscalefactor); dot((12.28494755303328,-4.436442947650338),linewidth(3.pt) + dotstyle); label("$C$", (12.408779649922664,-4.8026907289974305), E * labelscalefactor); dot((3.9989472101105914,-4.435855365532881),linewidth(3.pt) + dotstyle); label("$D$", (4.0157832944158045,-4.876098860386704), N * labelscalefactor); dot((6.348725581974177,1.1437640840619447),linewidth(3.pt) + dotstyle); label("$F$", (6.413782253132051,1.339122930571734), NE * labelscalefactor); dot((2.461641184232423,-0.7840281002725),linewidth(3.pt) + dotstyle); label("$E$", (2.0582331240351954,-0.5939578626791187), NE * labelscalefactor); dot((9.898373346167874,2.9042052056718477),linewidth(3.pt) + dotstyle); label("$Q$", (9.98631131407666,3.051979329654768), NE * labelscalefactor); dot((3.499224083084014,2.0065899221476933),linewidth(3.pt) + dotstyle); label("$X$", (3.13488571774453,1.9753267359454323), NE * labelscalefactor); dot((6.694437484503966,-4.436046509850702),linewidth(3.pt) + dotstyle); label("$M$", (6.511659761651081,-4.973976368905734), NE * labelscalefactor); dot((-1.0880065799612757,-2.544469221882402),linewidth(3.pt) + dotstyle); label("$D'$", (-1.2940715427415967,-2.282344884632395), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] One can conclude $X \in \ell$ more easily. Just note $X$ is the circumcenter of $\triangle DD'Q$, as it lies on perpendicular bisector of both the segments $DD',DQ$. Cosequently, it lies on perpendicular bisector of segment $QD'$, which just coincides with $\ell$ as $ED'=ED=FQ$. $\blacksquare$