Let $I$ and $I_a$ denote the incenter and A-excenter of $\triangle ABC.$ $X$ is the foot of the A-altitude and $M$ is the midpoint of $BC.$ It's well-known that $U \equiv AM \cap EF \cap DI$ and likewise $V \equiv AM \cap QR \cap PI_a$ (see for instance http://www.artofproblemsolving.com/community/c6h147114). Therefore if $J \equiv PU \cap AX$ and $W \equiv PV \cap EF,$ we have $\tfrac{JK}{JA}=\tfrac{PW}{PV}=\tfrac{PK}{PT}$ $\Longrightarrow$ $AT \parallel PU.$ Thus if $Y \equiv AT \cap BC,$ we obtain $\tfrac{MP}{MY}=\tfrac{MU}{MA}=\tfrac{MD}{MX}$ $\Longrightarrow$ $X,Y$ are isotomic points WRT $DP$ $\Longrightarrow$ $AY \equiv AT$ is the isotomic of $AX$ WRT $\triangle ABC.$ Similarly $AY \equiv AS$ is the isotomic of $AX$ $\Longrightarrow$ $A,S,T$ are collinear.

My solution. Let $DI,I_aP$ intersects $EF,QR$ at $X,Y$, reps. It is well-known that $XY$ passes through the midpoint $M$ of $BC$. By $\text{Thales}$ theorem, we have $\frac{XS}{SK}=\frac{XD}{KL}=\frac{PY}{KL}\ \text{(since DXPY is a parallelogram)}=\frac{TY}{TL}$. Hence $A,S,T$ are collinear. [asy][asy] import graph; size(8.4cm); 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 = 3.14, xmax = 11.54, ymin = -5.52, ymax = 3.96; /* image dimensions */ /* draw figures */ draw((4.46,-0.94)--(9.28,-1.04)); draw(circle((6.288143016600615,0.42322652864094923), 1.4008533524213052), linetype("2 2") + red); draw(circle((7.400405155693898,-4.883204624045323), 3.8813651230564985), linetype("2 2") + green); draw((4.937511889804902,0.7949598662911452)--(7.375812292974099,1.3060434180484843)); draw((3.658191525173438,-3.8532374585365057)--(10.4140266886414,-2.4371710031327782)); draw((5.66,3.42)--(5.517187136209143,-3.463580034719329)); draw((6.91139589686403,1.2086987344689823)--(5.517187136209143,-3.463580034719329)); draw((5.608456310394349,0.9355941610076316)--(9.127260993230992,-2.7068853601508622)); draw((5.66,3.42)--(9.127260993230992,-2.7068853601508622), red); draw((5.66,3.42)--(3.658191525173438,-3.8532374585365057)); draw((10.4140266886414,-2.4371710031327782)--(5.66,3.42)); draw((7.480914078061333,-1.0026745659348824)--(7.400405155693898,-4.883204624045323)); draw((4.937511889804902,0.7949598662911452)--(6.288143016600615,0.42322652864094923)); draw((6.288143016600615,0.42322652864094923)--(7.375812292974099,1.3060434180484843)); draw((10.4140266886414,-2.4371710031327782)--(7.400405155693898,-4.883204624045323)); draw((7.400405155693898,-4.883204624045323)--(3.658191525173438,-3.8532374585365057)); draw((6.301788130104568,1.080920999531428)--(6.259085921938662,-0.9773254340651174)); draw((5.66,3.42)--(7.438211871797141,-3.0609209078689674)); /* dots and labels */ dot((5.66,3.42),linewidth(3.pt) + dotstyle); label("$A$", (5.74,3.54), NE * labelscalefactor); dot((4.46,-0.94),linewidth(3.pt) + dotstyle); label("$B$", (4.14,-1.18), NE * labelscalefactor); dot((9.28,-1.04),linewidth(3.pt) + dotstyle); label("$C$", (9.4,-1.1), NE * labelscalefactor); dot((6.288143016600615,0.42322652864094923),linewidth(3.pt) + dotstyle); label("$I$", (6.38,0.26), NE * labelscalefactor); dot((6.259085921938662,-0.9773254340651174),linewidth(3.pt) + dotstyle); label("$D$", (6.34,-0.86), NE * labelscalefactor); dot((7.375812292974099,1.3060434180484843),linewidth(3.pt) + dotstyle); label("$E$", (7.46,1.42), NE * labelscalefactor); dot((4.937511889804902,0.7949598662911452),linewidth(3.pt) + dotstyle); label("$F$", (4.68,0.68), NE * labelscalefactor); dot((7.400405155693898,-4.883204624045323),linewidth(3.pt) + dotstyle); label("$I_a$", (7.36,-5.22), NE * labelscalefactor); dot((7.480914078061333,-1.0026745659348824),linewidth(3.pt) + dotstyle); label("$P$", (7.56,-0.88), NE * labelscalefactor); dot((10.4140266886414,-2.4371710031327782),linewidth(3.pt) + dotstyle); label("$Q$", (10.5,-2.32), NE * labelscalefactor); dot((3.658191525173438,-3.8532374585365057),linewidth(3.pt) + dotstyle); label("$R$", (3.32,-3.9), NE * labelscalefactor); dot((5.608456310394349,0.9355941610076316),linewidth(3.pt) + dotstyle); label("$K$", (5.68,1.06), NE * labelscalefactor); dot((5.517187136209143,-3.463580034719329),linewidth(3.pt) + dotstyle); label("$L$", (5.38,-3.76), NE * labelscalefactor); dot((6.91139589686403,1.2086987344689823),linewidth(3.pt) + dotstyle); label("$S$", (7.,1.32), NE * labelscalefactor); dot((9.127260993230992,-2.7068853601508622),linewidth(3.pt) + dotstyle); label("$T$", (8.98,-3.1), NE * labelscalefactor); dot((6.301788130104568,1.080920999531428),linewidth(3.pt) + dotstyle); label("$X$", (6.38,1.2), NE * labelscalefactor); dot((6.87,-0.99),linewidth(3.pt) + dotstyle); label("$M$", (6.7,-1.32), NE * labelscalefactor); dot((7.438211871797141,-3.0609209078689674),linewidth(3.pt) + dotstyle); label("$Y$", (7.52,-2.94), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

baopbc wrote: It is well-known that $XY$ passes through the midpoint $M$ of $BC$. Well, can you explain more specific about this? I don't know how to prove this without calculating.