Note that $(XFPD)$ and $(PYDE)$ are cyclic.Hence $$\angle{XDP}=\angle{AFE}=\angle{BFD}=\angle{XFD}=\angle{XPD}$$so $XP=XD$ and similarly $YP=YD$ thus $XY$ is the perpendicular bisector of $PD$ and we are done.$\square$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9cm); 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 = -35.98168636001514, xmax = 50.44588353158434, ymin = -25.740932105039025, ymax = 28.262176925267873; /* image dimensions */ /* draw figures */ draw(circle((7.541241400347624,-0.4890071928282668), 12.798388035524882), linewidth(0.8) + red); draw((7.611984442028657,12.309185325254907)--(2.4480980517865234,-3.571153466242332), linewidth(0.8) + blue); draw((7.611984442028657,12.309185325254907)--(-4.265062779060309,-5.429646618161914), linewidth(0.8) + blue); draw((-4.265062779060309,-5.429646618161914)--(14.183659189513431,-11.428706558991856), linewidth(0.8) + blue); draw((14.183659189513431,-11.428706558991856)--(7.611984442028657,12.309185325254907), linewidth(0.8) + blue); draw((1.2595931164031204,-7.226125633120309)--(2.4480980517865234,-3.571153466242332), linewidth(0.8) + blue); draw(circle((1.6734608314841721,3.4397693535464975), 10.67392160136988), linewidth(0.8) + red); draw(circle((10.89782181577104,0.4402393831315239), 12.315380831453378), linewidth(0.8) + red); draw((-2.393146132622136,-2.633866304660168)--(13.147911235851073,-7.687426917485401), linewidth(0.8) + blue); draw((-1.327923844358364,-1.0429153888179326)--(11.327789879645724,-1.1128708674697294), linewidth(0.8) + blue); draw((-2.393146132622136,-2.633866304660168)--(1.2595931164031204,-7.226125633120309), linewidth(0.8) + blue); draw((1.2595931164031204,-7.226125633120309)--(13.147911235851073,-7.687426917485401), linewidth(0.8) + blue); draw((2.4480980517865234,-3.571153466242332)--(11.327789879645724,-1.1128708674697294), linewidth(0.8) + blue); draw((-1.327923844358364,-1.0429153888179326)--(2.4480980517865234,-3.571153466242332), linewidth(0.8) + blue); draw((2.4480980517865234,-3.571153466242332)--(-4.265062779060309,-5.429646618161914), linewidth(0.8) + blue); draw((-1.327923844358364,-1.0429153888179326)--(1.2595931164031204,-7.226125633120309), linewidth(0.8) + blue); draw((2.4480980517865234,-3.571153466242332)--(14.183659189513431,-11.428706558991856), linewidth(0.8) + blue); draw((11.327789879645724,-1.1128708674697294)--(1.2595931164031204,-7.226125633120309), linewidth(0.8) + blue); /* dots and labels */ dot((7.611984442028657,12.309185325254907),dotstyle); label("$A$", (7.853473271279759,12.897275611205659), NE * labelscalefactor); dot((-4.265062779060309,-5.429646618161914),dotstyle); label("$B$", (-4.06562296867285,-4.840147229292631), NE * labelscalefactor); dot((14.183659189513431,-11.428706558991856),dotstyle); label("$C$", (14.40615177286508,-10.884428260927399), NE * labelscalefactor); dot((2.4480980517865234,-3.571153466242332),linewidth(4pt) + dotstyle); label("$H$", (2.6565213562293324,-3.145488996124005), NE * labelscalefactor); dot((11.327789879645724,-1.1128708674697294),linewidth(4pt) + dotstyle); label("$E$", (11.581721384250717,-0.6599902541433528), NE * labelscalefactor); dot((-1.327923844358364,-1.0429153888179326),linewidth(4pt) + dotstyle); label("$F$", (-1.1282153645139128,-0.6035016463710652), NE * labelscalefactor); dot((1.2595931164031204,-7.226125633120309),linewidth(4pt) + dotstyle); label("$D$", (1.4702605930113004,-6.760759893550408), NE * labelscalefactor); dot((-2.393146132622136,-2.633866304660168),linewidth(4pt) + dotstyle); label("$X$", (-2.1450103044150834,-2.1851826639951164), NE * labelscalefactor); dot((13.147911235851073,-7.687426917485401),linewidth(4pt) + dotstyle); label("$Y$", (13.389356832963909,-7.212668755728709), NE * labelscalefactor); dot((3.26196741799112,-1.0682863830755183),linewidth(4pt) + dotstyle); label("$P$", (3.503850472813641,-0.6035016463710652), NE * labelscalefactor); dot((2.260780267197122,-4.147206008097908),linewidth(4pt) + dotstyle); label("$G$", (1.1313289463775769,-5.122590268154069), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Not as nice as Pluto1708's solution, but still worthy of posting $\color{black}\rule{25cm}{1pt}$ Since we have $DX$ and $DY$ are tangents to their respected circles we must have that: $$\angle XDY = \angle XDA + \angle YDA = \angle DCA + \angle DBA = \angle B + \angle C = 180 - \angle A$$this in turn implies that $(AXDY)$ is cyclic. In turn this implies that $\angle YXA = \angle YDA = \angle DBA = \angle CBA$, from which is obvious that $XY \parallel BC$. This implies that there exists a homothety centered at $A$ which takes $B$ to $X$, $C$ to $Y$, since $X$ is on $AB$ and $Y$ is on $AC$. Let $G$ be the point of intersection between $XY$ and the A-altitude of $ABC$. To show that $GP=GD$ is the equivalent of proving that $P$ is the orthocenter of $AXY$, since of the reflected orthocenter lemma. Let the homothetic coefficient be denote with $k$, and we have that: $$k = \frac{AB}{AX} = \frac{AX+XB}{AX} = 1+ \frac{XB}{AX}$$to calculate $k$ we just find expressions for $XB$ and $AX$. $\color{red}\rule{25cm}{1pt}$ Calculation for $XB$ Notice the angles of $BXD$, they are easily calculated and we get that $\angle BDX=90-\angle C$, $\angle DBX = \angle B$, this implies that $\angle BXD = 90-\angle B+\angle C$. From the Law Of Sines on $BXD$ we have that: $$\frac{BX}{\sin 90 -\angle C}=\frac{BD}{\sin 90 - \angle B + \angle C} \implies BX=\frac{\cos \angle C}{\cos \angle B - \angle C}BD$$$\color{red}\rule{25cm}{1pt}$ Calculation for $AX$ Notice the angles of $AXD$ they are also easily calculated they are $\angle XAD = 90-\angle B$, $\angle XDA = \angle C$, this implies that $\angle DXA = 90-\angle C+\angle B$. Now we apply the Law Of Sines of $AXD$ and $BXD$ to get that: $$AX = \frac{\sin \angle C}{\cos \angle B} DX = \frac{\sin \angle C \sin \angle B}{\cos \angle B \cos \angle B - \angle C} BD$$$\color{red}\rule{25cm}{1pt}$ From our calculations we have that: $$k=1+\text{ctg} (\angle C) \text{ ctg}(\angle B)$$ Now we are going to show that $\frac{AH}{AP} = k$ $\color{red}\rule{25cm}{1pt}$ Calculation for $AH$ We have from the Law Of Sines on $AHB$ that: $$AH = \frac{\cos \angle A}{\sin \angle C} AB$$$\color{red}\rule{25cm}{1pt}$ Calculation for $AP$ From the Law Of Sines on $APE$ and $AEB$, we have that: $$AP=\frac{\sin \angle B}{\cos \angle B -\angle C}AE=\frac{\sin \angle B \cos \angle A}{\cos \angle B -\angle C}AB$$$\color{red}\rule{25cm}{1pt}$ Thus when we now plug in what we indeed have that $\frac{AH}{AP}=k$, thus we have that $P$ is the orthocenter of $AXY$ and thus we have that $GP=GD$

[asy][asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(12cm); 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 = 2.994, xmax = 23.718, ymin = -10.99, ymax = 6.412; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen wwqqzz = rgb(0.4,0.,0.6); pen ffqqff = rgb(1.,0.,1.); draw((11.559634549789402,-3.935866397786168)--(11.87076046643414,-3.9366812548118055)--(11.871575323459778,-3.6255553381670684)--(11.56044940681504,-3.6247404811414308)--cycle); /* draw figures */ draw(circle((14.5,-3.1), 7.529276193632426), blue); draw((11.58,3.84)--(7.695508204334062,-6.323180324262812), red); draw((7.695508204334062,-6.323180324262812)--(21.28751516505823,-6.358778587769432), red); draw((21.28751516505823,-6.358778587769432)--(11.58,3.84), red); draw(circle((9.637754102167028,-1.2415901621314052), 5.440117434717719), qqwuqq); draw(circle((16.433757582529125,-1.2593892938847167), 7.040080527987236), wwqqzz); draw((9.416145277196781,-1.8213958530929832)--(21.28751516505823,-6.358778587769432)); draw((7.695508204334062,-6.323180324262812)--(14.80896445898262,0.4476289323269972)); draw((11.58,3.84)--(11.553355571799969,-6.3332842528108335)); draw((8.729718573158161,-3.617326631043795)--(11.553355571799969,-6.3332842528108335), ffqqff); draw((11.553355571799969,-6.3332842528108335)--(18.702978147777895,-3.643447182878647), ffqqff); draw((9.416145277196781,-1.8213958530929832)--(14.80896445898262,0.4476289323269972)); draw((8.729718573158161,-3.617326631043795)--(18.702978147777895,-3.643447182878647)); draw((9.416145277196781,-1.8213958530929832)--(11.553355571799969,-6.3332842528108335)); draw((11.553355571799969,-6.3332842528108335)--(14.80896445898262,0.4476289323269972)); draw((8.729718573158161,-3.617326631043795)--(11.567543241830116,-0.9161967094720418)); draw((11.567543241830116,-0.9161967094720418)--(18.702978147777895,-3.643447182878647)); /* dots and labels */ dot((11.58,3.84),linewidth(3.pt) + dotstyle); label("$A$", (11.266,4.19), NE * labelscalefactor); dot((7.695508204334062,-6.323180324262812),linewidth(3.pt) + dotstyle); label("$B$", (7.196,-7.03), NE * labelscalefactor); dot((21.28751516505823,-6.358778587769432),linewidth(3.pt) + dotstyle); label("$C$", (21.474,-7.118), NE * labelscalefactor); dot((11.553355571799969,-6.3332842528108335),linewidth(3.pt) + dotstyle); label("$D$", (11.178,-7.096), NE * labelscalefactor); dot((14.80896445898262,0.4476289323269972),linewidth(3.pt) + dotstyle); label("$E$", (15.05,0.978), NE * labelscalefactor); dot((9.416145277196781,-1.8213958530929832),linewidth(3.pt) + dotstyle); label("$F$", (8.758,-1.464), NE * labelscalefactor); dot((11.56302336939229,-2.641958912032246),linewidth(3.pt) + dotstyle); label("$H$", (11.134,-2.256), NE * labelscalefactor); dot((8.729718573158161,-3.617326631043795),linewidth(3.pt) + dotstyle); label("$X$", (8.076,-3.994), NE * labelscalefactor); dot((18.702978147777895,-3.643447182878647),linewidth(3.pt) + dotstyle); label("$Y$", (18.878,-3.224), NE * labelscalefactor); dot((11.567543241830116,-0.9161967094720418),linewidth(3.pt) + dotstyle); label("$P$", (11.794,-0.474), NE * labelscalefactor); dot((11.56044940681504,-3.6247404811414308),linewidth(3.pt) + dotstyle); label("$G$", (10.98,-4.126), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] $\textit{Proof}$ First, let $G$ is the midpoint of $DP$ and $H$ be the orthocenter of $\bigtriangleup ABC$. It's suffice to show that $X,G,Y$ collinear. Well known that $(AH,PD)=-1$ so we have $AH.AG=AP.AD (1)$ We have (since $XD$ tagents $(ADC)$ at $D$) $\angle XDP=\angle XDA=\angle ACB=\angle AFE\implies XFPD\quad\text{concyclic}$ Similarly we also have $YEPD$ is concyclic quadrilateral too. By $(1)$ we have $$AF.AX=AP.AD=AH.AG=AE.AY$$which implies that $XFHG,YEHG$ concyclic so we have $\angle XGH=\angle YGH=90^o$ Which implies $X,G,Y$ are collinear, the end the proof.$\quad\blacksquare$

My solution: one can see that $P,F,X,D$ are cyclic implying $XD^2=XF\cdot XA=XP^2$ so we're done.

MarkBcc168 wrote: Let $\triangle ABC$ be a triangle with altitudes $AD,BE,CF$. Let the lines $AD$ and $EF$ meet at $P$, let the tangent to the circumcircle of $\triangle ADC$ at $D$ meet the line $AB$ at $X$, and let the tangent to the circumcircle of $\triangle ADB$ at $D$ meet the line $AC$ at $Y$. Prove that the line $XY$ passes through the midpoint of $DP$. Claim. $P$ is the orthocenter of triangle $\triangle AXY$. Proof. Let $Q=XP\cap AY$ and $R=YP\cap AX$.

DYFP is cyclic and this is the proof.

DXEP is cyclic and this is the proof.

EXBQ is cyclic and this is the proof.

FYCR is cyclic and this is the proof.

In the proofs that $EXBQ$ and $FYCR$ are cyclic, we use that $H$ is the incenter of the orthic triangle. From the conditions above we get $$ \measuredangle YQX=\measuredangle BQX=\measuredangle BEX=\measuredangle BEC=90^o \Longrightarrow YA\perp QX $$$$ \measuredangle YRX=\measuredangle YRC=\measuredangle YFC=\measuredangle BFC=90^o \Longrightarrow YR\perp AX $$Thus the Claim is proved. Now, $$ \measuredangle YAX=\measuredangle BAD+\measuredangle DAC=\measuredangle CDX+\measuredangle YDB=180^o-\measuredangle XDY $$Hence $D\in (AXY)$. In other words, $D$ is the intersection between $(AXY)$ and the $A$ altitude in $\triangle AXY$. This directly implies that $D$ is the reflection of the orthocenter of $\triangle AXY$ with respect to $XY$. That is, $D$ is the reflection of $P$ over $AD\cap XY$. Therefore, $XY$ passes through the midpoint of $DP$, and so we are done.

Attachments:

Notice $$\measuredangle YDX=\measuredangle ADX+\measuredangle YDA=\measuredangle YXA+\measuredangle AYX=\measuredangle CAB$$so $AXDY$ is cyclic. Hence, $$\measuredangle FPD=\measuredangle CEP=\measuredangle ECD=\measuredangle BXD=\measuredangle FXD$$and $DPFX$ is cyclic. Therefore, $$\measuredangle XPD=\measuredangle XFD=\measuredangle ACD=\measuredangle EFA=\measuredangle PDX$$and $PX=XD.$ Similarly, $Y$ lies on the perpendicular bisector of $\overline{PD}.$ $\square$

https://artofproblemsolving.com/community/c6h2968866_pq_bisects_dg__feet_of_altitudes_related Hope this is just a coincidence