Solution I submitted in the test : Humpty Dumpty Points [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(12); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.937259489861071, xmax = 23.819220167148476, ymin = -9.374847898528708, ymax = 5.5033919299760505; /* image dimensions */ pen ffefev = rgb(1.,0.9372549019607843,0.8980392156862745); pen ffwwqq = rgb(1.,0.4,0.); pen qqqqcc = rgb(0.,0.,0.8); pen evefev = rgb(0.8980392156862745,0.9372549019607843,0.8980392156862745); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); filldraw((7.,2.7)--(4.54,-3.64)--(14.84,-3.6)--cycle, ffefev, linewidth(0.8) + ffwwqq); filldraw((7.341583871988063,-2.1335679926174)--(7.232837759663594,-2.394670047481612)--(7.493939814527805,-2.5034161598060813)--(7.602685926852275,-2.2423141049418693)--cycle, evefev, linewidth(1.2) + qqwuqq); filldraw((7.023485477867514,-3.3475105508848086)--(6.74064489821984,-3.3486089609028578)--(6.74174330823789,-3.631449540550532)--(7.024583887885564,-3.6303511305324827)--cycle, evefev, linewidth(1.2) + qqwuqq); /* draw figures */ draw((7.,2.7)--(4.54,-3.64), linewidth(0.8) + ffwwqq); draw((4.54,-3.64)--(14.84,-3.6), linewidth(0.8) + ffwwqq); draw((14.84,-3.6)--(7.,2.7), linewidth(0.8) + ffwwqq); draw(circle((5.77709273721083,-2.2963798317884283), 1.8263936040417654), linewidth(0.8) + qqqqcc); draw(circle((10.93551380598617,-4.444805041439427), 3.9948351652209673), linewidth(0.8) + qqqqcc); draw(circle((9.69,-3.62), 5.150038834805033), linewidth(0.8)); draw((7.,2.7)--(7.024583887885564,-3.6303511305324827), linewidth(0.8)); draw((3.3696980425581566,-0.47932156099977413)--(10.92,-0.45), linewidth(0.8)); draw((-0.26060391488368606,-3.658643121999548)--(4.54,-3.64), linewidth(0.8)); draw((7.024583887885564,-3.6303511305324827)--(7.602685926852275,-2.2423141049418693), linewidth(0.8)); draw((3.3696980425581566,-0.47932156099977413)--(14.60482687618004,-5.158628212785114), linewidth(0.8)); draw(circle((3.369698042558158,-0.4793215609997743), 4.825679008227144), linewidth(0.8) + yqqqyq); draw((3.3696980425581566,-0.47932156099977413)--(-0.26060391488368606,-3.658643121999548), linewidth(0.8)); draw((1.6712215637847392,-2.1263152383565282)--(1.5131013959619173,-1.9457660413832856), linewidth(0.8)); draw((1.5959927317125544,-2.192198641616037)--(1.4378725638897327,-2.0116494446427944), linewidth(0.8)); draw((3.3696980425581566,-0.47932156099977413)--(7.,2.7), linewidth(0.8)); draw((5.068174521331575,1.16767211635698)--(5.226294689154397,0.9871229193837371), linewidth(0.8)); draw((5.14340335340376,1.2335555196164887)--(5.301523521226581,1.0530063226432458), linewidth(0.8)); /* dots and labels */ dot((7.,2.7),dotstyle); label("$A$", (7.08,2.9), NE * labelscalefactor); dot((4.54,-3.64),dotstyle); label("$B$", (4.18,-4.12), NE * labelscalefactor); dot((14.84,-3.6),dotstyle); label("$C$", (15.1,-3.48), NE * labelscalefactor); dot((5.77,-0.47),linewidth(4.pt) + dotstyle); label("$E$", (5.48,-0.2), NE * labelscalefactor); dot((10.92,-0.45),linewidth(4.pt) + dotstyle); label("$F$", (11.,-0.3), NE * labelscalefactor); dot((7.024583887885564,-3.6303511305324827),linewidth(4.pt) + dotstyle); label("$D$", (7.18,-4.02), NE * labelscalefactor); dot((7.602685926852275,-2.2423141049418693),linewidth(4.pt) + dotstyle); label("$K$", (7.6,-1.82), NE * labelscalefactor); dot((3.3696980425581566,-0.47932156099977413),linewidth(4.pt) + dotstyle); label("$Z$", (2.94,-0.24), NE * labelscalefactor); dot((-0.26060391488368606,-3.658643121999548),linewidth(4.pt) + dotstyle); label("$D'$", (-0.48,-4.22), NE * labelscalefactor); dot((7.012291943942782,-0.46517556526624165),linewidth(4.pt) + dotstyle); label("$P$", (7.1,-0.3), NE * labelscalefactor); dot((5.136018082752459,-1.2149742834259665),linewidth(4.pt) + dotstyle); label("$X$", (4.86,-1.06), NE * labelscalefactor); dot((14.60482687618004,-5.158628212785114),linewidth(4.pt) + dotstyle); label("$Y$", (14.32,-4.82), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note $K$ is $A$-dumpty point in triangle $ABC$. Add $E,F$ to be midpoints of $AB,AC$ respectively. We will use the following two lemmas in our proof : Lemma 1 : $K$ is $D$- Humpty point in triangle $DEF$. Proof : We angle chase to prove $KEBD$, $KFCD$ are cyclic with $EF$ being common tangent. Note its well-known that $AEFOK$ all lie on a circle. Using this, $\angle FEK = \angle FAK = \angle CAK = \angle ABK = \angle EBK \implies EF$ tangent to $(EBK)$. But $EF$ is tangent to $(EBD)$ too. Hence, $EKDB$ is cyclic with $EF$ being tangent to it. Similarly $EF$ is tangent to $FKDC$. This proves the claim. $\square$ Lemma 2 : In triangle $ABC$. Let $D$ be foot of $A$ onto $BC$, and let $D'$ on $BC$ such that $(DD';BC)=-1$. If $L$ is the $A$- Humpty point in triangle $ABC$, then $\angle ALD' = 90$. Proof : Invert about $(BC)$. The points in pairs $(D,D')$ and $(A,L)$ swap. Let $M$ be midpoint of $BC$. As $\angle ADM=90$, hence $\angle D'LM=90 \implies \angle D'LA = 90$. $\square$ We now go back to the problem. Define $Z = XY \cap EF$. Also add $D'$ such that $(DD';BC)=-1$. Claim 1 : $Z$ is midpoint of $AD'$. Proof : Add $AD \cap EF = P$, which is the foot of $D$ onto $EF$. By lemma 1, $K$ is the $D$-humpty in $DEF$. But we have been given that $\angle DKZ = 90$. Hence by applying lemma 2 on triangle $DEF$, we must have $(ZP;EF)=-1$. Homothety about $A$ with ratio 2 maps $P,E,F$ to $D,B,C$. As $(ZP;EF)=(D'D;EF)$, this homothety maps $Z$ to $D'$, proving the claim. $\square$ Claim 2 : $ZA^2 = ZD^2 = ZX \cdot ZY$. Proof : First notice that $(ADD')$ is orthogonal to $BC$. This is because $(ALDD')$ is cyclic by lemma 2, and on inverting about $(BC)$; the circle $(ALDD')$ inverts to itself. As $Z$ is the midpoint of $AD'$ and $\angle ADD'=90$, notice that $Z$ is the centre of $(ADD')$. Hence \begin{align*} ZA^2 = ZD^2 &= \text{Pow}_{(BC)}(Z) \hspace{0.5cm} (\text{as circle } (Z;A,D) \text{ orthogonal to } (BC)) \\ &=ZX\cdot ZY \hspace{1cm} \square \end{align*} To finish : Notice $Z$ the centre of $A-$ appolonius circle wrt triangle $AXY$. (This is because $Z$ lies on $XY$ and satisfies $ZA^2=ZX\cdot ZY$) But by last claim, $D$ lies on this applonious circle,. Hence $\frac{AX}{AY} = \frac{DX}{DY}$ and we are done!

Nice config problem! Here is my solution. First, we set up the stage by defining a few points and making basic observations. Let $L$ be the reflection of $A$ across $K$. Since $K$ is the Dumpty point, it follows that $L\in\odot(ABC)$ and $ABLC$ is harmonic quadrilateral. Let $A_1$ be the point on $\odot(ABC)$ such that $AA_1\parallel BC$. Notice that by Symmedian property, $L, M, A_1$ are collinear. Let $A'$ be the reflection of $A$ across $BC$. Since $BA'CA_1$ is parallelogram, it follows that $A'$, $L$, $M$ are collinear. Let $H$ be the orthocenter of $\triangle ABC$, and let $H'$ be its reflection across $BC$. Now, we begin the meat of the solution. First, note that by $DK$ is the $A$-midline of $\triangle AA'L$, so $DK\parallel A'L$. Now, we let $T=H_aL\cap BC$. Notice that lines $H_aL$ and $HT$ are symmetric across $BC$, but $H_aL\perp A'M$, so we have $HT\perp AM$. Thus, by Brokard's theorem on $BCEF$ ($E$ and $F$ are feet of altitudes in $\triangle ABC$), we have that $T$ lies on the polar from $A$ to $\odot(BC)$. Thus, if $P$ is the midpoint of $AT$, then we note the following: $P$ lies on the radical axis of $\odot(A,0)$ and $\odot(BC)$ as a consequence from the above paragraph; $PA=PD$; we have $\angle H_aLA_1=90^\circ$, so $H_aL\perp A'M\parallel DK\perp XY$, so $P$, $X$, $Y$ are collinear. These three facts imply that $PA^2 = PD^2 = PX\cdot PY$. Thus, $XP:PY = XA^2 : YA^2 = XD^2 : YD^2$, done.

Solved with CyclicISLscelesTrapezoid It suffices to show that there is a circle of Apollonius wrt $A, D$ passing through $X, Y$; AKA the circle centered on line $AD$ passing through $X, Y$ intersects line $AD$ at two points which form a harmonic bundle with $A, D$. We claim that the perpendicular bisector of $XY$ intersects $AD$ at the reflection of $A$ over $D$. This is because by angle chasing $K$ is the intersection of the $A$-symmedian and $(BOC)$, if $O$ is the circumcenter of $\triangle{ABC}$, and if $AK \cap (ABC) = K'$ and $M$ is the midpoint of $BC$ then $BC$ bisects $\angle{AMK'}$, and $\angle{AMO} = \angle{AK'O} \implies AOMK'$ is cyclic, so performing cross-overlaid inversion about $\triangle{ABC}$ we get that the inverse of $O$, which is the reflection of $A$ over $BC$, lies on $MK'$. By a $\frac{1}{2}$x homothety at $A$, this implies $DK \parallel MA'$. Now it suffices to show that $A'X = A'Y = A'D \cdot \sqrt{2}$, because this is the only possible radius which will make the circle centered at $A'$ orthogonal to $(AD)$. If $P, Q$ are the points on $BC$ such that $DP = DQ = DA$, then this is equivalent to $PQXY$ being cyclic. Now let $T = XY \cap BC$; we want to show that $TP \cdot TQ = TX \cdot TY$, or $TP \cdot TQ = TB \cdot TC$, which is equivalent to $T$ lying on the radical axis of $(PQ), (ABC)$. This is the line through $A$ perpendicular to $DO$, so it suffices to show that $AT \perp DO$, or $\angle{ODM} = \angle{TAD}$. Thus, it suffices to show that $\frac{DM}{MO} = \frac{DA}{DT}$, or $\frac{DM}{DA} = \frac{MO}{DT}$, but since $\angle{KDT} = \angle{AMD}$, the former fraction is equal to $\frac{DK}{KT}$, so it suffices to show that $\frac{DK}{MO} = \frac{KT}{DT}$, or $\frac{DK}{MO} = \frac{AD}{AM}$, which is obvious since $\triangle{AKO} \sim \triangle{ADM} \implies \triangle{ADK} \sim \triangle{AMO}$. $\square$

My problem Pranav1056 wrote: In triangle $ABC$, let $D$ be the foot of the perpendicular from $A$ to line $BC$. Point $K$ lies inside triangle $ABC$ such that $\angle KAB = \angle KCA$ and $\angle KAC = \angle KBA$. The line through $K$ perpendicular to like $DK$ meets the circle with diameter $BC$ at points $X,Y$. Prove that $AX \cdot DY = DX \cdot AY$ It suffices to prove that the following three lines concur: line $XY$; the perpendicular bisector of $AD$; the radical axis of the point circle $A$ and the circle with diameter $BC$. Indeed, let $O$ be the desired concurrency point; then $OA^2=OD^2=OX \cdot OY$, hence $O$ is the centre of the Appolonian circle consisting of points $P$ such that $\frac{PX}{PY}=\frac{AX}{AY}$, hence $D$ also lies on the circle, proving the claim. To show the concurrency, we dilate at $A$ with factor $+2$, to get the three new lines: the line passing through $F$ perpendicular to the line $A'F$; where $A'$ is the reflection of $A$ in line $BC$ and $F$ is the reflection of $A$ in $K$. the line $BC$; and the polar line to $A$ in the circle with diameter $BC$. Let $E$ and $F$ be the feet of the $B$ and $C$ altitudes in triangle $ABC$ and the lines $EF$ and $BC$ meet at $T$. Then, since $(BC; DT)=-1$, $A$ lies on the polar of $T$ so $T$ lies on the polar of $A$ (in the circle with diameter $BC$). Finally, observe that it is known that $K$ is the midpoint of the $A$-symmedian chord and so $F$ is the point on the circumcircle of $ABC$ where the $A$-symmedian meets it again. Now the third line is simpply the reflection in line $BC$ of the line through the orthocentre $H$ perpendicular to the $A$-median in triangle $ABC$; simply because it is known that the reflection of $F$ in line $BC$ is the $A$-HM point. The proof is complete. Remark: After $D$ inversion, the problem becomes equivalent to the fact: In triangle $ABC$ with incentre $I$, the incircle touches $BC$ at point $D$. The circle with diameter $AD$ meets the circle with diameter $BC$ at points $X$ and $Y$. Prove that $IX=IY$. This itself is a degenerate case for a more general statement true for circumscribed quadrilaterals (detaching $D$ from line $BC$). This latter result seems to be known, communicated to me by Zack Chroman.

Refering to this diagram above, it suffice to prove $TM^2-MB^2=TA^2$, but $TM^2+TA^2=2(TN^2+AN^2)$ so we want $$2(TN^2+AN^2-TA^2)=PQ^2$$Let projection from $T$ to $AN$ be $L$, then we want $2(LN^2+AN^2-LA^2)=PQ^2$, and $L$ is known to be the $A$-humpty point of triangle $APQ$. But $$LN^2+AN^2-LA^2=LN^2+(AN+LA)\cdot (AN-LA)=LN^2+(AN+LA)\cdot LN=2AN\cdot LN=2NP^2=\frac{1}{2}PQ^2$$and we are done. PS: This "bash" (if it is still considered so..) is unexpectedly short and direct

Alternate solution anantmudgal09 wrote: Remark: After $D$ inversion, the problem becomes equivalent to the fact: In triangle $ABC$ with incentre $I$, the incircle touches $BC$ at point $D$. The circle with diameter $AD$ meets the circle with diameter $BC$ at points $X$ and $Y$. Prove that $IX=IY$. This itself is a degenerate case for a more general statement true for circumscribed quadrilaterals (detaching $D$ from line $BC$) inverted problem wrote: Given an circumscribed quadrilateral $ABCD$, the circles with diameter $AC,BD$ intersect at $X,Y$. Prove that $IX=IY$ where $I$ is the center of the circle. Notice that it suffices to prove that $M-I-N$ where $M$ and $N$ are the midpoints of $AC$ and $BD$. Now we use Leon Anne’s theorem to prove this, it suffices to check $[AID]+[BIC] = [AIB] + [CID]$ , however this is obvious by Pitot's theorem. Done

Amazing problem! Here is a proof by Rijul Saini! Note that $K$ is the A-Dumpty point of triangle $ABC$. Now, we prove a lemma which is the crux of this problem: Dumpty-Harmonic Lemma: Let $ABC$ be a triangle and let $D$ be the foot of $A$ onto $BC$. Let $D'$ be the harmonic conjugate of $D$ with respect to $BC$, then, if $P$ is the midpoint of $AD'$, we have $\angle DKP=90^{\circ}$, where $K$ is the $A$-Dumpty point of triangle $ABC$. Proof. Let $M$ be the midpoint of $AD$, and let $D_1$ be that of $DD'$. it suffices to show that $K$ lies on $(PMDD_1)$, so it suffices to show that $\angle PKM=\angle PD_1M =\angle D'AD$. Let $S$ be the reflection of $A$ over $K$ (which is well-known to lie on $(ABC)$, as well as the fact that $AK$ is the $A$-symmedian). Let $X=AD\cap (ABC)$ and $S'=SD\cap (ABC)$. Then, we claim that $S-X-D'$ and $A-S'-D'$. In order to show that, note that $ABCS$ is harmonic, and so, if we invert at $D$ with radius $\operatorname{Pow}$$(D,(ABC))$, we see that $BXCS'$ is harmonic as well. If we project from $A$ onto $BC$, we get that $A-S'-D'$, and if we project from $S$ onto $BC$, we get that $S-X-D'$. Back to the lemma, from midpoints theorem, we see that \[\angle MKP=\angle D'SD=\angle S'AD=\angle D'AD\]and so we are done with the lemma. Back to the main problem, with the same terminologies as that of this lemma, we see that $P-X-Y$. Now, we know that the polar of $A$ with respect to the circle with diameter $BC$ passes through $D$, and it is well known that the radical axis of a point and a circle is the dilation of the polar of that point with respect to the circle by half, and so, $P$ lies on the radical axis of point circle $A$ and the circle with diameter $BC$. So, as $PA=PD$, we have that \[PA^2=PX\cdot PY=PD^2\]and the length equality follows immediately.

Claim: Let $\tau$ be the circle with diameter $BC$. Consider all pairs $X_1, Y_1\in \tau$ of distinct points for which $AX_1\cdot DY_1=DX_1\cdot AY_1$. We claim that $X_1Y_1$ passes through a fixed point $J$ which lies on the perpendicular bisector of $\overline{AD}$. Proof. Let $Z_1, Z_2$ denote the points on $AD$ where the (internal, external) angle bisectors of $\angle DX_1A$ intersect. Note that $(AD;Z_1Z_2)=-1$ and $Z_1X_1Z_2$ is the Apollonius circle. Since $\frac{AX_1}{DX_1}=\frac{AY_1}{DY_1}$ we have that $Y_1$ also lies on this Apollonius circle. So, $X_1, Y_1$ are the intersections of the Apollonius circle and $\tau$. It is well-known that for all pairs $Z_1, Z_2$ such that $(AD;Z_1Z_2)=-1$, all points on the perpendicular bisector of $\overline{AD}$ have constant power w.r.t the Apollonius circle of $Z_1,Z_2$. Define $J$ to be the intersection of $X_1Y_1$ with the perpendicular bisector of $\overline{AD}$ for some pair $X_{10},Y_{10}$. Since it lies on the radical axis, the power of $J$ w.r.t. $\tau$ and power of $J$ w.r.t. Apollonius circle of $Z_{10}, Z_{20}$ is same. Then, as $Z_1,Z_2$ vary, the power of $J$ w.r.t. Apollonius circle of $Z_1,Z_2$ remains same, which is also its power w.r.t. Apollonius circle of $Z_{10}, Z_{20}$ which is the power of $J$ w.r.t. $\tau$. Hence, it always lies on the radical axis, which is $X_1Y_1$. $\blacksquare$ Let $AD,BE,CF$ be the altitudes in $ABC$. Let $V$ be the mid-point of $\overline{AD}$. Let $S=2K-A$, $L=EF\cap BC$, $A'=2D-A$ and $O$ be the center of $(ABC)$. Let $A,N$ be intersections of $AD$ with $(ABC)$. Claim: $L,D,S,A'$ are cyclic. Proof. $K$ is the $A-$Dumpty point of $ABC$ which means $AK$ (or $AS$) is the symmedian as $(AS;BC)$ is a harmonic bundle. We show that $MDNS$ is cyclic. Note that $\angle MSB=\angle CBA$ which means $\pi-\angle CBA=\angle SBM + \angle BMS$ and hence \[\pi-\angle DMS=\pi-\angle BMS=\angle SBM+\angle CBA=\angle SBC+\angle CBA=\angle SBA=\angle SNA=\angle SND\]which means $MDNS$ is cyclic, which means $\angle NSM=\angle NDM=90^\circ$. It is well-known that $L,N,S$ are collinear. Note that $A,O,M,S$ is cyclic because $M$ is $C-$Dumpty point of $ACS$. Perform $\sqrt{bc}$ inversion to get that $M,S,A'$ are collinear. Hence, $N$ is the orthocenter of $MLA'$ and hence $LDSA'$ is cyclic. Taking homothety at $A$ with factor $\frac12$, we see that $A'$ goes to $D$, $S$ goes to $K$ and $D$ goes to $V$ and $L$ goes to some point on the perpendicular bisector of $\overline{AD}$. However, we want to show that $DKVJ$ are concyclic [because then $\angle DKJ=\angle DVJ=90^\circ$ and we'd be done], but we have shown that $A'SDL$ is cyclic, so we only need to show that $J$ is the mid-point of $\overline{AL}$. The idea for proving that is similar to the first claim, but just replace $(AD;Z_1Z_2)$ with $(LD;BC)$. Note that this means $J$ lies on the perpendicular bisector of $\overline{LD}$. Hence, $JA=JD=JL$ and so $J$ is the mid-point of $\overline{AL}$. $\blacksquare$

Attachments:

As in #$9$ we have the lemma: Dumpty-Harmonic Lemma: Let $ABC$ be a triangle and let $D$ be the foot of $A$ onto $BC$. Let $D^*$ be the harmonic conjugate of $D$ with respect to $BC$, then, if $P$ is the midpoint of $AD^*$, we have $\angle DKP=90^{\circ}$, where $K$ is the $A$-Dumpty point of triangle $ABC$. We will prove that $D_1DMK$ is cyclic where $D_1$ is the midpoint of $DD'$, $M$ is the midpoint of $AD$. Invert with radius $\sqrt{DB\cdot DC}$ centered at $D$, followed by a reflection about $D$. Clearly $B\leftrightarrow C$. Let $AD\cap (ABC)=A'$. By Power of a Point, $A$ is sent to $A'$. Let $E,F$ be midpoints of $AB,AC$ respectively. It is well known that $K=(EBD)\cap(DCF)\neq D$. $E$ is the centre of $(ABD)$ and $F$ centre of $(ADC)$. This means $E'$ is the reflection of $D$ about $A'C$ and $F'$ the reflection of $D$ about $A'B$. Then, $K=BF'\cap CE'$. Consider triangle $BK'C$. Evidently, $A'$ is the intersection of the bisectors at $B,C$ so $A'$ is the incenter of $\triangle BK'C$. $M$ is the midpoint of $AD$ so $M'$ is the reflection of $D$ about $A'$. But this is the antipode of $D$ on the incircle of $\triangle BK'C$. We want to prove that $D_1'M'K'$ are collinear. Note that $(D,D^*;B,C)=-1$ implies $(P_\infty,D^{*}\text{}';B,C)=-1$ as inversion preserves cross ratios. This means $D^{*}\text{}'$ is the midpoint of $BC$. Hence $D_1'$ is the reflection of $D$ about the midpoint of $BC$ hence $BD=D_1'C$. But as $D$ is the incircle tangency point of $BK'C$, it is well known from this length relation that $D_1'$ is the excircle tangency point. But it is well known that a vertex, incircle tangency antipode, and excircle tangency point are collinear. Hence we deduce that in fact $D_1'M'K'$ are collinear and we are done. Let $L$ be the $A-HM$ point of $\triangle{ABC}$. Let $Q$ be the midpoint of $BC$. By definition Power of a Point gives $QB^2=QL\cdot QA$ or that $L$ is the inverse of $A$ with respect to $(BC)$. Then the polar of $A$ with respect to $(BC)$ is $LH$ where $H$ is the orthocenter of $\triangle {ABC}$. The radical axis of a point and circle is the midline of the point and its polar. It is well known $LH$ also passes through $D'$ so a dilation at $A$ shows $P$ lies on the radical axis of $A$ and $(BC)$. This means $PX\cdot PY=PA^2=PD^2$. Hence the circle centered at $P$ through $A$ and $D$ is the $A$-apollonian circle of $\triangle AXY$, and since it passes through $A,D$ we are done.

Attachments:

Note that $K$ is the $A$-Dumpty point of $\triangle ABC$. We will define some more points. $\bullet$ $\triangle MNP$ is the medial triangle of $\triangle ABC$. $\bullet$ $X_A$ is the $A$-Ex point of $\triangle ABC$. $\bullet$ $S=\overline{PN} \cap \overline{XY}$. $\bullet$ $\omega$ is the circle with diameter $\overline{BC}$. Claim: $S \in \overline{AX_A}$. Proof: Redefine $S$ as the $A$-ex point of $\triangle APN$. So we need to prove that $\measuredangle SKD=90^{\circ}$. Reflect over $\overline{PN}$ and it is well known that $K$ goes to $A$-humpty point of $\triangle APN$ (call it $K'$); and it is also well known that $\measuredangle SK'A=90^{\circ}$ (for example by Brokard).$\square$ By Apollonian circle shenanigans we just need to prove that $(AX_AD)$ and $\omega$ are orthogonal, because then we are done by Pole-Polar. This is true since $(B,C;D,X_A)=-1$ and so $D$ and $X_A$ lie on each others polars w.r.t. $\omega$ and hence the two circles we mentioned above are orthogonal.

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