If $(ADE)$,centered at $O$,passes through a fixed point, say $X$, other than $A$, then the locus of $O$ must be the perpendicular bisector of $AX$ as $P$ varies. Observe that $\angle BPD=180-2\angle B,\angle CPE=180-2\angle C$, thus $\angle DPE=180-[360-2(\angle B+\angle C)]=180-2\angle A$. Note that $\angle DOE=2\angle A$, thus $P,D,O,E$ are concyclic, furthurmore, $PO$ bisects $\angle DPE$ since $DO=OE$. Let $PO$ meet $DE$ at $K$. Since $DOK\sim POK$, thus $OP=\frac {DO*DP}{DK}$(*). By extended sine rule and angle bisector theorem we have $DO=\frac {DE}{2sin\angle A}$ and $DK=\frac {DE*DP}{DP+PE}=\frac {DE*DP}{BC}$(since $DP=BP,PE=CP$).Now we plug these results into (*) and after some simplifying we have $OP=\frac {BC}{2sin\angle A}$ which is a fixed value. Furthermore, $\angle OPB=\angle DPB+\angle DPO=270-2\angle B-\angle A$, which is also fixed. Hence as $P$ moves along $BC$, the magnitude and direction of $OP$ stays the same. which means that the locus of $O$ is a line parallel to $BC$ and the fixed point on $(O)$ is the intersection of the $A$-altitude with $(O)$(other than $A$).

we can also prove that the circle $(O)$ passes through the orthocenter of triangle $ABC$ let $X,Y$ be midpoints of $BD,CE$ respectively and let $AH$ intersects the circumcircle of $ABC$ at $Z$ and let $G=AH \cap BC$ so we have that circumcircles of $ABC,ADE,AXYG$ are concurrent at a point other than $A$ ,let this point be $K$ now we have that if $S$ be the intersection of $AH$ and $(O)$ then $G$ is midpoint of $SZ \Longrightarrow S \equiv H$ so we are done

Darn this took way too long. We claim that the orthocenter $H$ of $\bigtriangleup ABC$ lies on the circumcircle of $\bigtriangleup ADE$ regardless of the location of the point $P$ on $BC$. [asy][asy] import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.300000000000005,xmax=7.960000000000012,ymin=-5.500000000000006,ymax=6.300000000000007; pair A=(-0.6800000000000008,4.980000000000006), B=(3.160000000000004,-0.7400000000000008), C=(-1.760000000000002,-1.060000000000001), P=(1.723475284670574,-0.8334325018100450), D=(2.353907271759469,0.4607422931082958), H=(-0.3487737647763798,-0.1126033665631657), M=(1.779593460306762,1.316230574751391); draw(A--C); draw(C--B); draw(B--A); draw((-1.465637979375667,0.5862468560842407)--P); draw(P--D); draw((-1.465637979375667,0.5862468560842407)--D); draw(circle((0.5090884022823820,2.500265814908376),2.750093245174520)); draw((-1.465637979375667,0.5862468560842407)--H,linetype("2 2")); draw(H--D,linetype("2 2")); draw((-1.465637979375667,0.5862468560842407)--H,linetype("2 2")); draw((-1.465637979375667,0.5862468560842407)--H,linetype("2 2")); draw(H--(-1.552124096897579,0.1025652358691018),linetype("2 2")); draw(M--H,linetype("2 2")); dot(A,ds); label("$A$",(-0.6000000000000003,5.100000000000006),NE*lsf); dot(B,ds); label("$B$",(3.480000000000006,-1.160000000000001),NE*lsf); dot(C,ds); label("$C$",(-2.220000000000003,-1.440000000000002),NE*lsf); dot(P,ds); label("$P$",(1.720000000000003,-1.320000000000002),NE*lsf); dot((-1.465637979375667,0.5862468560842407),ds); label("$E$",(-1.380000000000001,0.7000000000000008),NE*lsf); dot(D,ds); label("$D$",(2.440000000000004,0.5800000000000006),E*lsf); dot(H,ds); label("$H$",(-0.4600000000000001,-0.5600000000000007),NE*lsf); dot((-1.552124096897579,0.1025652358691018),ds); label("$N$",(-2.380000000000003,0.02000000000000002),NE*lsf); dot(M,ds); label("$M$",(1.860000000000003,1.440000000000002),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] If we can show that $\angle EAH = \angle EDH$ or $\angle DAH = \angle DEH$, then the conclusion follows. Let the projection of $H$ to $AB, BC$ be $M$ and $N$ respectively. It isn't hard to determine that \[ \frac{MH}{NH} = \frac{\cos B } {\cos C} \] with some trigonometry on $\bigtriangleup AMH$ and $\bigtriangleup ANH$. Additionally, since $BC = BP + CP$, we have that \[ \frac{ BC-2BP } { 2CP - BC } =1 .\] Noting that $BM = 2BP \cos B, CN = 2CP \cos C$ (because $\bigtriangleup BPM, \bigtriangleup CPN$ are isosceles), we can further find that \[ \frac{MD}{NE} = \frac{ BC \cos B - 2BP \cos B } { 2CP \cos C - BC \cos C } \] \[ = \frac{ \cos B }{ \cos C} \cdot \frac{BC-2BP}{2CP-BC} = \frac{ \cos B }{ \cos C}.\] Hence, by SAS similarity, we have that $\bigtriangleup MDH$ and $\bigtriangleup NEH$ are similar. Since they share a common vertex and are directly similar, $H$ is the center of a spiral similarity that takes $E$ to $D$ and $N$ to $M$. As a result, $\bigtriangleup EDH$ is similar to $\bigtriangleup NMH \implies \angle EDH = \angle NMH = 90^{\circ} - \angle C = \angle EAH$, and we're done. For me, the motivation for this solution was to determine the point by drawing a big, accurate diagram; once it was identifiable as the orthocenter, dropping the perpendiculars should have been the natural reaction to get some manageable triangles. Instead I spent an hour trying to figure out the significance of the restriction on $D$ and $E$, which remained unclear to me until the right triangles were made. Oops.

We can prove that the circle passes through the orthocenter $H$ of the tringle $ABC$ Let $M$ be the midpoint of $BC$ and $S,T$ the foots of altitudes from $C ,B$ .Observe that $SM||PD$ and $PE||MT$ $\frac{SD}{SB}=\frac{PM}{MB}=\frac{PM}{MC}=\frac{ET}{TC} \Rightarrow \frac{SD}{ET}=\frac{SB}{TC}=\frac{SH}{HT} \Rightarrow \Delta SHD \sim \Delta EHT \Rightarrow \angle SHT =\angle DHE =180-\angle BAC$ $\Rightarrow ASHT$ is cyclic Done!!!..

Tsk, fixed point orthocenter again? This seems to be a popular trope on USA TST's.

Not only is H sometimes the fixed point, it also can serve other useful purposes in TST problems.

We can also use barycentric coordinates. Also note that the proposition is also true if we extend sides $AB$, $BC$, $CA$ into lines and $P$ moves along line $BC$.

Nice! A couple of optimizations: 1. We have $PC = PE \iff EC = 2 PC \cos C$, so using this (and the Law of Cosines to compute $\angle C$) saves an application of the distance formula in determing $q$ and $r$. 2. It's probably better to put $D=(q,1-q,0)$ and $E=(r,0,1-r)$ at the beginning (instead of $E=(1-r,0,r)$) so that the resulting expressions are symemtric in $q$ and $r$. No big deal here.

Let the antipodes of $A$ wrt $(ABC),(ADE)$ be $M,M'$ respectively. Then by considering projections on $AB,AC$ we see that the midpoint of $MM'$ is $P$. Thus the distance from $M'$ to $BC$ is the same regardless of the position of $P$; in particular, setting $P$ to the midpoint of $BC$ gives $M'$ to be the orthocenter $H$ (parallel lines give a parallelogram and whatnot). When $M'\neq H$, $M'H\parallel BC$ and so $\angle AHM'=90^\circ=\angle ADM'=\angle AEM'$, implying $A,D,E,H,M'$ are concyclic, so $H$ is the fixed point we want.

If you have already figured out the fixed point is $H$, just let it be. $(PDE)$ intersects $BC$ again at $Q$. WLOG assume $PC < \frac{BC}{2}$, in which case $DQPE$ is a cyclic quadrilateral. Let $H$ be the incenter of $\Delta QDE$. Consider $\Delta QDE: \angle QDE = \angle EPC = 180^0-2C; \angle QED = \angle DPB = 180^0-2B$ $ \Rightarrow \angle DQE = 180^0-2A$. Hence $\angle DHQ = 90^0 + \frac{\angle DEQ}{2} = 180^0-B \Rightarrow DHQB$ is cyclic $\Rightarrow \angle HBD = \angle HQD = 90^0-A \Rightarrow BH \bot AB$. Similarly $CH \bot AC$, so we are done. A very similar problem is this one: Given $\Delta ABC$ and $P$ is a point between $B, C. M \in AB, N \in AC$ such that $MP=MB, NP=NC$. Prove that $(AMN)$ goes through circumcircle of $\Delta ABC$. Does anyone know the source of that problem? I heard it was from IMO SL but can't find it

First,for some point $P$ let $O$ be the circumcenter of $ADE$.Now,from an easy angle chase we have that $DPEO$ is cyclic,so again angle chase we get that for all $P$,$PO$ is parallel to some fixed line.Now,it will be enough to prove that $PO$ is fixed,cause then all points will be on the line,now we have to prove that if $PD+PE$ is fixed and $<DPE$ is fixed,$PO$ is fixed.Now,this is quite well known,but still here is proof of this:Let $F$ be on $PD$ such that $DF=OF$,now we that spiral similarity with center $D$ pictures $DOF$ to $DEP$,so we get that $DFP$ is similar to $PDE$ and from this is obvious that $PO$ is fixed,so we are finished.

$DE$ is simply tangent to a fixed parabola, and when $P$ is point s.t. $PC = PA$ and $PB=PA$ this parabola is tangent to $AB, AC$ so done. Note we can define this, by taking $P$ being $B, C$, and indeed it is the orthocentre.

My solution: Let $ Q \equiv \odot (PDE) \cap BC $ . Since $ \angle EDQ=\angle EPB=180^{\circ} -2\angle CBA ,\angle QED=\angle CPD=180^{\circ} -2\angle ACB $, so we get $ \triangle QDE $ is similar to the orthic triangle of $ \triangle ABC $ , hence the Miquel point of $ \{Q, D, E \} $ is the orthocenter $ H $ of $ \triangle ABC $. i.e. $ \odot (ADE) $ pass through a fixed point $ H $ Q.E.D

Let $PQR$ be points on $BC$ (motivated by affinity). $D, E, D', E', D'', E''$ be their corresponding points. Because $DD'/D'D'' = EE'/E'E''$, if $ADE \cap AD''E'' = Q$ then $Q$ is also centre of spiral similarity of $AD'E' \mapsto ADE \mapsto AD''E''$, so it just suffice to show for two points, $P=B,C$.

Let CY, BX be heights and let S be PD cut CY, T is BX cut PE. Then notice PS=PC, PT=PB therefore PS*PD=PC*PB=PE*PT and thus DSET is cyclic. But it is easily seen that HSDA, THEA are cyclic, where H is orthocenter, since <AHS = 180-<B = <ADS, and similarly for THEA. By a well known fact, since HATE, HADS, TEDS are all cyclic, this implies HATEDS is cyclic, so circumcircle of ADE always goes through H.

Consider the inversion $\mathcal{I} : X \mapsto X'$ with pole $A.$ Define $\omega \equiv \odot (AB'C')$ and let $A_1$ be the antipode of $A$ w.r.t. $\omega.$ Let $Y', Z'$ be the reflections of $B', C'$ in $AA_1$, respectively, and denote $K' \equiv B'Z' \cap C'Y'.$ Let the line through $P'$ perpendicular to $AP'$ cut $AB'$ at $Q'$, and let $P'D'$ cut $\omega$ for a second time at $Y^*.$ We will prove that the image of $\odot (ADE)$, i.e. the line $D'E'$ passes through the fixed point $K'.$ It is clear that $\mathcal{I}$ sends $BC \mapsto \omega$, so $P$ is sent to a variable point on the circumcircle of $\triangle AB'C'.$ Meanwhile, from $PB = PD$, we obtain $\measuredangle DBP = \measuredangle PDB \implies \measuredangle ABP = \measuredangle PDA$, where the angles are directed. Hence, under inversion, we have \[-\measuredangle AP'B' = -\measuredangle D'P'A \implies 90^{\circ} - \measuredangle AP'B' = 90^{\circ} - \measuredangle D'P'A \implies \measuredangle B'P'Q' = \measuredangle Q'P'D'.\] Combining this relation with $P'A \perp P'Q'$, it is well-known (Lemma 5) that $(A, Q; B', D')$ is a harmonic division. Furthermore, note that $AP' \perp P'Q', \; AP' \perp P'A_1 \implies P', Q', A_1$ are collinear. Therefore, we may take perspective at $P'$ onto $\omega$ to find that $AB'A_1Y^*$ is a harmonic quadrilateral. Therefore, $\tfrac{B'A}{B'A_1} = \tfrac{Y^*A}{Y^*A_1} \implies Y^*$ is the reflection of $B'$ in $AA_1 \implies Y^* \equiv Y'.$ Hence, $P', D', Y'$ are collinear. Similarly, we find that $P', E', Z'$ are collinear. Then by Pascal's Theorem on cyclic hexagon $B'AC'Y'P'Z'$ it follows that $K' \in D'E'$, as desired. $\square$

My solution: $H$ is the orthocenter of $\triangle{ABC}$ $M, N\in CA, AB: BM = BC = CN \Rightarrow BM\parallel{PE}, CN\parallel{PD}$ We have: $(MA, MB) \equiv (MC, MB) \equiv (CB, CM) \equiv (HA, HB) (\because CB\perp{HA}, CM\perp{HB})$ (mod $\pi$) $\Rightarrow A, M, H, B$ are concyclic $\Rightarrow (MH, MC) \equiv (MH, MA) \equiv (BH, BA) \equiv (BH, BN)$ (mod $\pi$) Similarly, we have: $(CH, CM) \equiv (NH, NB)$ (mod $\pi$) $\Rightarrow \triangle{HBN} \sim \triangle{HMC}$ (1) On the other hand: $\frac{BD}{DN} = \frac{BP}{PC} = \frac{ME}{EC}$ (2) (1), (2) $\Rightarrow \triangle{HBD} \sim \triangle{HME}$ $\Rightarrow (DH, DA) \equiv (DH, DB) \equiv (EH, EM) \equiv (EH, EA)$ (mod $\pi$) $\Rightarrow \odot{(ADE)}$ passes through the fixed point $H$ Q.E.D

TelvCohl wrote: My solution: Let $ Q \equiv \odot (PDE) \cap BC $ . Since $ \angle EDQ=\angle EPB=180^{\circ} -2\angle CBA ,\angle QED=\angle CPD=180^{\circ} -2\angle ACB $, so we get $ \triangle QDE $ is similar to the orthic triangle of $ \triangle ABC $ , hence the Miquel point of $ \{Q, D, E \} $ is the orthocenter $ H $ of $ \triangle ABC $. i.e. $ \odot (ADE) $ pass through a fixed point $ H $ Q.E.D After noticing that the two triangles are similar, why does it follow that $ H $ is the Miquel point of $ \{Q, D, E \} $ ?

We use linearity of power of a point. Where $g(P, \omega)$ denotes the power of $P$ with respect to $\omega$, define the function \[ f(P) = g(P, \omega) - g(P, H_0) \] where $H_0$ is the singularity centered at $H$ and $\omega = (ADE)$. In order to solve this problem, it suffices to show that $f(H) = 0$ regardless of where $P$ is. Observe that if we choose $P$ to be the midpoint of $BC$, $D$ and $E$ are the respective foot of perpendiculars from $C$ and $B$ respectively, so in the scenario $f(H) = 0$. Now this allows us to only focus on nonconstant terms in our linearity of power of a point expansion. Let $H_c$ be the foot of the perpendicular from $C$ to $AB$. By linearity of power of a point, we have that $f(H) = \frac{HC(AH_c-AD)(AH_c)+(HH_c)(CA)(CE)}{HC+HH_C}$ The only variable in this expression is the value of $AD$, thus we may just compute everything in terms of the one degree of freedom, which may be represented by $BP$. Let $BP = x$. Then the coefficient of $x$ in $HC(AH_c-AD)(AH_c)$ is $\frac{-2b^2\cos{B}\cos{A}\cos{C}}{\sin{B}}$. The coefficient in $(HH_c)(CA)(CE)$ is the same after lengthy computations. Hence, $f(H) \equiv 0$ because it is zero at one point and has slope zero, so $H$ lies on $(ADE)$ for all $P$ and we are done.

I actually came up with this problem independently but then quickly found out it existed already. Let the orthocenter of $\triangle ABC$ be $H$. Let the altitude from $A$ cut $BC,\odot ABC$ at $X,Y$ respectively. Let the midpoints of $BD,CE$ be $M,N$ respectively. By the gliding principle we have that $\triangle BYC \sim \triangle MXN \sim\triangle DHE$, thus we have $\angle DHE = \angle BYC$ which means $ADEH$ is cyclic.

An application of so called method of moving points. Suppose $P$ moves at a constant speed along $BC$. Then the projections of $P$ on $BA$, resp. $CA$ also move at constant speeds (possibly different) along $BA$, resp. $CA$. The same is true for the points $D$ and $E$ (see the drawing attached, which I borrowed from #18 @tranquanghuy7198). Now, consider a point $X$ moving at a constant speed along the perpendicular bisector of $AH$ ($H$ is the orthocenter of $\triangle ABC$). It follows the projections $X_B$ and $X_C$ of $X$ on $AB$, resp. $AC$ also move at constant speeds (different). The same is true for the points $D'$ and $E'$ determined as $AD':=2AX_B; AE':=2AX_C$. The key point. We can choose the motion of the point $X$ in such way so that for $P=B$ to have $D=D', E=E'$ and for $P=C$ to hold $D=D', E=E'$. It's because in the particular cases $P=A, P=B$ the corresponding circumcircles pass through $H$. Since, we have a linear motion, it follows $D=D', E=E'$ along all the motion and the result follows. Remark. I first heard about this method from one of my high school teachers, Borislav (Bobby) Mihaylov (he was the author of IMO 1995 p1). In his book (Problems in "elementary" geometry, Sofia 1995) he devoted a chapter on the method of a point moving at a constant speed. As an example, consider two lines $\ell_1,\ell_2$ and two points $P_1\in \ell_1,P_2\in\ell_2$ each of them moving at a constant speed along $\ell_1$, resp. $\ell_2$ (the speeds may be different). Consider another two points $P'_1\in\ell_1,P'_2\in \ell_2$ doing the same along the lines $\ell_1,\ell_2$. If there exists two moments for which $P_1=P_1, P_2=P'_2$ then $P_1\equiv P'_1$ and $P_2\equiv P'_2$ along the entire motion. Saying it in other words, consider a linear mapping $f:\ell_1\to \ell_2$, one may assume we have one dimensional coordinate systems on $\ell_1$ and $\ell_2$ and the coordinates $x$ and $y$ of the points $P$ and $f(P)$ satisfy $y=ax+b$ for some constants $a,b$. If $g:\ell_1\to \ell_2$ is another linear mapping and $f(P_1)=g(P_1), f(P_2)=g(P_2)$ for some points $P_1\ne P_2$ then $f\equiv g$. All such linear mapping can be obtained applying homothety, rotation and translation and consider restriction of it on some line (circus). Bobby Mihaylov's book considers only such kind of transformations (which lead to a motion at a constant speed). If you add additionally an inversion you obtain the class of conformal mapping described by Mobious transforms $w=\frac{az+b}{cz+d}, z\in\mathbb{C}$. They preserve the cross ratio and are connected with the projective geometry. But, in this case, in order to claim two Mobius transforms $f,g$ coincide one needs three points they are equal at. Some additional info, I met recently.

Complex bashing $A,B,C$ lie on a unit circle with orthocenter $H=a+b+c.$ It sufficies to prove that \[\frac{e-d}{h-d} \div \frac{e-a}{h-a} \in\mathbb{R}\]Since $P$ on $BC$, we have $p = b+c-bc\overset{-}{p}$ $\bullet$ Easy to find, that $D = a+p-ab\overset{-}{p}$ $\bullet$ Analogously, $E = a+p-ac\overset{-}{p}$ \[\frac{e-d}{h-d} \div \frac{e-a}{h-a} = \frac{ba\overset{-}{p}-ac\overset{-}{p}}{ba\overset{-}{p}+b+c-p}\cdot \frac{b+c}{p-ac\overset{-}{p}}\]\[\overset{-------------}{(\frac{e-d}{h-d} \div \frac{e-a}{h-a})}=\frac{cp-bp}{cp+ab+ac-abc\overset{-}{p}}\cdot \frac{a(b+c)}{abc\overset{-}{p}-bp}\]It remains to prove, that \[\frac{p}{cp+ab+ac-abc\overset{-}{p}}= \frac{b\overset{-}{p}}{ba\overset{-}{p}+b+c-p}\]or \[abp\overset{-}{p}+bp+cp-p^2=bcp\overset{-}{p}+ab^2\overset{-}{p}+abc\overset{-}{p}-ab^2c\overset{-}{p}^2 \implies\]\[(p-b-c+bc\overset{-}{p})(ab\overset{-}{p}-p)=0\]which is true, since $p = b+c-bc\overset{-}{p}.$ We are done!

Solution with mOvInG pOiNtS We claim that the fixed point is the orthocenter $H$ of $\triangle ABC$. Let $B',C'$ denote the feet of altitude from $P$ onto $AB,AC$ respectively. Animate $P$ projectively on $BC$. Note that we have the following projective maps: $$ D \mapsto B' \mapsto P \mapsto C' \mapsto E$$ Where $D \mapsto B'$ is a consequence of homothety at $B$. And $B' \mapsto P$ is projective because $PB' \perp AB$. Now consider the map which sends every point $X$ on $B$ to some point $Y$ on $AC$ such that $\angle XHY = \pi-A$. Note that this is rotation about a fixed point so it's a projective map. We want to show that these maps coincide. So we only need to check for 3 different locations of $P$. If $P$ is the midpoint of $BC$, then $D,E$ are the feet of altitudes from $B,C$, so the conclusion is obvious. If $P=B$ ( or $=C$), we have that $D=B$, $BB'$ is a altitude and that $E$ is the reflection of $C$ over $B'$. By PoP at $B'$ we have : $$ (B'E)(B'A)=(B'C)(B'A)=(B'B)(B'H)$$ So $(AEDH)$ is concyclic as desired. We're done. $\square$

Let $M, N$ be the midpoints of $BD, CE$ respectively, $H$ be the orthocenter of $ABC$, and $BH, CH$ meet $CA, AB$ at $Y, Z$ respectively. Trivially, we know $PM \perp BD$ and $PN \perp CE$. Now, we will show $(ADE)$ always passes through $H$, which is clearly fixed. Claim: $HDZ \sim HEY$. Proof. Clearly, we have $$\angle HZD = 90^{\circ} = \angle HYE.$$Now, observe $$2 = 2 \left(\frac{BC}{BC} \right) = 2 \left(\frac{BP}{BC} + \frac{CP}{CB} \right)$$so $$1 - \frac{2 \cdot BP}{BC} = \frac{2 \cdot CP}{CB} - 1.$$Now, notice $$DZ = BZ - BD = BZ - 2 \cdot BM = BZ - 2 \left( \frac{BP}{BC} \cdot BZ \right) = BZ \left(1 - \frac{2 \cdot BP}{BC} \right)$$and $$EY = CE - CY = 2 \cdot CN - CY = 2 \left(\frac{CP}{CB} \cdot CY \right) - CY = CY \left(\frac{2 \cdot CP}{CB} - 1 \right)$$so $$\frac{DZ}{EY} = \frac{BZ}{CY} \cdot \frac{1 - \frac{2 \cdot BP}{BC}}{\frac{2 \cdot CP}{CB} - 1} = \frac{BZ}{CY} = \frac{HZ}{HY}$$where the last equality follows from $HBZ \sim HCY$. Now, the desired conclusion is immediate from SAS similarity. $\square$ Now, we have $$\angle DHE = \angle DHZ + \angle ZHE = \angle EHY + \angle ZHE = \angle ZHY = 180^{\circ} - \angle ZAY = 180^{\circ} - \angle DAE$$which finishes. $\blacksquare$ Remark: My (original) solution sucks. It doesn't account for configuration issues and involves a solid amount of computation. A Pretty Spiral Solution - My "Summary": $\bullet$ Let $X$ be the projection of $A$ onto $BC$, $(ADE)$ meet $(ABC)$ again at $F$, and $H_a$ denote the reflection of $H$ over $BC$, which lies on $(ABC)$ by the Orthocenter Reflection Lemma. $\bullet$ $F$ is clearly the Miquel Point of $BCED$, so there exists a spiral similarity at $F$ taking $B, M, D$ to $C, N, E$, as $M, N$ are the midpoints of $BD, CE$ respectively. $\bullet$ Now, the Spiral Similarity Lemma yields $FDE \overset{+}{\sim} FBC$, so $AFMN$ is also cyclic by the Spiral Center Lemma. $\bullet$ Now, notice $$\measuredangle AMP = \measuredangle ANP = 90^{\circ} = \measuredangle AXP$$so $AFMXPN$ is actually a cyclic hexagon. $\bullet$ Now, notice $$\measuredangle MXN = \measuredangle MAN = \measuredangle BAC = \measuredangle BH_aC$$and $$\measuredangle XMN = \measuredangle XAN = \measuredangle H_aAC = \measuredangle H_aBC$$so $XMN \overset{+}{\sim} H_aBC$. $\bullet$ Hence, the spiral similarity at $F$ taking $MN$ to $BC$ also takes $X$ to $H_a$. Now, the Spiral Similarity Lemma implies $FBM \overset{+}{\sim} FCN \overset{+}{\sim} FH_aX$. $\bullet$ Hardest Idea: By definition, we know $D, E$ are the reflections of $B, C$ over $M, N$ respectively. But the Orthocenter Reflection Lemma implies that $H$ is the reflection of $H_a$ over $X$, so $$FBMD \overset{+}{\sim} FCNE \overset{+}{\sim} FH_aXH.$$$\bullet$ Now, the Spiral Similarity Lemma yields $$FDEH \sim FBCH_a$$so $$\measuredangle DHE = \measuredangle BH_aC = \measuredangle BAC = \measuredangle DAE$$which finishes. $\blacksquare$

We will use barycentric coordinates with reference triangle $\triangle ABC.$ Let $P = (0, t, 1 - t)$ for some real $0 \le t \le 1.$ Furthermore, let $E = (u, 0, 1 - u)$ and $D = (v, 1 - v, 0)$ for reals $0 \le u, v \le 1.$ Notice that we have $PC = at, EC = bu, DB = cv.$ Now, since $\triangle PEC$ is isosceles, notice that we have \[\dfrac{EC/2}{PC} = \dfrac{bu}{2at} = \cos(C) \implies u = \dfrac{2at\cos(C)}{b}.\]Similarly, we can compute that \[v = \dfrac{2a(1 - t)\cos(B)}{c}.\]Now, we know that the equation of $(ADE)$ is of the form \[-a^yz - b^2xz - c^2xy + (i_1x + i_2y + i_3z)(x + y + z) = 0\]for some constants $i_1, i_2$ and $i_3.$ Plugging the point $A = (1, 0, 0)$ gives $i_1 = 0,$ putting in the point $D = (v, 1 - v, 0)$ gives the value $i_2 = c^2v,$ and $E = (u, 0, 1 - u)$ gives $i_3 = b^2u.$ Thus, after substituting the values of $u$ and $v,$ the full equation is \[-a^2yz - b^2xz - c^2xy + (2ac(1 - t)\cos(B)y + 2ab(1 - t)z)(x + y + z).\]We now claim that the fixed point is the orthocenter $H = (\tan(A) : \tan(B) : \tan(C)).$ Indeed, it suffices to prove that \[-a^2\tan(B)\tan(C) - b^2\tan(C)\tan(A) - c^2\tan(A)\tan(B) + (2act\sin(B) + 2ab(1 - t)\sin(C))(\tan(A) + \tan(B) + \tan(C)) = 0.\]Since $2ac\sin(B) = 2ab\sin(C) = 4S,$ where $S$ is the area of $\triangle ABC,$ we only need to prove that \[-a^2\tan(B)\tan(C) - b^2\tan(C)\tan(A) - c^2\tan(A)\tan(B) + 4S\tan(A)\tan(B)\tan(C) = 0\]since $\tan(A) + \tan(B) + \tan(C) = \tan(A)\tan(B)\tan(C)$ for any acute triangle. Dividing by $\tan(A)\tan(B)\tan(C),$ we see that it only suffices to show that \[4S = \dfrac{a^2}{\tan(A)} + \dfrac{b^2}{\tan(B)} + \dfrac{c^2}{\tan(C)}.\]Now, notice that \begin{align*} & \tan A \\ &= \dfrac{\sin A}{\cos A} \\ &= \dfrac{a}{2R\cos A} \\ &= \dfrac{2S}{bc\cos A} \\ &= \dfrac{2S}{bc \cdot \frac{b^2 + c^2 - a^2}{2bc}} \\ &= \dfrac{4S}{b^2 + c^2 - a^2}. \end{align*}Thus, it suffices to prove that \[a^2(b^2 + c^2 - a^2) + b^2(c^2 + a^2 - b^2) + c^2(a^2 + b^2 - c^2) = 16S^2,\]but the left-hand side factors as $(a + b + c)(a + b - c)(a - b + c)(c + b - a),$ which can be seen to be $16S^2$ by Heron's Formula. Thus, $H$ always lies on $(ADE),$ so we're done!

If I am not wrong we can just use the fixed point as our weapon to solve the problem.Also I solved this problem back in hostel where we were not allowed any GGB so any construction of new point is fairly motivated I guess.Also this is a completely new solution(In this thread).

A small story on how I stumbled upon this unique solution which no one has posted in this thread yet. Yesterday I misread the problem as $BD=PD$ and $CE=PE$ before going to bed and so found that the fixed point was the circumcenter of $\triangle ABC$ and then thought some miquel stuff would help seeing the config, then constructing $BC \cap \odot(PDE)$ helped a lot and i was able to finish with miquel. Today morning when I visited this thread to type the sol I realised I misread the problem and tried to solve the correct problem. Somehow the same method worked and here I am with this solution. Let $BC \cap \odot(PDE) = X$ and let $H$ be the incenter of $\triangle XDE$ Notice that the angles of $\triangle XDE$ are $\angle DEX = 180-2 \cdot \angle A$ and similarly $\angle EDX = 180-2\cdot \angle C$,$\angle DEX = 180 - 2 \cdot \angle C$, this means that $H = \odot(ADE) \cap \odot(BDX) \cap \odot(CEX)$ Now this cyclicity gives us $$\angle HAB = \angle HAD = \angle HED = \frac{\angle DEX}{2}=90 - \angle ABC \implies AH \perp BC \text{ and similarly } BH \perp AC, CH \perp AB \implies H \text{ is the orthocenter of } \triangle ABC$$which is a fixed point, the end $\blacksquare$

Let $A=(1:0:0),B=(0:1:0),C=(0:0:1)$ and $BP=m,PC=n$ to find that $P=(0:N:M),D=(m:c-m:0),E=(n:0:b-n).$ This means that $$(AED):-a^2yz-b^2xz-c^2xy+(cmy+bnz)(x+y+z)=0.$$This shows that $X=(a:b:c)$ is a fixed point. However, the correct fixed point should be the orthocenter, and $X$ is the incenter.

Yay,TST solve, I guess (although I had to use hints to find orthocenter); Let $P = (0:t:s)$. From trig, we get $D= (2sCos(B):c-2sCos(B):0)$ and $E=(2tCos(c): 0 : b-2tCos(C))$. Next, we find that $(ADE)$ is $$-a^2yz-b^2xz-c^2xy+(2scCos(B)y+2tbCos(C)(x+y+z)=0.$$Simplifying further, we get $$-a^2yz-b^2xz-c^2xy+(\frac{2sS_By+2tS_Cz}{a})(x+y+z)=0.$$And then just plugging in the orthocenter which is $(S_{BC}:S_{AC}:S_{AB})$, we get the desired result.

Hey, they never said you had to find the specific point We invoke barycentric coordinates with $\triangle ABC$ as the reference triangle. We let $P = (0, m, n)$ where $m+n = 1$. We can see that $D = (2n\cos B:0: c-2n\cos B)$ and $E = (2m\cos C:0: b-2m\cos C)$. We can see now that considering $(ADE)$ we have $u = 0$, $v = 2cn \cos B$, and $w = 2bm \cos C$. Now we can see that we simply need $y = b \cos C$ and $z = c \cos B$ and then $m$ and $n$ are eliminated and we have a single variable expression in $x$. $\blacksquare$

If we have two linearly moving points, isn't it well-known that the map between them must be a spiral similarity? This would finish the problem instantly, as $D$ and $E$ are moving linearly, and the center of described similarity would lie on $(ADE)$ by Miquel.

Edit: The below solution is for a different problem (I misread the problem, my bad). For those interested, this is the problem: In acute triangle $ABC$, $\angle A < \angle B$ and $\angle A < \angle C$. Let $P$ be a variable point on side $BC$. Points $D$ and $E$ lie on sides $AB$ and $AC$, respectively, such that $BP = BD$ and $CP = CE$. Prove that as $P$ moves along side $BC$, the circumcircle of triangle $ADE$ passes through a fixed point other than $A$. A solution using "Inversion Bash", followed with some standard length computations. Perform $\sqrt{AI\cdot AL}$-swap, where $I$ is the incenter and $L=AI\cap BC$. Suppose $D$ goes to $D'$ and $E$ goes to $E'$. By Menelaus' Theorem, we need to show that $\frac{AD'}{D'C}\div\frac{AE'}{E'B}$ is fixed (in fact it is $\frac cb$, and we show this.) Let $BC=a, CA=b, AB=c$. Then, $AD+AE=b+c-a$ remains fixed. Let $AI\cdot AL=r^2$. Let $AD=x$. Then, $AD'=\frac{r^2}{x}$ and $AE'=\frac{r^2}{b+c-a-x}$. Note, $D'C=AC-AD'=b-\frac{r^2}{x}$ while $E'B=AE'-AB=\frac{r^2}{b+c-a-x}-c$ Hence, we have \[\frac{AD'}{D'C}\div\frac{AE'}{E'B}=\frac{1-\frac{c(b+c-a-x)}{r^2}}{\frac{bx}{r^2}-1}=\frac cb,\]So, we want $b-\frac{bc(b+c)-bc(a+x)}{r^2}=\frac{bcx}{r^2}-c$ So, we want $b+c=\frac{bc(b+c-a)}{r^2}$ and hence $r^2=\frac{bc(b+c-a)}{b+c}$. But this is true because \[AI=\sqrt{\frac{(s-a)bc}{s}} \text{ and } AL=\frac2{b+c}\sqrt{s(s-a)bc} \Longrightarrow AI\cdot AL=\frac{2bc(s-a)}{b+c}=\frac{bc(b+c-a)}{b+c}=r^2\]We are done. $\blacksquare$ Yay!

Nice and easy