We use barycentrics WRT $\triangle ABC.$ $D(0:1:k), E(0:k:1)$ for some $k.$ Infinity point of the line $AE$ is $A_{\infty}(k+1:-k:-1).$ Equation of line passing through $D$ and $A_{\infty}$ is then $\ell_A \equiv (k-1)x+ky- z=0.$ Isogonal $\lambda_A \equiv kc^2y-b^2z=0$ of $AE$ WRT $\angle BAC$ cuts $\ell_A$ at $P(k(c^2-b^2):(k-1)b^2:k(k-1)c^2)$ Eliminating $k$ and setting $(x:y:z) \to \left ( \frac{_{a^2}}{^x}: \frac{_{b^2}}{^y}:\frac{_{c^2}}{^z} \right)$ gives the equation of the perpendicular bisector of $\overline{BC},$ namely $\tau_A \equiv (c^2-b^2)x+a^2(z-y)=0$ $\Longrightarrow$ Isogonal conjugate of $P$ lies on $\tau_A$ $\Longrightarrow$ $\angle PBA=\angle PCA.$

Let $ F $ be a point in the plane such that $ ABFC $ is parallelogram . Then using $ BD =CE $ , we have $ DF || AE $ or $ DF $ passes through $ P $ . Let $ h_1 , h_2 , h_3 , h_4 $ be the distances of $ P $ from lines $ AB , BF , FC , CA $ respectively . From $ \angle BAP = \angle CAE = \angle BFD $ and $ \angle BAC = \angle BFC $ , we have $ h_1 : h_4 = h_2 : h_3 $ , which implies $ h_1 :h_2 = h_4 : h_3 $ . Since $ \angle ABF = \angle ACF $ , we have $ \angle ABP = \angle ACP $ as desired .

Sort, but what's ELMO? And can you tell me if the following scheme of solution is right : There is a unique point P on a secant AQ ( when it's different of the angle bissector of an isosceles triangle) of a triangle ABC such that angle ABP=angle PCA So we can reconstruct the exercise : let E be on BC with angle EAC=x. Let Q be on BC with angle QAB =x. Let P be thé unique point on AQ such that angle PBA =angle PCA= y. Let D be on BC such that PD is parallel to AE. We havé to show that BD=CE. Let thé intersection of PC and AR be R. By angle Chasing and Law of sines we have BD = BP sin(BAC - x + y)/sin(ACB + x) And CE = CR sin (x+y) /sin (ACB+x) But BPA is similar to CRA so BP / CR= AP/AR =sin(x+y) / sin (BAC-x+y) Hence the result.... Is this right ?

apluscactus wrote: Sorry, but what's ELMO ? I 've now seen the description of thé compétition...

Let $ABFC$ and $ABGP$ are parallelograms. Because $\Delta AEC \cong \Delta FDB$, $\angle BFP = \angle BFD = \angle CAE = \angle BAP = \angle BGP$, so $BGFP$ are concyclic. Since $\Delta APC \cong \Delta BGF$, we have $\angle ABP = \angle BPG = \angle BFG = \angle ACP$.

Attachments:

Let $\angle PAB=\angle EAC=\theta$. Let $M$ be the midpoint of $BC$ and $Q$ be a point such that $BPCQ$ is a parallelogram. Then $MD=ME$ and $MP=MQ \Longrightarrow PDQE$ is a parallelogram $\Longrightarrow PD \parallel QE \Longrightarrow$ $A$, $E$, $Q$ are collinear. Now, choose $T$ such that $APBT$ is a parallelogram. Observe that $ACQT$ is also a parallelogram. Hence, $\angle PAB=\angle ABT=\theta=\angle QAC=\angle AQT \Longrightarrow ATBQ$ is cyclic $\Longrightarrow BQT=\angle BAT$. Bu then $\angle BQT=\angle PCA$ and $\angle BAT=\angle PBA$. Hence $\angle PBA=\angle PCA \text{ } \square$ [asy][asy] import graph; size(250); unitsize(1cm); import math; import olympiad; pen main = linewidth(0.87) + fontsize(10); defaultpen(main); pair A,B,C,D,E,M,P,Q,F,G; B=(0,0); A=(2.45,6); C=(8,0); M=(B+C)*0.5; real r=1.67; D=(M.x-r,M.y); E=(M.x+r,M.y); dot(A,4bp+black); label("$A$",A,N); dot(B,4bp+black); label("$B$",B,SW); dot(C,4bp+black); label("$C$",C,SE); dot(M,4bp+blue); label("$M$",M,S); dot(D,4bp+blue); label("$D$",D,NE); dot(E,4bp+blue); label("$E$",E,SE); draw(A--B--C--cycle,black); path g=circumcircle(A,B,C); pair[] L=intersectionpoints(A--(A+10*(E-A))--cycle,g); F=L[1]; pair R=B+F-C; pair[] K=intersectionpoints(F--R--cycle,g); pair G=K[1]; pair S=D+E-A; P=extension(S,D,A,G); dot(P,4bp+blue); label("$P$",P,NW); draw(A--P--cycle,blue+linewidth(0.67)); draw(P--D--cycle,blue+linewidth(0.68)); Q=B+C-P; dot(Q,4bp+blue); label("$Q$",Q,SW); draw(P--Q--cycle,blue+linewidth(0.6)); draw(A--Q--cycle,blue+linewidth(0.66)); draw(B--P--C--Q--cycle,red+linewidth(0.63)); pair T=A+B-P; draw(A--T,blue+linewidth(0.44)); draw(Q--T,green+linewidth(0.7)); dot(T,4bp+blue); label("$T$",T,NW); draw(B--T--cycle,green+linewidth(0.7)); markscalefactor=0.07; filldraw(anglemark(B,A,P),cyan); filldraw(anglemark(E,A,C),cyan); filldraw(anglemark(A,B,T),cyan); filldraw(anglemark(A,Q,T),cyan); draw(circumcircle(A,T,B),purple+linewidth(0.62)); [/asy][/asy]

Since nobody has posted this yet, here's why I think Luis González's and apluscactus's approaches work so well: We want to show that $P,Q$ are isogonal conjugates, where $Q$ is the intersection of $AE$ and the perpendicular bisector $\ell$ of $BC$. Let $Q'$ be the isogonal conjugate of $Q$, and suppose the in-ellipse of $\triangle{ABC}$ with foci $Q,Q'$ touches $BC,CA,AB$ at $X,Y,Z$, respectively. Observe that $X$ is just the point along $BC$ that minimizes $QX+Q'X$, or equivalently, the point on $BC$ collinear with $Q'$ and the reflection of $Q$ over $BC$. First note that if $P,Q$ were isogonal conjugates, we would have $\angle{PDB}=\angle{QED}=\angle{QDE}$, so $D$ would equal $X$. (This is not strictly necessary but it helps motivate what follows.) Since $Q\in\ell$, we have $\angle{CQE}=\angle{BQD}$. By this property of ellipses, $\angle{BQX}=\angle{CQE}$, so $D=X$. But then $\angle{Q'DB}=\angle{QDC}=\angle{QEB}$, so $Q'D \parallel QE$ implies $P=Q'$, as desired. (Alternatively, define $D'\in BC$ such that $Q'D'\parallel AE$; then $\angle{AQ'B}+\angle{D'Q'C} = 180^\circ$, so $D'=X$, and then $\angle{QD'E} = \angle{Q'D'B} = \angle{QED'}$ finishes the proof.) Of course, a lot of this can be phrased without ellipses, but this is just a "higher-level" perspective.

Let $AP\cap BC=M$, $X=PC\cap AE$, and $P'$ the point on $AM$ such that $\triangle BP'A\sim\triangle CXA$. Let $D'$ be the point on $BC$ such that $P'D' || AE$. Note that $\angle AXP'=\angle D'P'X=\angle BP'M$ from the above information. $\angle BP'M=\angle D'P'C$ implies that $\angle BP'D'=\angle MP'C$. We will first prove that $\frac{AC}{AB}=\frac{BD'}{PD'}\cdot\frac{P'M}{BM}$. By the law of sines, $\frac{BD}{PD}=\frac{\sin\angle BP'D}{\sin\angle P'BD'}$ and $\frac{PM}{BM}=\frac{\sin\angle P'BD'}{\sin\angle BP'M}$. As $\angle BP'D'=\angle MP'C$ and $\angle BP'M=\angle D'P'C$, we have \[\frac{BD}{PD}\cdot\frac{PM}{BM}=\frac{\sin\angle MP'C}{\sin\angle P'BD'}\cdot\frac{\sin\angle P'BD'}{\sin\angle CP'D'}=\frac{\sin\angle MP'C}{\sin\angle D'P'C}=\frac{\sin\angle XP'A}{\sin\angle P'XA}=\frac{AX}{AP'}=\frac{AC}{AB}\] The last steps come from the similar triangles and parallel lines. Then using the law of sines on triangles $BAM$ and $ACE$ with $P'D' || AE$ \[\frac{CE}{BM}=\frac{AC}{AB}\cdot\frac{AE}{AM}=\frac{AC}{AB}\cdot\frac{P'D'}{P'M}=\frac{BD'}{PD'}\cdot\frac{P'M}{BM}\cdot\frac{P'D'}{P'M}=\frac{BD'}{BM}\] Therefore $CM=BD'$. But then $D=D'$ and $P=P'$, and we have the desired equality.

Here is my approach:Let $R$ be the intersection point of $BP$ and $AE$ and $S$ be the intersection point of $AE$ and $CP$.Now,from Thales theorem and using $BD=CE$ we get $BP/PR=CS/PS$ and since angles $SAC$ and $PAB$ are equal,it is obvious that $ABR$ is similar to $APC$,so we are finished.

I tried drawing a line through E,P such that AP is the symmedian, but couldn't do it.

The problem can be solved by reflecting! (Let $R(\triangle)$ denote the circumradius of a triangle). (I did NOT use trig bash, the sines are only used for technicalities). Let $M$ be the midpoint of $BC$ and $P', A'$ the reflections of $P, A$ across $M$. Notice $AE || PD || P'D$ so $AEP'$ are collinear. Similarly $A'DP$ are collinear. Let $P''$ be the reflection of $P$ across the midpoint of $AC$. Then $P'A', AP, CP''$ are equal and parallel segments and so $\angle CP''P = \angle CA'P' = \angle BAP = \angle CAP'$, so $P'', C, P', A$ form a cyclic quadrilateral. Thus, $R(AP'C)=R(AP''C)=R(APC)$. However, $R(ABP)=R(A'P'C)=\frac{P'C}{2\text{sin}(\angle P'A'C)} = \frac{P'C}{2\text{sin}(\angle P'AC)}= R(P'AC)=R(APC)$, which implies $\frac{AP}{\text{sin}(\angle ABP)} = \frac{AP}{\text{sin}(\angle ACP)}$. So $\angle ABP = \angle ACP$ or $180-\angle ACP$. However, the second choice would imply $APBC$ is cyclic, but it isn't even convex! Therefore, the first choice is correct, so $\boxed{\angle ABP=\angle ACP}$. (originally posted at: http://artofproblemsolving.com/community/c6h1106437_equiangular_line).

An easy synthetic solution. Note that $CE/CD$ =$CE/BE$=$[QCA]/[QAB]$=$CQ/CP$=$[QCA]/[PCA]$ this implies $[QAB]$ =$[PCA]$.This implies $AP.AC.SinCAP$ =$AQ.AB.SinBAQ$. Since $\angle CAP$ =$\angle BAQ$ we have $\triangle APB$ similar to $\triangle AQC$.Done.(here $Q$ is the intersection of $AE$ and $CP$.)

Let $A_1'$ be the point such that $ABA_1'P$ is a parallelogram and let $A'$ be the point such that $ABA'C$ is a parallelogram. First we claim that $BPA_1'A'$ is cyclic. This is pretty easy angle chasing; note that $\angle BA_1'P=\angle BAP$ and $\angle BA’P=\angle BA’D=180^{\circ}-\angle A’BD-\angle BDA’=180^{\circ}-\angle C-\angle AEC=\angle CAE$ so by isogonals, we get the cyclicity. Now, $\angle PBA=\angle BPA_1’$, and since $\triangle BA_1'A'\sim \triangle APC$, $\angle PCA=\angle A_1’A’B=\angle A_1'PB$ and we're done.

Lemma: Let $B, C, O$ be three fixed points in the plane. Then the locus of points $A$ for which $O$ lies on the nine-point circle of $\triangle ABC$ is the rectangular hyperbola $\mathcal{C}$ with center $O$ passing through $B, C.$ Proof: Let $H$ be the orthocenter of $\triangle ABC$ and let $D$ be the reflection of $H$ in $O.$ Since $H$ is the center of positive homothety that sends the nine-point circle of $\triangle ABC$ to the circumcircle, we deduce that $D \in \odot (ABC).$ Therefore, it is well-known that the midpoint $O$ of $\overline{HD}$ is the center of the rectangular hyperbola passing through $A, B, C, D.$ Hence, $A \in \mathcal{C}$, as desired. $\blacksquare$ Now, let $\tau$ be the perpendicular bisector of $\overline{BC}$, let $M$ be the midpoint of $\overline{BC}$, let $H$ be the orthocenter of $\triangle ABC$ and let $A', P', H'$ be the reflections of $A, P, H$ in $M.$ We will show that $M$ lies on the nine-point circle of $\triangle APB.$ Note that $DPEP'$ is a parallelogram because its diagonals bisect one another. Hence, $DP \parallel EP' \implies A, E, P'$ are collinear. Therefore, $\measuredangle BAP = \measuredangle P'AC = \measuredangle PA'B$, where the last step follows from the homothety $\mathcal{H}(M, -1).$ Thus, if $X, Y, Z$ are the midpoints of $\overline{AB}, \overline{BP}, \overline{PA}$, the homothety $\mathcal{H}\left(A, \tfrac{1}{2}\right)$ implies that $\measuredangle XAZ = \measuredangle ZMX.$ But recalling that $AXYZ$ is a parallelogram, it follows that $\measuredangle ZYX = \measuredangle ZMX$, implying that $M$ lies on the nine-point circle of $\triangle APB.$ Thus by the lemma, we deduce that $P$ lies on the rectangular hyperbola $\mathcal{C}$ with center $M$ passing through $A, B.$ By symmetry in $M$, we see that $C \in \mathcal{C}$, and therefore $\mathcal{C}$ is a circumhyperbola of $\triangle ABC.$ Then since $\mathcal{C}$ is rectangular, it follows that $H \in \mathcal{C}$, and by symmetry in $M$, we deduce $H' \in \mathcal{C}.$ But it is well-known that $H'$ is the antipode of $A$ WRT $\odot (ABC).$ Therefore, the fourth intersection $H'$ of $\mathcal{C}$ with $\odot (ABC)$ is the isogonal conjugate of the point at infinity on $\tau.$ Hence, $\mathcal{C}$ is the isogonal conjugate of $\tau.$ Thus, if $Q$ is the isogonal conjugate of $P$, we have $Q \in \tau \implies \angle QBC = \angle QCB \implies \angle PBA = \angle PCA$ as desired. $\square$

My synthetics solution: Let $\angle EAC=\angle PAB=\alpha$ $\angle PAE=\beta,\angle DPC=x,\angle PDC=y,\angle PBC=z.$ Then we find all angles in triangle. We must to show that $x+\beta+z=y.$ We know by Sine of Law $\frac{BD}{Sin(y-z)}=\frac{PD}{Sin(z)},\frac{PD}{Sin(x+y)}=\frac{PC}{Sin(y)}$ $\frac{PC}{Sin(\alpha+\beta)}=\frac{AC}{Sin(x+\beta)},\frac{AC}{Sin(x)}=\frac{FC}{Sin(\alpha)},$ $\frac{FC}{Sin(y)}=\frac{EC}{Sin(x)}=\frac{BD}{Sin(x)}.$ We can get easily $\frac{Sin(\alpha+\beta)\cdot Sin(x+y)\cdot Sin(y-z)}{Sin(\alpha)\cdot Sin(\beta+x)\cdot Sin(z)}=1.$ And we know from trig Ceva with respect point $P$ $\frac{Sin(\alpha+\beta)\cdot Sin(y-\alpha-\beta-z)\cdot Sin(x+y)}{Sin(\alpha)\cdot Sin(z)\cdot Sin(x-\alpha)}=1.$ Then we find easily $Cos(2\beta+x-y+\alpha+z)=Cos(x-y-\alpha+z).$ $1) 2\beta+x-y+\alpha+z=x-y-\alpha +z.$ This is false. $2)2\beta+x-y+\alpha+z=-x+y+\alpha-z,$ then we find $x+\beta+z=y.$ So we are done.

Hmm, nobody seems to have posted this solution yet, so I be posting it. Basically it seems like the problem is begging to be reflected over the midpoint of \(BC\), so we do just that. Let \(A', P'\) be the reflections of \(A, P\) over the midpoint of BC. It is easy to see that the problem now is equivalent to proving that \(P, P'\) are isogonal with respect to angle \(ABA'\) (or \(ACA'\), by symmetry). But then by the isogonality lemma (\(X and Y\) are isogonal with respect to angle \(BAC\) if and only if \(BX \cap CY\) and \(CX \cap BY\) are too), then we just have to prove that the point at infinity at \(AE\) is isogonal to the point at infinity at \(AP\), or that equivalently, the angle bisector of \(\angle DPA\) is parallel to the angle bisector of \(\angle ABA'\) or that, equivalently, the angle bisector of \(\angle BAC\) is perpendicular to that of \(\angle ABA'\) (since \(AP\) and \(AE\) are isogonal), which is easy to prove by simple angle chasing. I hope that that was easy to follow. I do think that it is quite easy to do so with a good enough diagram, no?

[asy][asy] unitsize(0.2inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); 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 = -12.918754138617931, xmax = 13.80470590754751, ymin = -21.249368309582554, ymax = 11.375522496777725; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-6.38,5.95)--(-8.48,-4.99), linewidth(2) + wrwrwr); draw((-8.48,-4.99)--(2.58,-4.87), linewidth(2) + wrwrwr); draw((2.58,-4.87)--(-6.38,5.95), linewidth(2) + wrwrwr); draw((-6.91293811346049,-3.911999697726297)--(-6.38,5.95), linewidth(2) + wrwrwr); draw((-6.38,5.95)--(0.3580915823374584,-4.894107505435761), linewidth(2) + wrwrwr); draw((-6.91293811346049,-3.911999697726297)--(-6.2580915823374585,-4.96589249456424), linewidth(2) + wrwrwr); /* dots and labels */ dot((-6.38,5.95),dotstyle); label("$A$", (-6.282428227153514,6.164447787775469), NE * labelscalefactor); dot((-8.48,-4.99),dotstyle); label("$B$", (-8.398035480808277,-4.769901281113881), NE * labelscalefactor); dot((2.58,-4.87),dotstyle); label("$C$", (2.6699308883119093,-4.636283980883054), NE * labelscalefactor); dot((-6.2580915823374585,-4.96589249456424),dotstyle); label("$D$", (-6.171080476961158,-4.747631731075409), NE * labelscalefactor); dot((0.3580915823374584,-4.894107505435761),linewidth(4pt) + dotstyle); label("$E$", (0.442975884464789,-4.725362181036938), NE * labelscalefactor); dot((-6.91293811346049,-3.911999697726297),linewidth(4pt) + dotstyle); label("$P$", (-6.816897428076823,-3.723232429305735), NE * labelscalefactor); dot((-3.0091873715640443,0.5251027458193853),linewidth(4pt) + dotstyle); label("$P'$", (-2.9197261713443625,0.7084080283500299), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We use barycentrics wrt $ABC$. Let $D=(0,m,n)$ and $E=(0,n,m)$ where $m+n=1$. Suppose $P=(x,y,z)$. We have that $PD\parallel AE$, so $[PDE]=[PDA]$. Therefore, \[\begin{vmatrix}x&y&z\\0&m&n\\0&n&m\end{vmatrix}=\begin{vmatrix}x&y&z\\0&m&n\\1&0&0\end{vmatrix},\]so \[x(m^2-n^2)=yn-zm,\]so since $m+n=1$, we have \[(m-n)x = ny - mz.\]Note this now also holds if $P=(x:y:z)$. However, we have that $AP$ isogonal to $AE$, so $P=(t:b^2/n:c^2/m)$, so \[(m-n)t = n\frac{b^2}{n}-m\frac{b^2}{m}=b^2-c^2,\]so \[P=\left(\frac{b^2-c^2}{m-n}:\frac{b^2}{n}:\frac{c^2}{m}\right).\]Note then that the isogonal conjugate $P'$ of $P$ has coordinates \[P' = \left(\frac{a^2(m-n)}{b^2-c^2}:n:m\right).\]It suffices to show that $\angle P'BC=\angle P'CB$, or that $P'$ is on the perpendicular bisector of $BC$. But this has equation $a^2(z-y)+x(c^2-b^2)=0$, so it is clear that $P'$ is on it, as desired.

Nice problem. My solution: Let $M$ be the midpoint of $BC$, and $P'$ be the reflection of $P$ in $M$. As $D$ and $E$ are isotomic points in $BC$, $M$ must be the midpoint of $DE$. But, $M$ is the midpoint of $PP'$ also, and so $PDP'E$ is a parallelogram. This gives $P'E \parallel DP \Rightarrow A,E,P'$ are collinear. Also $AP$ and $AP'$ are isogonal in $\angle BAC$. By the Converse of Parallelogram Isogonality Lemma, we get $\angle ABP=\angle ACP$. $\blacksquare$

Pick a point $Q$ on $\overrightarrow{AE}$ such that $\angle QCE = \angle BPD$ and $Q$, $A$ are on the opposite sides of each other. Then observe that $\angle QEC = \angle DEA = \angle PDB$, and $BD = CE$, so $\Delta EQC \cong \Delta DPB$. Since $PD = EQ$ and $PD || EQ$, from this easily follows that $PBCQ$ is a parralelogram. Now, let $X_{\infty} = BP \cap CQ$ and $Y_{\infty} = BQ \cap CP$. Since $(AP, AQ), (AB, AC)$ are isogonal w.r.t $(AB, AC)$ , by the isogonal line lemma, $(AX, AY)$ are isogonal too. But then $\angle PBA = \angle X_{\infty}AB = \angle Y_{\infty} AC = \angle PCA$, as desired.

By Steiner's definition of a conic, it is clear that $P$ moves on a conic through $A$ and $A'$, where $A'$ is the reflection of $A$ through the midpoint $M$of $BC$. $D=B$ and $D=C$ implies that $B$ and $C$ lies on the conic as well. $D=BC_{\infty}$ then yields that the intersection $T$ of the tangent at $A$ to the circumcircle $\Gamma$ of $ABC$ and the line parallel to $BC$ through $A'$ lies on the conic $c$ as well. Let the intersection of the tangent to $c$ at $A$ intersect $\Gamma$at $K$. We have $(A,K;B,C)$ $ \stackrel{\text{A}}{=} (T,A;B,C) \stackrel{\text{A'}}{=} (BC_{\infty},M;B,C)=-1$ implying that $AK$ is a symmedian in $\angle{BAC}$. By property $30$ in the famous handout by Luis Gonzàlez it is clear that $c$ is the hyperbola described in the problem and thus a rectangular hyperbola. By Theorem 4.1 in "Geometry of Conics" it is clear that $M$ lies on the Euler-circle of $ABP$. A homotethy with factor $2$ and center $B$ now yields that $A$,$P$,$C$ and $H$ are concyclic where $H$ is the orthocenter of $ABP$. To conclude notice that $\measuredangle{PBA}=\measuredangle{AHP}=\measuredangle{ACP}$

Employ barycentric coordinates. Let $P=(u,v,w)$ and $P^*=(\frac{a^2}{u} : \frac{b^2}{v} : \frac{c^2}{w})$ its isogonal conjugate WRT $ABC$. The desired condition is equivalent to $P^*$ lies on the perpendicular bisector of $BC$. Now, observe that $E=(0 : \frac{b^2}{v} : \frac{c^2}{w})= (0 : b^2w : c^2v)$. Analogously, $D= (0 : c^2v : b^2w)$. Then, since the point at the infinity along $AE$ is $(-(b^2w+c^2v) : b^2w : c^2v)$, and this point is collinear with $DP$, since $DP \parallel AE$. $$\implies 0= \begin{vmatrix}0&c^2v&b^2w\\u&v&w\\-(b^2w+c^2v)&b^2w&c^2v\end{vmatrix}$$ $\implies 0 = u(b^4w^2-c^4v^2)-(b^2w+c^2v)(c^2vw-b^2vw) \iff 0 = u(b^2w-c^2v)-(c^2vw-b^2vw) \iff \frac{u^2}{vw}(b^2w-c^2v)-u(c^2-b^2)=0$ This means that $(\frac{a^2}{u} : \frac{b^2}{v} : \frac{c^2}{w})=P^*$ lies on the perpendicular bisector of $BC$, so we are done. $\blacksquare$

Let $Q$ be the reflection of $P$ across the midpoint of $DE$. Then $PDQE$ is a parallelogram, and since $PD || AE$ by definition, $Q$ lies on $AE$. Therefore, $AP$ and $AQ$ are isogonal, so by the converse of the first isogonality lemma, we finish. $\square$

Same as many of the solutions above - posting for storage. Solved with L567. Let $M$ be the midpoint of $BC$ and let $F$ be the reflection of $P$ across $M$. Notice that by $BM = MC$ and $DM = EM$ that $PBFC$ and $PDFE$ are both parallelograms meaning that $F$ lies on $AE$ as $AE \parallel PD \parallel AF$. Then $$\angle FAC = \angle EAC = \angle PAB$$meaning that $AF$ and $AP$ are isogonal. We get that $$\angle PBA = \angle PCA$$by the Converse of the Parallelogram Isogonality Lemma. $\blacksquare$

Solution. Let $M$ be the midpoint of $BC$, and $X$ reflection of $P$ in $M$. $BD=CE\implies M$ is also the midpoint of $DE\implies PDXE$ is a parallelogram$\implies \overline{A-E-X}$ are collinear. But since $AP$ and $AX$ are isogonal w.r.t $\angle A$, by the Converse of the Parallelogram Isogonality Lemma, we have that $\angle PBA=\angle PCA$.$\blacksquare$

Construct a parallelogram $BPCQ$. Note that $\triangle BPD\cong \triangle CQE$, because $BP=CQ, BD=CE, \measuredangle DBP=\measuredangle ECQ$. So $\measuredangle BDP=\measuredangle CEQ$. Now $\measuredangle CEQ+\measuredangle CEA=\measuredangle BDP+\measuredangle CDP=0$, so we have $A, E, Q$ are collinear. Let $BP\cap CQ=X_{\infty}$ and $BQ\cap CP=Y_{\infty}$, where $X_{\infty}, Y_{\infty}$ are the point along $PB$ and $CQ$ respectively. It is given that $AP$ and $AE\equiv AQ$ are isogonal. So by the isogonal leama we have $AX_{\infty}, AY_{\infty}$ are isogonal. And therefore we get $\measuredangle PAB=\measuredangle ACP$ because $PB\parallel AX_{\infty}, PC\parallel AY_{\infty}$ $\square$

Let $P'$ be the reflection of $P$ in $M,$ the midpoint of $\overline{BC}.$ Notice $DPEP'$ is a parallelogram so $\overline{EP'}\parallel\overline{PD}\parallel\overline{AE}$ so $P'$ lies on $\overline{AE}.$ The converse of the Parallelogram Isogonality Lemma finishes. $\square$

Let $Q$ be on $AE$ such that $QB=QC$. We have $$\angle DQC+\angle BQA=\angle BQE+\angle BQA = 180^\circ$$So, $Q$ must have an isogonal conjugate in quadrilateral $ABDC$. Since $P,Q$ are isogonal in $\angle BAC$ and $\angle PDB = \angle QED = \angle QDE$ $P$ must be the isogonal conjugate. So, $$\angle PBA=\angle QBC=\angle QCB=\angle ACP$$as needed.

So cute!! Let $X = AC \cap PD$ and $Q = PX \cap AB$, then, let $\alpha = \angle EAC$, so $$ \angle CAE = \angle AXQ = \angle QAP \longrightarrow PA \parallel (AXQ) \rightarrow PA^2 = PQ\cdot PX \rightarrow $$$$\boxed{\frac{AP}{PX} = \frac{PQ}{AP}}$$ Also, see that if $$\angle PBA = \angle PCA \iff \triangle QPB ~ \triangle CPA \iff \frac{BQ}{QP} = \frac{AC}{AP} \iff \frac{BQ}{AC} = \frac{QP}{AP} \iff$$$$\boxed{\frac{BQ}{AC} = \frac{AP}{PX}}$$ But, by law of sines in $\triangle BQD$ $$\frac{BQ}{\sin{\angle BDQ}} = \frac{BD}{\sin{\angle BQD}}$$And $\angle BDQ = \angle BEA = C + \alpha$ where $C = \angle BCA$, and $\angle BQD = \angle BAE = A - \alpha$ where $A = \angle BAC$, so $$BQ = BD \cdot \frac{\sin (C + \alpha)}{\sin (A - \alpha)}$$ Also, by law of sines in $\triangle AEC$ $$\frac{EC}{\sin \angle EAC} = \frac{AC}{\sin \angle AEC}$$ But $\angle EAC = \alpha$ and $\angle AEC = 180 - C - \alpha$, so $\sin \angle AEC = \sin (C + \alpha)$, so $$AC = EC \cdot \frac{\sin (C+\alpha)}{\sin \alpha}$$ Then, we have $$\frac{BQ}{AC} = \frac{BD \cdot \frac{\sin (C + \alpha)}{\sin (A - \alpha)}}{EC \cdot \frac{\sin (C+\alpha)}{\sin \alpha}} = \frac{\sin \alpha}{\sin (A - \alpha)} \longrightarrow$$$$\boxed{\frac{BQ}{AC} = \frac{\sin \alpha}{\sin (A - \alpha)}}$$ For the right side, we have, by law of sines in $\triangle AXP$ $$\frac{AP}{PX} = \frac{\sin \angle PXA}{\sin \angle XAP}$$ And $\angle PXA = \alpha$ and $\angle PAX = 180 - (A-\alpha)$, then $\sin \angle PAX = \sin (A-\alpha)$, so $$\boxed{\frac{AP}{PX} = \frac{\sin \alpha}{\sin (A-\alpha)}}$$ And we're done$_{_\blacksquare}$

We invoke barycentric coordinates with reference triangle $\triangle ABC$. We let $d = (0,d,1-d)$ and $e = (0,1-d,d)$. We can see that the line parallel to $AE$ through $D$ hits $AB$ at $(1-d:2d-1:0)$. This means that the line $DP$ is described by $(2d-1)x - (1-d)y + dz = 0$. We can also see that $AP$ is isogonal to $AE$. This means it is paramtrized by $\left(t:\frac{b^2}{1-d}:\frac{c^2}{d} \right)$. Thus, we can see that we have $P = \left(\frac{b^2-c^2}{2d-1}:\frac{b^2}{a-d}:\frac{c^2}{d}\right)$. The condition is equivalent to the isogonal conjugate of $P$ being on the perpendiculat bisector of $P$. This conjugate is given by $\left(\frac{a^2(2d-1)}{b^2-c^2}:1-d:d\right)$. We can see that this satisfies the perpendilcar bisector formula ($0 = a^2(z-y) + x(c^2-b^2)$). $\blacksquare$

Let $F=AP\cap BC$ and let $P'$ be the isogonal conjugate of $P$ with respect to $ABC$. We will of course use barycentric coordinates with respect to $ABC$. Set $D=(0,d,1-d)$, then $E=(0,1-d,d)$ and therefore $F=\left (0::\dfrac{b^2}{1-d}:\dfrac{c^2}{d}\right )$. We also have $P_{\infty AE}=(-1:1-d:d)$, so the line parallel to $AE$ through $D$ has equation$$0=\begin{vmatrix}x&y&z\\0&d&1-d\\-1&1-d&d\end{vmatrix}=(2d-1)x+(d-1)y+dz.$$Since $P$ in $AF$, we have $P=\left (t:\dfrac{b^2}{1-d}:\dfrac{c^2}{d}\right )$ for some $t\in \mathbb{R}$, so $P=\left (\dfrac{c^2-b^2}{1-2d}:\dfrac{b^2}{1-d}:\dfrac{c^2}{d}\right )$. Therefore $P'=\left (\dfrac{a^2}{c^2-b^2}(1-2d):1-d:d\right )$. The perpendicular bisector of $BC$ has equation $(b^2-c^2)x+a^2(y-z)=0$, hence $P'$ is in the perpendicular bisector of $BC$. We conclude that$$\angle PBA=\angle P'BC=\angle P'CB=\angle PCA.$$

Amazing Problem!! We consider $Q$ as the isogonal conjugate of $P$ in $\triangle ABC,$ and we wish to show that $QB=QC.$ Now let $X\in BC$ such that $XA\parallel PB$ and $Y\in BC$ such that $YA\parallel CP.$ By the isogonal conjugate condition we get that $QB$ adn $QC$ are tangent to $(AXB)$ adn $(YAC)$ respectively. So $Pow_Q(AXB)=QB^2$ and $Pow_Q(AYC)=QC^2,$ so if we show that $Q$ lies on the radical axis of both of these circumferences we will be done. Clearly $A$ is in the radical axis, so we sill show $E$ also is. For this notice that we want to show that $$EB\cdot EX=EC\cdot EY\iff \frac{EX}{EY}=\frac{EC}{EB}.$$But now notice that clearly $\triangle BPC\sim \triangle AXY,$ and in this similarity $D\to E$ because of $AE\parallel DP,$ so $$\frac{EX}{EY}=\frac{DB}{DC},$$but $DB=CE, DC=EB$ implies that $\frac{DB}{DC}=\frac{EC}{EB},$ which ends the problem.

NO SHOT THIS WORKS WHATTTT We prove the converse; suppose $X$ and $Y$ lie on $\overline{AC}$ and $\overline{AB}$ respectively, such that $BCXY$ is cyclic. Then, if we let $P$ be the intersection of diagonals $\overline{CY}$ and $\overline{BX}$, let $E$ be the point on $\overline{BC}$ such that $\angle PAB = \angle EAC$ and let $D$ be the point on $\overline{BC}$ such that $\overline{PD} \parallel \overline{AE}$, we will show that $BD = EC$. Proof: stretch about the bisector of $\angle BAC$ until $\angle BYC = \angle BXC = 90^{\circ}$. (Stretching about this specific line preserves the concyclicity of $BYXC$.) So, in this special case, $P$ is the orthocenter of $\triangle ABC$. In this case, line $AE$ is a diameter of $(ABC)$, so line $HD$ passes through the reflection of $A$ across the midpoint of $\overline{BC}$. By symmetry, $BD = EC$. The converse is easy to handle with phantom points or whatever.

Solved with swynca. Let $PBLC$ be a parallelogram. $PL$ bisects $BC$ so it also bisects $DE$. Since $MP=ML$ and $MD=ME$, we see that $A,E,L$ are collinear. DDIT at quadrilateral $PBLC$ gives $(AP,AL),(AB,AC),(A (PB)_{\infty},A(PC)_{\infty})$ is an involution which must be reflection over the angle bisector of $\measuredangle CAB$. Thus, $\measuredangle ABP=\measuredangle PCA$ as desired.$\blacksquare$