It's easy to see by monotonicty that there is only one point satisfying $\angle APD=\angle BPC$. Thus, we may redefine $P=ME\cap AB$. We'll show that $\angle APD=\angle BPC$. Since $AD^2+BC^2=AB^2$, we may choose $X\in(AB)$ so that $AX=AD$ and $BX=BC$. The first claim is that $P$ is the foot from $X$ to $AB$. Redefine $P$ in this manner, we'll show that $P,E,M$ collinear. Note that \[\frac{AP}{BP}=\frac{AX\cos\angle XAB}{BX\sin\angle XAB}=\frac{AX^2}{BX^2}.\]But \[\frac{AX}{BX}=\frac{AD}{BC}=\frac{2R\sin\angle ABD}{2R\sin\angle BAC}=\frac{AE}{BE},\]where the last step is law of sines on $AEB$. Thus, \[\frac{AP}{BP}=\frac{AE^2}{BE^2},\]so $EP$ is the $E$-symmedian of $AEB$. There are many ways to see that this implies that $P,E,M$ collinear, but perhaps the easiest is to invert at $E$. Let $E'\in(AEB)$ so that $(EE';AB)=-1$ (read: the harmonic conjugate of $E$ in $AB$ in the complex projective line). Clearly, $E,P,E'$ collinear. Let $\phi$ be inversion at $E$ swapping $(A,C)$ and $(B,D)$. Note that $\phi((ABE))=CD$, and \[-1=\phi((EE';AB))=(\phi(E')\infty,CD),\]so $\phi(E')=M$, so $E',E,M$ collinear, so $E,P,M$ collinear, as desired. We'll now show that $\angle APD=\angle BPC$. To do this, let $D'=DP\cap(ABCD)$ and $C'=CP\cap(ABCD)$. We claim that $AD=AD'$ and $BC=BC'$. We'll show the former, and the latter follows by symmetry. Let $\omega=(ABCD)$ and $\omega_A$ the circle centered at $A$ with radius $AD$. The power of $P$ with respect to $\omega_A$ is \[AP^2-AD^2=AP^2-AX^2=-PX^2,\]and the power of $P$ with respect to $\omega$ is $-|PA|\cdot|PB|=-PX^2$, so $P$ is on the radical axis of $\omega_A$ and $\omega$. Thus, the radical axis is $DP$, so $D'$ is on the radical axis. But since $D'\in\omega$, we must also then have that $D'\in\omega_A$, so $AD=AD'$, as desired. The problem is clearly equivalent now to $AB$ being the external angle bisector of $CC'$ and $DD'$. But $DD'\parallel\ell_A$ (the tangent to $\omega$ at $A$), and $CC'\parallel\ell_B$ (the tangent to $\omega$ at $B$), so the external bisector of $CC',DD'$ is parallel to the external bisector of $\ell_A$ and $\ell_B$. But the external bisector of those two tangents is clearly parallel to $AB$, and as $AB,CC',DD'$ are concurrent, it's clear that $AB$ is the external bisector of $CC'$ and $DD'$. $\blacksquare$

If $M$ is the midpoint of $CD$, let $Q$ be where $EM$ and $AB$ meet. The goal is to show that $P=Q$. Note that since $P$ is unique (try moving $P$ along segment $AB$), it suffices to show $\angle{AQD}=\angle{BQC}$. Observe that since $M,E,Q$ collinear, $EQ$ and $EM$ are corresponding isogonal lines in similar triangles $ECD$ and $EAB$. So $EQ$ is a symmedian and thus $\frac{AQ}{QB}=\frac{AE^2}{EB^2}$. But by the Law of Sines, \[\frac{AE}{EB}=\frac{\sin\angle{ABE}}{\sin\angle{EAB}}=\frac{\sin\angle{ABD}}{\sin\angle{CAB}}=\frac{AD}{BC}\]so $\frac{AQ}{QB}=\frac{AD^2}{BC^2}$. Since $AQ+QB=AB$, it follows that $AQ=\frac{AD^2}{AB}$. Then $AD$ is tangent to $(BQD)$ so $\angle{QDA}=\angle{ABD}$. Then \[\angle{AQD}=\pi-\angle{QDA}-\angle{DAQ}=\pi-\angle{ABD}-\angle{DAB}=\angle{BDA}.\]Similarly, $\angle{BQC}=\angle{ACB}$. But $\angle{BDA}=\angle{ACB}$ by cyclic quad, so $\angle{AQD}=\angle{BQC}$ as desired.

Let $\omega_A$ denote the circle with center $A$ and radius $AD$. Let $\omega_B$ denote the circle with center $B$ and radius $BC$. We claim that these two circles necessarily intersect. Proof: Assume for the sake of contradiction that they don't. Then $AD + BC < BC \implies AD^2 + BC^2 + 2AD \cdot BC < AB^2 \implies 2AD \cdot BC < 0$ which is a contradiction. Hence, they intersect at a point, say $X$. Then by the Pythagorean theorem, $\angle AXB = 90^\circ$ so $\omega_A$ and $\omega_B$ are orthogonal. Let $P'$ be the projection of $X$ onto $\overline{AB}$. Upon inverting about $\omega_A$ we see that $P'D$ and $AC$ meet on $\omega_B$ at $M$ and by symmetry, $P'C$ and $BD$ meet on $\omega_A$ at $N$. Since $B$ is mapped to $P'$ and $M$ is mapped to $C$, $AXYB$ is a cyclic quadrilateral to $\angle AND = \angle BMC$ hence $\angle AP'D = \angle BP'C$ where we have used the fact that quadrilaterals $ABP'N$ and $BCP'M$ are cyclic. Moreover, since $XY$ and $CD$ are both antiparallel to $AB$ they are parallel so by Ceva's theorem $P'E$ passes through the midpoint of $BC$. It remains to show that $P' = P$. Clearly, $P'$ satisfies the angle condition for $P$ so we need to prove that $P$ is unique. To do this, we show a unique construction for $P$. Let the perpendicular bisector of $CD$ meet $AB$ at $T$ and then the circumcircle of $\triangle TCD$ meet $AB$ again at $P$.

Here was the original solution I found, with some suggestions from Evan to make it simpler.

SYMMEDIAN!!!!! :rage:

[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(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -30.10774683338543, xmax = 20.292988546313076, ymin = -14.748881041259867, ymax = 20.17411985764882; /* image dimensions */ pen xfqqff = rgb(0.4980392156862745,0,1); pen qqffff = rgb(0,1,1); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((-5.21,0.62), 6.5651208575055), linewidth(0.4) + xfqqff); draw(circle((-11.773257257435112,0.4635837400426075), 7.761816012333698), linewidth(0.8) + qqffff); draw(circle((1.3532572574351138,0.7764162599573926), 10.590441902267118), linewidth(0.8) + qqffff); draw((-7.335393968675698,6.8315627946250945)--(-7.037079805882728,-5.685758578996604), linewidth(0.4) + wrwrwr); draw(circle((-5.371985513785333,7.41694425145151), 9.451216976541685), linewidth(0.8) + red); draw((-14.821393813947864,7.601836421348473)--(3.2945646862479396,11.187409735708462), linewidth(0.8) + wrwrwr); draw((-14.821393813947864,7.601836421348473)--(-11.773257257435112,0.4635837400426075), linewidth(0.8) + wrwrwr); draw((1.3532572574351138,0.7764162599573926)--(3.2945646862479396,11.187409735708462), linewidth(0.8) + wrwrwr); draw((-14.821393813947864,7.601836421348473)--(-4.407768443334378,-1.984959161087927), linewidth(0.8) + wrwrwr); draw((-14.821393813947864,7.601836421348473)--(1.3532572574351138,0.7764162599573926), linewidth(0.8) + wrwrwr); draw((3.2945646862479396,11.187409735708462)--(-11.773257257435112,0.4635837400426075), linewidth(0.8) + wrwrwr); draw((-11.773257257435112,0.4635837400426075)--(1.3532572574351138,0.7764162599573926), linewidth(0.8) + wrwrwr); draw((-9.032292019977847,-1.2967035850400774)--(3.2945646862479396,11.187409735708462), linewidth(0.8) + wrwrwr); draw((-7.186236887279213,0.5729021078142449)--(-5.763414563849959,9.394623078528468), linewidth(0.8) + wrwrwr); draw((-9.119258911737576,2.352444423319879)--(-4.532772070486577,3.2602178002833475), linewidth(0.8) + wrwrwr); /* dots and labels */ dot((1.3532572574351138,0.7764162599573926),linewidth(4pt) + dotstyle); label("$B$", (1.8941041775996061,0.03126540659287662), NE * labelscalefactor); dot((-11.773257257435112,0.4635837400426075),linewidth(4pt) + dotstyle); label("$A$", (-12.624446757901781,-0.05593309752424867), NE * labelscalefactor); dot((-7.335393968675698,6.8315627946250945),linewidth(4pt) + dotstyle); label("$X$", (-7.17454025058144,7.18154274419715), NE * labelscalefactor); dot((-7.186236887279213,0.5729021078142449),linewidth(4pt) + dotstyle); label("$P$", (-7.000143242347189,-0.6663226263441258), NE * labelscalefactor); dot((3.2945646862479396,11.187409735708462),linewidth(4pt) + dotstyle); label("$C$", (3.463677251707864,11.541467950053415), NE * labelscalefactor); dot((-14.821393813947864,7.601836421348473),linewidth(4pt) + dotstyle); label("$D$", (-15.589195897884046,7.8791307771341526), NE * labelscalefactor); dot((-4.407768443334378,-1.984959161087927),linewidth(4pt) + dotstyle); label("$G$", (-4.166191858540612,-3.0206822375065085), NE * labelscalefactor); dot((-9.032292019977847,-1.2967035850400774),linewidth(4pt) + dotstyle); label("$H$", (-9.790495374095203,-2.06149869221813), NE * labelscalefactor); dot((-6.611342353611276,4.137338144623156),linewidth(4pt) + dotstyle); label("$E$", (-7.218139502640003,4.739984628917642), NE * labelscalefactor); dot((-9.119258911737576,2.352444423319879),linewidth(4pt) + dotstyle); label("$L$", (-9.659697617919516,3.0396137986336993), NE * labelscalefactor); dot((-4.532772070486577,3.2602178002833475),linewidth(4pt) + dotstyle); label("$N$", (-3.8173978420721104,2.9960145465751364), NE * labelscalefactor); dot((-7.037079805882728,-5.685758578996604),linewidth(4pt) + dotstyle); label("$Y$", (-7.305338006757128,-6.6830194104257705), NE * labelscalefactor); dot((-5.763414563849959,9.394623078528468),linewidth(4pt) + dotstyle); label("$M$", (-5.604967176473182,9.753898615652346), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $\Gamma_1$ and $\Gamma_2$ be the circle centered at $A, B$ with radius $AD, BC$ respectively. Let $X,Y$ be the intersection of these two circles. Notice that $X, Y$ lies on the circle with diameter $AB$ by Pythagorean theorem. Let $P'$ be the intersection of $AB$ and $XY.$ We claim that $P=P'.$ First, notice that $XY$ is the radical axis of circle $\Gamma_1$ and $\Gamma_2.$ Since $AB$ is the line connecting the center, we have $XY\perp AB.$ Let $G=DP'\cap \Gamma_1$ and $H=CP'\cap\Gamma_2.$ By the power of a point, $DP'\times GP'=AX^2 -AP'^2=P'X^2=AP'\times BP'$ that $(A,D,B,G)$ cyclic. Similarly, we have $(A,C,B,H)$ cyclic. Therfore, points $G,H$ lies on the circle $ABCD.$ Now, $AD^2=AX^2 =AP'\times AB,$ that $\triangle ABD$ and $\triangle ADP'$ is similar. Thus, $\angle AP'D=\angle ADB.$ In the same way, we get $\angle CP'B=\angle ACB.$ Therefore, $\angle AP'D=\angle BP'C.$ Such point on the line $AB$ is unique(requires proof) so we have $P=P'.$ This implies that when we extend $DP$ and take the intersection with $ABCD$ which is $G$ in this case, we have $AG=AD$ and same for $H.$ Let $L,N$ be the intersection of $CA,DP$ and $BD,CP$ respectively. Notice that $AD=AG$ and $BC=BH.$ By angle chase, we have $\angle LAB=\angle HAB$ and $\angle NBA=\angle GBA.$ Therefore, $L,N$ is symmetric with $H,G$ respect to line $AB.$ Since $(A,B,G,H)$ is cyclic,$(A,B,N,L)$ is cyclic as well. Then, $\angle DCA=\angle DBA=\angle CLN$ that $CD\parallel LN.$ Let $M$ be the intersection of $PE$ and $CD.$ By Ceva's theorem, we have that $CM=DM.$

how any points would I get if my solution would be complete if I proved PE was symmedian, but I didnt prove it

wutt ignore

Let $M$ be the midpoint of $CD$ and define $P' = ME \cap AB$; by monotonicity, it suffices to show that $\angle AP'D = \angle BP'C$. Since $\triangle ABE \sim \triangle DCE$, $EP'$ is the symmedian of $\triangle ABE$. Therefore \[AP' = \frac{AB \cdot AE^2}{AE^2 + EB^2} = \frac{AB \cdot AD^2}{AD^2 + BC^2}.\]The last step is since $\triangle EAD \sim \triangle EBC$. By the length condition, we then have $AP' = \frac{AD^2}{AB}$. But then \[\frac{AP'}{AD} = \frac{AD}{AB},\]so $\triangle AP'D \sim \triangle ADB$. Similarly, $\triangle BP'C \sim \triangle BCA$. So \[\angle AP'D = \angle ADB = \angle ACB = \angle BP'C.\]

Let $\Omega$ be the circumcircle of $ABCD$, let $\Gamma$ be the circle with diameter $AB$, and let $\omega_A$, $\omega_B$ be the circles centered at $A$, $B$ with radii $AD$, $BC$, respectively. [asy][asy] size(9cm); defaultpen(fontsize(10pt)); pen pri=blue; pen sec=lightblue; pen tri=purple+pink; pen qua=purple; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair O,A,B,C,Y,D,P,EE,M,Z,T,U,V; O=(0,0); A=dir(148); B=reflect( (0,0),(0,1))*A; C=dir(-70); path omega=circle(O,1); path gamma=circle((A+B)/2, length(A-B)/2); path wb=circle(B,length(C-B)); Y=intersectionpoint(gamma,wb); path wa=circle(A,length(Y-A)); D=intersectionpoints(omega,wa)[1]; P=extension(A,B,C,reflect(A,B)*D); EE=extension(A,C,B,D); M=(C+D)/2; Z=reflect(A,B)*Y; T=2*A*B/(A+B); U=2*foot(O,D,P)-D; V=2*foot(O,C,P)-C; filldraw(gamma,tfil,tri+dashed); draw(A--T--B,qua); draw(D--U,tri); draw(C--V,tri); filldraw(wa,sfil,sec+dashed); filldraw(wb,sfil,sec+dashed); draw(A--C,sec); draw(B--D,sec); filldraw(omega,fil,pri); draw(P--M,pri+Dotted); draw(A--B--C--D--cycle,pri); dot("$A$",A,dir(165)); dot("$B$",B,dir(15)); dot("$C$",C,S); dot("$D$",D,SW); dot("$P$",P,N); dot("$E$",EE,dir(60)); dot("$Y$",Y,dir(105)); dot("$Z$",Z,dir(265)); dot("$T$",T,N); dot("$U$",U,dir(75)); dot("$V$",V,dir(150)); dot(M); [/asy][/asy] The condition $AD^2+BC^2=AB^2$ means $\omega_A$, $\omega_B$ are orthogonal, so their intersection points $Y$, $Z$ lie on $\Gamma$. Let $P$ be the midpoint of $\overline{YZ}$. Then $P$ lies on $\overline{AB}$, the radical axis of $\Omega$, $\Gamma$, so $P$ is the common radical center of $\Omega$, $\Gamma$, $\omega_A$, $\omega_B$. Let the tangents to $\Omega$ at $A$, $B$ meet at $T$, and let $\overline{DP}$, $\overline{CP}$ meet $\Omega$ again at $U$, $V$. Since $\overline{DU}$ is the radical axis of $\Omega$, $\omega_A$, we have $\overline{AT}\parallel\overline{DU}$. Similarly $\overline{BT}\parallel\overline{CV}$, so $\measuredangle APD=\measuredangle BAT=\measuredangle TBA=\measuredangle CPB$. By monotonicity, $P$ is the point described in the problem statement. Finally we have \[\frac{PA}{PB}=\left(\frac{YA}{YB}\right)^2=\left(\frac{AD}{BC}\right)^2=\left(\frac{EA}{EB}\right)^2,\]so $\overline{EP}$ is the $E$-symmedian of $\triangle EAB$, which bisects $\overline{CD}$.

Here is a solution by inversion. By hypothesis, the circle $\omega_a$ centered at $A$ with radius $AD$ is orthogonal to the circle $\omega_b$ centered at $B$ with radius $BC$. For brevity, we let $\mathbf{I}_a$ and $\mathbf{I}_b$ denote inversion with respect to $\omega_a$ and $\omega_b$. We let $P$ denote the intersection of $\overline{AB}$ with the radical axis of $\omega_a$ and $\omega_b$; hence $P = \mathbf{I}_a(B) = \mathbf{I}_b(A)$. This already implies that $\measuredangle DPA \overset{\mathbf{I}_a}{=} \measuredangle ADB = \measuredangle ACB \overset{\mathbf{I}_b}{=} \measuredangle BPC$ so $P$ satisfies the angle condition. [asy][asy] size(10cm); pair A = dir(200); pair B = -conj(A); pair D = dir(140); pair O = origin; pair P = extension(D, foot(D, A, O), A, B); pair Z = 100*foot(P,O,B)-99*P; pair C = IP(P--Z, unitcircle); filldraw(unitcircle, invisible, deepcyan); draw(A--B--C--D--cycle, deepcyan); filldraw(CP(A, D), invisible, red); filldraw(CP(B, C), invisible, orange); pair X = IP(CP(A, D), CP(B, C)); pair Y = OP(CP(A, D), CP(B, C)); draw(X--Y, yellow); draw(A--C, deepcyan+dotted); draw(B--D, deepcyan+dotted); pair K = extension(D, P, A, C); pair L = extension(C, P, D, B); draw(D--P, red+dashed); draw(C--P, orange+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$D$", D, dir(D)); dot("$P$", P, dir(P)); dot("$C$", C, dir(C)); dot(X); dot(Y); dot("$K$", K, dir(K)); dot("$L$", L, dir(L)); /* TSQ Source: !size(12cm); A = dir 200 B = -conj(A) D = dir 140 O := origin P = extension D foot D A O A B Z := 100*foot(P,O,B)-99*P C = IP P--Z unitcircle unitcircle 0.1 palecyan / deepcyan A--B--C--D--cycle deepcyan CP A D 0.1 lightred / red CP B C 0.1 orange / orange X .= IP CP A D CP B C Y .= OP CP A D CP B C X--Y yellow A--C deepcyan dotted B--D deepcyan dotted K = extension D P A C L = extension C P D B D--P red dashed C--P orange dashed */ [/asy][/asy] Claim: The point $K = \mathbf{I}_a(C)$ lies on $\omega_b$ and $\overline{DP}$. Similarly $L = \mathbf{I}_b(D)$ lies on $\omega_a$ and $\overline{CP}$. Proof. The first assertion follows from the fact that $\omega_b$ is orthogonal to $\omega_a$. For the other, since $(BCD)$ passes through $A$, it follows $P = \mathbf{I}_a(B)$, $K = \mathbf{I}_a(C)$, and $D = \mathbf{I}_a(D)$ are collinear. $\blacksquare$ Finally, since $C$, $L$, $P$ are collinear, we get $A$ is concyclic with $K = \mathbf{I}_a(C)$, $L = \mathbf{I}_a(L)$, $B = \mathbf{I}_a(B)$, i.e.\ that $AKLB$ is cyclic. So $\overline{KL} \parallel \overline{CD}$ by Reim's theorem, and hence $\overline{PE}$ bisects $\overline{CD}$ by Ceva's theorem.

Somewhat of a sketch: Let the circles \(\omega_1\) and \(\omega_2\) be the circles centered at \(A,B\) with radii \(AD,BC\), respectively. Let the circles meet at \(X,Y\) and let \(P'=XY \cap BC\). Let \(T\) be the reflection of \(E\) over \(M\). Note that by similar triangles, \(AP' \cdot AB= AX^2=AB^2\) so \(\triangle P'AD \sim \triangle DAB\). Then we see that \(\angle AP'D=\angle ADB=\angle ACB = \angle BP'D\) because of a similar argument on B and C, so \(P'\) satisfies the given conditions. It's not hard to prove that \(P\) is unique; just note that \(\angle AQD-\angle BQC\) monotonically varies as Q varies on AB. Then note that \(\angle DCT=\angle CDE=\angle CDB=\angle CAB=\angle PCB\), and similarly \(\angle CDT = \angle PDA\). We finish by Jacobi's on triangle \(PCD\).

By Trig Ceva in $\triangle PCD,$ it suffices to show \[\frac{\sin PCA}{\sin ACD}\cdot\frac{\sin BDC}{\sin PDB}\cdot\frac{PC}{PD}=1.\]But $\frac{\sin PCA}{\sin PDB}=\frac{AP}{BP}\cdot\frac{BD}{AC}$ by Law of Sines, and $\frac{\sin BDC}{\sin ACD}=\frac{BC}{AD}.$ Law of Sines again tells us $\frac{PC}{PD}=\frac{BC}{AD}\cdot\frac{BD}{AC},$ and putting it all together it suffices to show $\frac{AP}{BP}=\frac{AD^2}{BC^2}.$ Adding $1$ to both sides and using the metric condition, it suffices to show $BP\cdot BA=BC^2,$ or equivalently $\angle DPA = \angle ACB.$ For this, use complex numbers. Let $(ABCD)$ be the unit circle with $A=a,B=\overline{a},P=p,C=p+r\omega,D=p+s\overline{\omega}$ for $r,s$ real and $\omega$ on the unit circle. The metric condition tells us \[(a-d)(\overline{a}-\overline{d})+(\overline{a}-c)(a-c)=-(a-\overline{a})^2\]\[\iff -a(\overline{p}+s\omega+p+r\omega)-\overline{a}(p+r\overline{\omega}+\overline{p}+s\overline{\omega})=-(a+\overline{a})^2.\]But $a+\overline{a}=p+\overline{p}$ since $P\in AB,$ so this rearranges to $(r+s)(a\omega+\overline{a\omega})=0,$ meaning that $\cos(\text{arg}(a\omega))=0.$ But $\text{arg}(a)=\tfrac{\angle AOB}{2}$ and $\text{arg}(\omega)=\tfrac{\pi}{2}-\angle DPA,$ so $\angle ACB=\tfrac{\angle AOB}{2}=\angle DPA,$ as desired.

When you coord bash the hardest question on the Day 1 test (JMO)... and get a full solution with conclusion

ItsAmeYushi wrote: When you coord bash the hardest question on the Day 1 test (JMO)... and get a full solution with conclusion I would like to see this.

I assume "side $\overline{AB}$" refers to the segment and $ABCD$ is convex. I claim that there is at most one point $P$ on segment $\overline{AB}$ with $\angle APD=\angle BPC$. Suppose we had two points $P$ and $P'$. By law of sines on $\triangle APD,\triangle BPC$ and $\triangle AP'D,\triangle BP'C$, it follows that $\frac{PD}{PC}=\frac{P'D}{P'C}$, hence $P'$ lies on the $P$-Apollonian circle of $\triangle CDP$. But we can actually identify the other intersection of this circle with $\overline{AB}$: it's just $\overline{AB} \cap \overline{CD}$, which doesn't lie on segment $\overline{AB}$. Therefore we can redefine $P$ to be the point on $\overline{AB}$ such that $\overline{EP}$ is a symmedian in $\triangle ABE$. Because $\triangle EAB \sim \triangle ECD$ this implies that $\overline{EP}$ bisects $\overline{CD}$. I will show that $\angle APD=\angle BPC$. By the symmedian condition and $\triangle EAD \sim \triangle EBC$, we have $$\frac{AP}{BP}=\frac{AE^2}{BE^2}=\frac{AD^2}{BC^2} \implies BC^2=\frac{BP}{AD}AD^2,$$so by using the given length condition, $$AD^2+\frac{BP}{AP}AD^2=AB^2 \implies \frac{AB}{AP}AD^2=AB^2 \implies AD^2=AP\cdot AB,$$hence $\overline{AD}$ is tangent to $(BDP)$. Likewise, $\overline{BC}$ is tangent to $(ACP)$. Now let $\overline{PC}$ intersect $(BDP)$ again at $C'$ and $\overline{PD}$ intersect $(ACP)$ again at $D'$. We wish to prove $\widehat{AD'}=\widehat{C'B}$ (the LHS is an arc in $(APC)$, the RHS is an arc in $(BDC)$). This follows from $\measuredangle ACB=\measuredangle ADB \implies \widehat{AC}=\widehat{DB}$ and $\measuredangle CPD'=\measuredangle CPD' \implies \widehat{CD'}=\widehat{C'D}$. $\blacksquare$

Let $\omega$ be the circle with diameters $\overline{AB}$, and $Q$ a point on this circle that is neither $A$ nor $B$. Define $\omega_A$ to be the circle with radius $\overline{AQ}$ centered on $A$, and $\omega_B$ to be the circle with radius $\overline{BQ}$ centered on $B$. We now define $\Omega$ to be the circumcircle of cyclic quadrilateral $ABCD$. It is evident then that $C = \omega_B \cap \Omega$ and $D = \omega_A \cap \Omega$. Claim. Let $P'$ be the foot of the perpendicular from $Q$ to $\overline{AB}$. Then $P' = P$. Proof. We invert around $\omega_B$. $P'$ gets sent to $A$ and $C$ gets sent to itself. By the inversion angle condition, we know that $\angle BP'C = \angle ACB$. Analogously, if we invert around $\omega_A$, we get that $\angle AP'D = \angle ADB$. But since $ABCD$ is cyclic, $\angle ACB = \angle ADB$, which proves that $\angle AP'D = \angle BP'C$. We now introduce points $C'$ and $D'$ which are simply reflections of their respective points across $\overline{AB}$. Because $\angle BPC = \angle APD$, we know that $C,P,D'$ are collinear and $C',P,D$ are also collinear. Now, let $X$ be the intersection of $\omega_A$ and line $BD$. Claim. $C,X,P,D$ are collinear. Proof. We invert around $\omega_B$ once more. We know that $C$ gets sent to itself, $P$ gets sent to $A$, and $X$ gets sent to $D$ because circles $\omega_A$ and $\omega_B$ are orthogonal. But, $ACBD$ are cyclic, so this implies that $X$ lines on $\overline{CPD}$. Claim. $X$ lies on the circumcircle of $BAC'D'$. Proof. Invert around $\omega_B$. It is evident that our claim is true if and only if $C',P,D$ are collinear. But this collinearity has already been proven. It follows analogously that the circumcircle of $BAC'D$, $\overline{AC}$, $\overline{C'PD}$, and $\omega_B$ all concur at a point $Y$. Now we can finish. $CC'DD'$ is an isosceles trapezoid. $C'D'XY$ are also cyclic, so $\overline{XY} \parallel \overline{BC}$. Now, it is clear that $E = \overline{CY} \cap \overline{DX}$ because of all the collinearities we showed above. We can now use Ceva's theorem to conclude that $\overline{PE}$ passes through the midpoint of $\overline{CD}$. $\blacksquare$

Really hope this is right cuz i didn't bother doing the last part. Let $Q$ be a point such that $(DQB)$ is tangent to $AD$. We claim that $Q = P$. Note that \[AD^2 = AQ\cdot AB\implies AB^2-AD^2 = AB^2-AQ\cdot AB = BC^2 = BQ\cdot AB\]Thus, $(AQC)$ is also tangent to $CB$. We have that \[\angle{AQD} = \pi - \angle{QAD}-\angle{QAD} = \pi - \angle{BAD}-\angle{DBA} = \angle{BDA}\]Similarly, \[\angle{BQC} = \angle{ACB} = \angle{BDA} = \angle{DQA}\]Thus, $Q = P$. Let $Y = CA\cap DP$ and $X = BD\cap CP$. It suffices to prove that \[\frac{CM}{MD}\cdot\frac{DY}{YP}\cdot \frac{PX}{XC} = 1\]Note, \[\frac{DY}{\sin\angle{DAY}} = \frac{YA}{\sin\angle{ADY}} = \frac{YA}{\sin\angle{DBA}}\]and \[\frac{YP}{\sin\angle{CAB}} = \frac{YA}{\sin{\angle{BPA}}} = \frac{YA}{\sin\angle{BCA}}\]Divding the two gives, \[\frac{DY}{YP} = \frac{\sin\angle{DAC}\cdot \sin\angle{BCA}}{\sin\angle{DBA}\cdot\sin\angle{CAB}}\]Doing the same to $\frac{XP}{CP}$ satisfies Ceva's theorem, and we're done.

i finally managed to make the solution i was trying on contest day work yayyy!!! we start by quantifying the desired special points on $AB$, independently of the length condition. Let $P'$ be the point on $AB$ such that $P'E$ bisects $CD$. Claim 1: $\frac{AP'}{BP'}=-\frac{AD^2}{BC^2}$

Claim 2: $\frac{AP}{BP}=\frac{AD \cdot AC}{BD \cdot BC} \cdot \frac{BD^2-BC^2}{AD^2-AC^2}$.

We check when these ratios are equal (and hence when $P$ and $P'$ coincide). Let $AB,BC,CD,DA=a,b,c,d$, and recall \[AC=\sqrt{\frac{(ac+bd)(ad+bc)}{(ab+cd)}}\quad\text{and}\quad BD=\sqrt{\frac{(ac+bd)(ab+cd)}{(ad+bc)}}.\]Hence, we have \[\frac{AP}{BP}=\frac{d}{b} \cdot \frac{(ac+bd)(ab+cd)-(ad+bc)b^2}{(ab+cd)d^2-(ac+bd)(ad+bc)}.\]We also just have \[\frac{AP'}{BP'}=-\frac{d^2}{b^2}.\]These are equal if and only if \[b((ac+bd)(ab+cd)-(ad+bc)b^2)=d((ac+bd)(ad+bc)-(ab+cd)d^2).\]Expanding both sides gives \[a^2b^2c+abc^2d+b^2cd^2-b^4c=a^2d^2c+abc^2d+b^2cd^2-d^4c\]and simplifying gives \[b^2(a^2-b^2)=d^2(a^2-d^2).\]So either $b^2=d^2$ or $a^2=b^2+d^2$. All our manipulations are reversible, so $a^2=b^2+d^2$ implies $\frac{AP'}{BP'}=\frac{AP}{BP}$, so $P=P'$, gg.

Let $M$ be the midpoint of $CD$, and let $ME$ intersect $AD$ at $P'.$ We will show that $\angle AP'D=\angle BP'E.$ WLOG $AB=1$, and let $AE=a$ and $BE=b$. Since $EP'$ is a symmedian of $\triangle EAB$, we have $$AP'=\frac{a^2}{a^2+b^2},BP'=\frac{b^2}{a^2+b^2}.$$ However, due to $\triangle ADE\sim\triangle BCE$, we have that $$AD=ka,BC=kb$$for some constant $k$. However, by the given condition we have $$k^2a^2+k^2b^2=1\rightarrow k^2=\frac{1}{a^2+b^2}.$$This means that $$AD=ka=\frac{a}{\sqrt{a^2+b^2}}.$$However, this means that $$AP'=AD^2,$$which is just saying that $$AP'\cdot AB=AD^2.$$Hence, $$\angle AP'D=\angle ADB.$$Similarly, we have $$\angle BP'C=\angle BCA,$$so since $ABCD$ is cyclic we are done.

Our solution hinges on the two orthogonal circles $\omega_A$ centered at $A$ with radius $AD$ and $\omega_B$ centered at $B$ with radius $BC$. The key is to notice that $P$ is the intersection of $AB$ and the radical axis of these two circles. This is simple through phantom points - suppose $Q$ is this point, and note \[AD^2 = AP \cdot AB, \quad BC^2 = BP \cdot BA \implies \measuredangle APD = \measuredangle BDA = \measuredangle BCA = \measuredangle CPB.\] Indeed, $P$ is unique through monotonicity, so $P = Q$. We continue with inversion. Let $X = AC \cap DP$ and $Y = BD \cap CP$. Inversion about $\omega_A$ sends $(ABCD) \mapsto DP$ and $\omega_B \mapsto \omega_B$, so $X \in \omega_B$. Similarily, we know $Y \in \omega_A$. Inversion about $\omega_A$ also sends $AXYB \mapsto CYP$, so $AXYB$ is cyclic. Consequently, Reim's tells us that $XY \parallel CD$, and Ceva's on $\triangle CDP$ finishes. $\blacksquare$

Our solution goes in the reverse direction through phantom points. Suppose $M$ is the midpoint of $CD$ and $Q = ME \cap AB$. Noting that our desired point $P$ is unqiue by monotonicity, it suffices to show $\angle AQD = \angle BQC$. This alternative formulation tells us $EQ$ is a symmedian of $\triangle EAB$, so we proceed with length chase: \[\frac{AQ}{QB} = \left(\frac{AE}{EB}\right)^2 = \left(\frac{DE}{EC}\right)^2 = \left(\frac{AD}{BC}\right)^2 = \left(\frac{AB}{BC}\right)^2-1\]\[\implies \frac{AB}{QB} = \left(\frac{AB}{BC}\right)^2 \implies BC^2 = BA \cdot BQ.\] Thus we have the angle equality $\angle BQC = \angle BCA$, and analogously $\angle AQD = \angle BDA$, so we're done. $\blacksquare$

had to use a hint buh . The length condition implies that if $Q$ is on $AB$ such that $AD^2=AB\cdot AQ$ then $BC^2=AB\cdot BQ$ so $AD$ is tangent to $(BDQ)$ and $BC$ is tangent to $(ACQ).$ Then angle chasing gives $\angle AQD=\angle ADB=\angle ACB=\angle PQC$ so $P=Q.$ Thus if $PD$ intersects the circumcircle again at $R$ and $BD\cap CP=X$ we see $\angle RBA=\angle RDA=\angle PBD=\angle XBA,$ and $\angle RPB=\angle APD=\angle BPC,$ so $X,R$ are reflections over $AB.$ Similarly if $PC$ intersects the circumcircle again at $S$ and $AC\cap PD=Y$ then $Y,S$ are reflections over $AB.$ Thus $AYXB$ is cyclic and by Reim's $XY\parallel CD$ and Ceva's finishes.

If $\angle{ADB}=\angle{APD}$, similar triangle would yield $\frac{AD}{AB}=\frac{AP}{AD}, AD^2=AP\cdot AB; \frac{BC}{BA}=\frac{BP}{BC}, BC^2=AB\cdot BP$, sum the two equations to get $AD^2+BC^2=AB^2\implies \angle{ADB}=\angle{APD}=\angle{BPC}$. However, if we vary point $P$ from the point satisfying $\angle{APD}=\angle{CPB}$, one angle would be larger and the other one would be smaller. Thus, $P$ is unique and we have to prove $PE$ bisects $CD$. Since $\angle{ADB}=\angle{ACB}=\angle{CPB}=\angle{APD}$, it imples $ADSP, BPRC$ are cyclic($S=BD\cap CP; R=AC\cap DP$). Note $\angle{CRB}=\angle{CPB}=\angle{APD}=\angle{ASD}=180-\angle{ASB}=180-\angle{ARB}\implies \angle{ARB}=\angle{ASB}\implies ARSB$ is cyclic. Finally, note $\angle{BAS}=\angle{BRS}=\angle{SDP}; \angle{BRP}=\angle{BCP}=\angle{BAC}=\angle{BDC}\implies RS||CD$. Thus $M$ is the midpoint of $CD$ by Ceva's theorem.

Let $P'$ be the point on side $AB$ so that $AP' \cdot AB = AD^2$, and by PoP we have $AD$ tangent to $(BPQ)$. We have $AD^2 = AB^2 - BC^2 \implies AP' \cdot AB = AB^2 - BC^2 \implies BC^2 = AB \cdot BP'$ so $AC$ is also tangent to $(BP'C)$. From the length conditions, we have $\triangle AP'D \sim \triangle ABD$ and $\triangle BP'C \sim \triangle BCA$. From this we get that $\angle BP'C = \angle BCA = \angle BDA = \angle APD$, so $P' = P$. Note that $AP \cdot AB = AD^2$ and $AB \cdot BP = BC^2$ implies that $\frac{AP}{BP} = \frac{AD^2}{BC^2} = \frac{AE^2}{BE^2}$ so $PE$ is the $E$-symmedian of $\triangle BEA$. This implies that $EM$ is a median of $\triangle CED$, as desired.

We prove the converse, something possible by phantom points. By Ratio Lemma on $\triangle ECD$ with the midpoint (say) M, we have $ED \sin(\angle PEB) = EC \sin(\angle PEA)$. Now, Ratio Lemma on $\triangle EAB$ gives $\frac{AP}{PB} = \frac{EA \sin(\angle PEA)}{EB \sin(\angle PEB)} = \frac{EA \cdot ED}{EB \cdot EC} = \frac{EA^2}{EB^2} = \frac{AD^2}{BC^2}$ with the last following by $\triangle EAD \sim \triangle EBC$. Let $\frac{AD^2}{AP} = \frac{BC^2}{BP} = x$. Now, $x(AP + BP) = AB^2 \implies x = AB$ or $AD^2 = AP \cdot AB$ and $BC^2 = BP \cdot AB$. This implies tangents. Now, $\angle DPA = \angle ADB = \angle ACB = \angle CPB$. Motivation: P is an unknown point so it makes sense to try and 'solve for its location' (i.e. find the ratio $\frac{AP}{PB}$). The use of the converse is more natural as it allows one to easily find the ratio. No similar triangles or any other length condition can be found directly by the angle condition so using the midpoint is more natural. This motivates the use of trig and allows for a relatively straightforward solution via length bashing.

Let $M$ be the midpoint of $\overline{CD}$, and suppose $\overline{ME}$ intersects $\overline{AB}$ at $X$. We are left to prove that $X=P$, i.e., $\angle AXD=\angle BXC$. Note that $\triangle EAB\stackrel{-}{\sim}\triangle EDC$ and $\overline{EM}$ is the $E$-median of $\triangle EDC$, so $\overline{EX}$ is the $E$-symmedian of $\triangle EAB$. Thus \[\frac{AX}{BX}=\left(\frac{AE}{BE}\right)^{2}=\left(\frac{AD\cdot\tfrac{\sin\angle ADE}{\sin\angle AED}}{BC\cdot\tfrac{\sin\angle BCE}{\sin\angle BEC}}\right)^{2}\]but, as $\angle ADE=\angle BCE$ (cyclic quadrilateral) and $\angle AED=\angle BEC$ (vertical angles), this is just $\left(\tfrac{AD}{BC}\right)^{2}$. It follows $AX=\tfrac{AD^{2}}{AB}$ and $BX=\tfrac{BC^{2}}{AB}$; rearranging gives that $\overline{AD}$ and $\overline{BC}$ are tangent to $(BXD)$ and $(AXC)$, respectively. Then \[\angle AXD=\pi-\angle ADX-\angle DAX=\pi-\angle DBX-\angle DAX=\angle ADB,\quad\text{and}\quad\angle BXC=\pi-\angle BCX-\angle CBX=\pi-\angle CAX-\angle CBX=\angle ACB\]but obviously, $\angle ADB=\angle ACB$ because $ABCD$ is cyclic, $\angle AXD=\angle BXC$ and $X=P$ as desired.

50th post!!! Let $M$ be the midpoint of $CD$ and let $ME\cap AB=P'$. So we need to prove $\angle AP'D=\angle CP'B$ $\textbf{Claim 1.}$ $\triangle EDC \sim \triangle EAB$ \begin{align*} \frac{\sin{\angle P'EA}}{\sin P'EB}=\frac{\sin{\angle MEC}}{\sin{\angle MED}}=\frac{ED}{EC}=\frac{EA}{EB} \end{align*}$\therefore EP'$ is the symmedian of $\triangle ABE$. Note that \begin{align} \frac{AP'}{P'B}&=\frac{EA\cdot \sin{\angle AEP'}}{EB\cdot \sin{\angle BEP'}}=\frac{AE^2}{BE^2}=\frac{DE^2}{CE^2}\\ \frac{AE}{BE}&=\frac{\sin{\angle ABE}}{\sin{\angle BAE}}=\frac{\sin{\angle ABD}}{\sin{\angle BAC}}=\frac{AD}{BC} \end{align}From (1) and (2) we have $$\boxed{\frac{AE^2}{BE^2}=\frac{AD^2}{BC^2}=\frac{DE^2}{CE^2}=\frac{AP'}{P'B}}$$So, $\triangle EAD \sim \triangle EBC$. $\textbf{Claim 2.}$ $\triangle BCA \sim \triangle BP'C$ Using the fact that $AP'+P'B=AB$ we get: $$\frac{AB}{BP'}=\frac{AB^2}{BC^2}\implies \frac{AB}{BC}=\frac{BC}{BP'}$$Similarly we also get $\triangle BDA\sim \triangle DPA'$. By $\textbf{Claim 2.}$ we have $$\angle AP'D=\angle ADB=\angle ACB=\angle CP'B$$$\blacksquare$

Here we go: Redefine $P$ to be a point on $AB$ such that: $AP \cdot AB = AD^2, BP \cdot BA = BC^2$ (which exists uniquely due to the given condition). It suffice to show that: $\angle APD = \angle BPC$ and $PE$ to be the symmedian of $\triangle EAB$. Claim: $\angle APD = \angle BPC$ Note that: $\triangle APD \sim \triangle ADB, \triangle BPC \sim BCA$. Thus: $$\angle APD = \angle ADB = \angle ACB = \angle BPC.$$ Claim: $PE$ is a symmedian of $\triangle EAB$. Note that; $AP \cdot AB = AD^2, BP \cdot BA = BC^2$. Dividing them gives us: $$\frac{AP}{PB} = \frac{AD^2}{BC^2} = \frac{AE^2}{BE^2}$$and therefore we are done.

Let $M$ be the midpoint of $\overline{CD}$, we define the phantom point $P' = ME \cap \overline{AB}$. To prove $P \equiv P'$ it suffices to show $\angle AP'D = \angle BP'C$. Suppose $N$ is the midpoint of $\overline{AB}$, clearly $\angle AEN = \angle DEM = \angle BEP'$ so $N$ and $P'$ are isogonal conjugates w.r.t. $E$ in $\triangle AEB$, then $EP'$ is the $E$-symmedian. By a well-known fact, \[\dfrac{AP'}{BP'} = \dfrac{AE^2}{BE^2} = \dfrac{AD^2}{BC^2} \implies \dfrac{AP' + BP'}{BP'} = \dfrac{AD^2 + BC^2}{BC^2} \]\[ \implies \dfrac{AB}{BP'} = \dfrac{AB^2}{BC^2} \implies \dfrac{BP'}{BC} = \dfrac{BC}{AB} \]but $\angle P'BC = \angle CBA$ so by SAS Similarity we have $\triangle BP'C \sim \triangle BCA$. Analogously, $\triangle AP'D \sim \triangle ADB$, then \[ \angle AP'D = \angle ADB = \angle ACB = \angle BP'C \]as wanted.