I needed over two hours for the solution, even though it is not so hard. Let $P_1$, $P_2$, $P_3$, $P_4$ be the orthogonal projections of the point P on the lines AB, BC, CD, DA, and let $Q_1$, $Q_2$, $Q_3$, $Q_4$ be the reflections of P in these lines, or, equivalently, the reflections of P in the points $P_1$, $P_2$, $P_3$, $P_4$. From < PBC = < DBA, you see that the line BD is the reflection of the line BP in the angle bisector of the angle ABC. Now you use the following (quite easy) lemma: If X is a point in the plane of an angle VUW, and V' and W' are the reflections of X in the lines UV and UW, then the perpendicular bisector of the line V'W' is the reflection of the line UX in the angle bisector of the angle VUW. Now, apply this lemma to the angle ABC and the point P, whose reflections in the lines BA and BC are $Q_1$ and $Q_2$, respectively, and you see that the perpendicular bisector of the line $Q_1 Q_2$ is the line BD. Hence, the points $Q_1$ and $Q_2$ are symmetric to each other with respect to the line BD. Similarly, the points $Q_4$ and $Q_3$ are symmetric to each other with respect to the line BD, too. Hence, since symmetry is a congruence transformation , we have $Q_1 Q_4 = Q_2 Q_3$. Now, since $Q_1$ and $Q_4$ are the reflections of P in $P_1$ and $P_4$, the points $P_1$ and $P_4$ are the midpoints of the segments $PQ_1$ and $PQ_4$, and thus $P_1 P_4 = \frac12 Q_1 Q_4$. Similarly $P_2 P_3 = \frac12 Q_2 Q_3$. Now, from $Q_1 Q_4 = Q_2 Q_3$, it follows that $P_1 P_4 = P_2 P_3$. Since $\measuredangle AP_4 P = 90^{\circ}$ and $\measuredangle AP_1 P = 90^{\circ}$, the points $P_4$ and $P_1$ lie on the circle with diameter AP. In other words, the circumcircle of triangle $AP_4 P_1$ has the segment AP as diameter. Therefore, by the Extended Law of Sines, we have $P_1 P_4 = AP \cdot \sin \measuredangle P_1 A P_4 = AP \cdot \sin A$, where A = < DAB and C = < BCD are two opposite angles of our quadrilateral. Similarly, $P_2 P_3 = CP \cdot \sin C$. Since $P_1 P_4 = P_2 P_3$, we conclude that AP = CP holds if and only if sin A = sin C. Now, sin A = sin C means either A = C, or A + C = $180^{\circ}$. In the case of A + C = $180^{\circ}$, the quadrilateral ABCD is cyclic, and we are happy. Remains to show that the case A = C cannot occur. In fact, a simple angle chase shows that in this case, we have < BPD = $180^{\circ}$, so that the point P (if it exists) lies on the diagonal BD, and from < PBC = < DBA we conclude that the diagonal BD bisects the angle ABC. This contradicts the assumption of the problem. Hence, AP = CP holds if and only if the quadrilateral ABCD is cyclic. Peter has another solution, applying the sine law one thousand times (later he admitted it were just four times or something like this ). Darij

First of all, let's show that the "only if" implies the "if". Assume we know that $AP=CP$ for all cyclic $ABCD$. Now assume $ABCD$ isn't cyclic. Then it was obtained from a cyclic quadrilateral by sliding $D$ along line $BD$. During this k\movement, $BP$ has remained the same, so $P$ moved along the line $BP$. However, when $ABCD$ was cyclic, $P$ was the intersection between $BP$ and the perpendicular bisector of $AC$, so it can't have this position anymore (because then $BP$ and the perpendicular bisector of $AC$ would intersect in two points). Now let's show the "only if" part: If $\ell$ is the perpendicular bisector of $AC$, then $D'=BP\cap \mbox{circle},\ B'=DP\cap \mbox{circle}$ are the symmetrics of $D,B$ respectively wrt $\ell$. This means that $BB'D'D$ is an isosceles trapezoid which has $\ell$ as an axis of symmetry. Then the intersection of its diagonals, which is $P$, must lie on $\ell$. I don't know if you did essentially the same, Darij, but that seemed longer.. Is there something wrong? I'm asking because this seems to be really easy

Though I am not a member of this year's IMO team, but I've tried Problem 5. Here is my solution: Let BP meet AC at point E, DP meet AC at point F. First we assume that ABCD is concyclic. Then it follows that \triangle BCE\sim\triangle BDA, which implies CE=\frac{BC\bullet AD}{BD}. Similarly, CF=\frac{CD\bullet AB}{BD}. Thus CE+CF=\frac{BC\bullet AD+CD\bullet AB}{BD}=AC(it follows from the Ptolemy's Theorem), so CE=AF. Note that PE=PF and \angle PFA=\angle PEC, because \angle PEF=\angle PFE=\angle ABD+\angle ADB(it can be shown easily), so \triangle PFA\cong\triangle PEC, indicating AP=CP. The remaining part of proving ABCD is concyclic if AP=CP can be shown easily using the conclusion above, though I have spent more than half an hour on it.

Here's my solution, bearing some resemblance to grobber's. Part 1: Assume $ABCD$ is cyclic. Extend $BP$ and $DP$ to $D'$ and $B'$ respectively on the circumcircle of $ABCD$. Then $AB=CB'$ and $AD=CD'$ since they subtend equal arcs. The circle also gives us $\angle PDA=\angle PD'C$ by equal arcs $BC$ and $B'A$. $\angle BPD=\angle B'PD'$ since they are vertical, and $\angle BAD=\angle B'AD'$ by subtraction. The equal angles and two adjacent equal sides force $ABPD \cong CB'PD'$ and $AP=CP$ by this congruence. Part 2: Assume $AP=CP$. Extend $BP$ and $DP$ to meet the circumcircle of $BCD$ at $D'$ and $B'$ respectively. Then $\angle PB'C=\angle DB'C=\angle DBC=\angle ABP$ and $\angle PD'C=\angle BD'C=\angle BDC=\angle ADP$. Again $\angle BPD=\angle B'PD'$ since they are vertical, and $\angle BAD=\angle B'AD'$ by subtraction. Now since $BB'D'D$ is cyclic by construction, $\angle PBD=\angle PB'D'$ and triangles $PBD$ and $PB'D'$ are similar in ratio $r=\frac{BD}{B'D'}$. We have $\angle DBA=\angle D'B'C$ by subtraction, and triangles $ABD$ and $CB'D'$ are also similar in ratio $r=\frac{BD}{B'D'}$. Therefore quadrilaterals $ABPD$ and $CB'PD'$ are similar in ratio $r$, so $\frac{AP}{CP}=r$. $\frac{AP}{CP}$ is $1$ by hypothesis, so $ABPD \cong CB'PD'$. This forces $ABPD$ and $CB'PD'$ to be reflections of each other across the bisector of angle $BPB'$. Since $BB'CD'D$ is cyclic, its reflection $B'BADD'$ is also cyclic and the original quadrilateral $ABCD$ is cyclic.

For the second part I did it in a slightly different manner from the way jmerry did it, but its really the same idea: Suppose AP=PC. Reflect ABD about the angle bisector of <APC. Let D go to D' on BP and B go to B' on DP; note A goes to C. Now BB'DD' is an iscocles trapezium because PD=PD' and PB=PB', and is thus cyclic, and further it follows the angle bisector of <APC is a diameter of the circle by symmetry. Further, <D'B'C=<ABD=<D'BC so C lies on the the circle also, and since A is the reflection of C about the angle bisector of <APC, A also lies on the circle and ABDC is cyclic as required. [EDIT: this proof is wrong in assuming D' and B' lie on BP and DP . ]

Grobber, your solution is ingenious... They have told me there are 6 proposed solutions! darij

To vinoth: I would have posted hours earlier if I could have made that reflection work. How can you be sure that D' is on BP and that B' is on DP? I tried extending to match lengths, and couldn't prove cleanly that C and A were reflections of each other. The angle information is in the wrong places.

Jmerry: I also used the Grobber reflection in my solution, and it makes perfect sense; I think you're looking at it backwards-first extend $B$ to $D'$, $D$ to $B'$, then transform to show that everything pops up in the right place. Clearly, under reflection in $l$, the perpendicular bisector of segment $AC$, $A \rightarrow C$. But we are given that $m<CDB'=m<ADB$, and so arc$AB$=arc$CB'$ (neither arc containing $D$). It follows immediately that $B \rightarrow B'$ and by an analogous argument that $D \rightarrow D'$. Consequently, $l$ is an axis of symmetry for isosceles trapezoid $BB'D'D$, implying $P = BD' \cap B'D\in l$ so that by definition, $PA = PC$.

That's a proof of the "only if" direction, but I was talking about the "if" direction. The angle-chasing is easy when you already have the circle.

Jmerry: The other direction is equally simple. Let $BD \cap circ(ADC) = B'$, and $P'$ be the associated intersection point for quadrilateral $A'BCD$. $AB'CD$ is cyclic, so you agree that $P'A = P'C$. Therefore, $l \cap DP' = P'$, where $l$ is the perpendicular bisector of $AC$, is the unique intersection point of those two lines. However, $m<CBP' \not= m<ABP'$, and therefore, $P\not=P'$ while $P \in P'D$. Whence, by the uniqueness argument above, $PA \not= PC$.

jmerry wrote: To vinoth: I would have posted hours earlier if I could have made that reflection work. How can you be sure that D' is on BP and that B' is on DP? I tried extending to match lengths, and couldn't prove cleanly that C and A were reflections of each other. The angle information is in the wrong places. yes, my proof was wrong. proving B' and " lie on DP and BP is equivalant to proving the angle bisector of external <BPD coincides with external bisector of <BPD, i.e. <APB+<DPC=180 ... which probably isnt very easy to prove. anyway, your proof and the 'shifting' proof work for proving the converse.

mecrazywong wrote: The remaining part of proving ABCD is concyclic if AP=CP can be shown easily using the conclusion above, though I have spent more than half an hour on it. mecrazywong wrote: easily This is the most difficult part of the pb, worth 4 pts !!!! I wait for your solution

mecrazywong, I'd be interested in seeing your completion of that argument: I had exactly your proof for the first part, but was unable to see any way to use it to get the other part: I drew in four similar triangles to get points E_1, E_2, F_1 and F_2 and used the equal lengths calcuation to get two congruent triangles, but could find no way to use them. mecrazywong wrote: The remaining part of proving ABCD is concyclic if AP=CP can be shown easily using the conclusion above, though I have spent more than half an hour on it.

grobber wrote: First of all, let's show that the "only if" implies the "if". Assume we know that $AP=CP$ for all cyclic $ABCD$. Now assume $ABCD$ isn't cyclic. Then it was obtained from a cyclic quadrilateral by sliding $D$ along line $BD$. During this k\movement, $BP$ has remained the same, so $P$ moved along the line $BP$. However, when $ABCD$ was cyclic, $P$ was the intersection between $BP$ and the perpendicular bisector of $AC$, so it can't have this position anymore (because then $BP$ and the perpendicular bisector of $AC$ would intersect in two points). what about P' remained the original position when A,B,C,D' are concyclic?

Damn! Let me think about that for a while..

Michael Lipnowski wrote: However, $m<CBP' \not= m<ABP'$, and therefore, $P\not=P'$ while $P \in P'D$. How can you deduce that $P\neq P'$?

liyi wrote: grobber wrote: First of all, let's show that the "only if" implies the "if". Assume we know that $AP=CP$ for all cyclic $ABCD$. Now assume $ABCD$ isn't cyclic. Then it was obtained from a cyclic quadrilateral by sliding $D$ along line $BD$. During this k\movement, $BP$ has remained the same, so $P$ moved along the line $BP$. However, when $ABCD$ was cyclic, $P$ was the intersection between $BP$ and the perpendicular bisector of $AC$, so it can't have this position anymore (because then $BP$ and the perpendicular bisector of $AC$ would intersect in two points). what about P' remained the original position when A,B,C,D' are concyclic? Thanks for pointing this out. It can be patched up, but all the magic is gone . I no longer like the solution, so I won't post it in detail. First of all, we may assume WLOG that $BD$ meets the $\perp$ bisector of $AC$ (call it $d$) on the same side of $AC$ as $D$ (otherwise we move $B$, not $D$). Assume the point of intersection between the diagonals is $S$ and that $SA<SC$ and the point of intersection between $BD$ and $d$ is $T$. $D$ can only have positions on the ray $(ST$, because otherwise the quadrilateral would become concave. The main idea is this: we show that if $D$ goes outside the segment $(ST)$ then $P$ won't be able to lie on $d$ because the bisector of $\angle ADC$ would interpose between $P$ and $d$. Then we show (not too hard, we just draw the Apollonius circles etc.) that on the segment $(ST)$, when $D$ goes towards $T$, the ratio $\frac {AD}{DC}$ increases to $1$. The important thing here is that it increases, so $\frac {AX}{XC}\ (*)$ also increases, where $X=DP\cap AC$. But if $D'$ was nearer to $T$ than $D$ and $P'=P$, then $X'$ would be nearer to $A$ than $X$, so the ratio $(*)$ would in fact decrease, and not increase, as it should, so we have a contradiction. We do something similar if $D'$ is nearer to $S$ than $D$. It looks Ok to me right now, but who knows..

well, it makes me not like it too... Denote by $\ell$ the perpendicular bisector of $AC$. Suppose $B'$ and $D'$ are the symmetric points of $B$ and $D$ by $\ell$. With $AP=PC$, we know that $CBB'D'DA$ is axial symmetric with the symmetry axis $\ell$.

sorry if im missing something trivial .. i dont understand how that reflection works. How do you know that <ABP = <CB'D?

$A,C$ are isogonal conjugates wrt $\triangle BPD,$ so $\angle APB+\angle CPD=\pi.$ Note that $\angle BAD\neq \angle BCD,$ since $P\notin BD$. Hence the following facts are equivalent: $$|AP|=|CP|,$$$$|AB|\cdot \sin \angle ABP=|CD|\cdot \sin \angle CDP,$$$$|AB|\cdot \sin \angle CBD=|CD|\cdot \sin \angle ADB,$$$$|BD|\cdot \sin \angle BAD=|BD|\cdot \sin \angle BCD,$$$$\angle BAD+\angle BCD=\pi.$$

This took too long... Suppose $ABCD$ is cyclic, inscribed in $\omega.$ Let $B_1=\overline{BP}\cap\omega$ and $D_1=\overline{DP}\cap\omega.$ Notice $\angle DBA=\angle CBB_1$ so $AD=BC_1$ and $\overline{B_1D}\parallel\overline{AC}.$ Similarly, $\overline{BD_1}\parallel\overline{AC}.$ Since $P$ is the center of cyclic trapezoid $BDB_1D_1,$ it lies on the perpendicular bisector of $\overline{BD_1}$ which is equivalent to the perpendicular bisector of $\overline{AC}.$ Suppose $PA=PC.$ Construct $B_2$ on $\overline{BP}$ such that $PB_2=PD.$ Similarly construct $D_2$ on $\overline{DP}$ such that $PD_2=PB.$ Notice $\measuredangle BPC=\measuredangle APD$ as $A,C$ are isogonal conjugates wrt $\triangle PBD$ so $$\measuredangle B_2PC=\measuredangle BPC=\measuredangle APD$$and $\triangle PDA\cong\triangle PB_2C.$ Hence, $$\angle BB_2C=\angle ADP=\angle BAC$$and $ABCB_2$ is cyclic isosceles trapezoid. Also, $\angle B_2CA=\angle CAD$ and $CB_2=AD$ so $\overline{AC}\parallel\overline{B_2D}.$ Similarly, $ABD_2C$ is cyclic isosceles trapezoid and so $BDB_2D_2$ is cyclic. $\square$

jesus Suppose $ABCD$ is cyclic. Let $PB$ intersect the circumcircle of $ABCD$ at $K.$ Let $DP$ intersect the circumcircle of $ABCD$ at $J.$ Note that $\angle PBD=\angle ABP$ so $CD=AK.$ This implies $KD||AC.$ Similarly, $BJ||AC.$ $DKJB$ is a cyclic trapezoid so $P$ is on the perpendicular bisector of $KD.$ Thus it must also lie on the perpendicular bisector of $AC.$ Suppose $P$ lies on the perpendicular bisector of $AC.$ Then, let $\ell$ be the perpendicular bisector of $AC.$ Let $K,J$ be reflections of $D,B$ over $\ell$ respectively. Note that since $P$ is on $\ell,$ $DP=PK,BP=PJ.$ Since $\triangle CDB$ is a reflection of $\triangle KAB$ over $\ell$ we have $\angle KBA=\angle CBD=\angle KJA$ so $KAJB$ is cyclic. Also, $\angle AKJ=\angle CDB=\angle ABJ.$ Thus, $AKDJ$ is cyclic. This implies $K,J$ are on $(ABD).$ $BD=KJ$ so $BDKJ$ is isosceles trapezoid. Thus, $\ell$ is a diameter of the circle $(ABD)$ so the reflection of $A$ over $\ell,$ $C$ is also on that circle. Thus, $ABCD$ cyclic.

Really good problem! This is probably one of the shorter solutions in this thread (I also notice that @above @2above have done similar (almost same) stuff). Also I found this solution really fast as I didn't want to use tools like protractors to construct isogonal conjugates and instead used isosceles trapezoid which miraculously lead to a clean solution. (i) Assume $ABCD$ is cyclic and $\odot(ABCD) = \Omega$ Let $BP \cap \Omega = X$ and $DP \cap \Omega = Y$ Notice that $ADXC$ and $ABYC$ are both isosceles trapezoids and therefore $$\widehat{BD} = \widehat{AD}+\widehat{AB}=\widehat{CX}+\widehat{CY}= \widehat{XY} \implies \angle XDY = \angle BXD \implies PD=PX \implies \triangle APD \cong \triangle CPX \implies AP = CP \ \blacksquare$$(ii) Assume $AP=CP$ Let $X$ be a point such that $ADXC$ is an isosceles trapezoid and let $XP \cap \odot(ADC) = B'$, this immediately gives us that $\angle PBC=\angle DBA$ It is obvious that $\triangle APD \cong \triangle CPX$ which implies $$\angle ADP = \angle CXP = \angle CXB' = \angle CDB' \implies \angle ADB = \angle CDP \implies B=B' \implies B \in \odot(ADC) \ \blacksquare$$

Notice that $A$ and $C$ are isogonal conjugates with respect to $\triangle BDP$, so the external angle bisector of $\angle BPD$ bisects $\angle APC$. Only if direction: Let $O$ be the circumcenter of $ABCD$. By angle chasing, $BPOD$ is cyclic. Since $OB=OD$, the external angle bisector of $\angle BPD$ passes through $O$. Since $OA=OC$ and $O$ lies on the angle bisector of $\angle APC$, either $PA=PC$ or $OAPC$ is cyclic. However, the latter is absurd because $P$ would have to be the intersection of the circumcircles of $AOC$ and $BOD$, which is outside of the circumcircle of $ABCD$ (and thus outside of $ABCD$). If direction: Fix point $A$. Notice that the function taking points $A'$ on $\overline{AB}$ to the corresponding point $P$ is injective, so we must have $A$ such that $ABCD$ is cyclic.

Let $l$ be the perpendicular bisector of $AC$ and $X=l\cap BD$. Part 1: If $\Gamma=(ABCD)$ is cyclic, then define $\{N, M\}=l\cap \Gamma$. If we have two distinct points $P_1, P_2\in l$ such that $\angle P_1DA=\angle BDA$ and $\angle P_2BA=\angle DBC$, then notice that $$(X, P_1; N, M)=-1=(X, P_2; N, M)\implies P_1=P_2$$so one direction is completed Part 2: Now assume $P\in l$. Redefine $\{N, M\}=l\cap (CDA)$. $B, D$ both belong to the $D$ - apolonius circle in $\triangle DPX$, which has diameter $NM$. So $\angle NBM=90=\angle NDA$ and thus $(NMDB)$ is also cylcic. Hence $(ABCD)$ is cyclic.

Let $A_1$ denote the reflection of $A$ in $BD$, the midpoints of $A_1C$ and $A_1P$ be $M$ and $N$ respectively, and the pedal triangle of $A_1$ wrt $BCD$ be $XYZ$. It's easy to see $A_1YBZ$, $A_1ZCX$, and $A_1XDY$ are cyclic with diameters $A_1B$, $A_1C$, and $A_1D$ respectively. Notice $$\angle PBC = \angle DBA = \angle DBA_1$$and $$\angle PDC = \angle BDA = \angle BDA_1.$$Now, because $ABCD$ is convex, it follows that $A_1$ and $P$ are isogonal conjugates wrt $BCD$. Thus, a well-known lemma implies $N$ is the center of $(XYZ)$. By a homothety centered at $A_1$, we know $PA = PC$ holds if and only if $NY = NM$, which is equivalent to $M \in (XYZ)$. Now, since $A_1ZCX$ is centered at $M$, $$\angle XMZ = 2 \angle XCZ = 2 \angle BCD.$$In addition, we have $$\angle XYZ = \angle XYA_1 + \angle A_1YZ = \angle XDA_1 + \angle A_1BZ$$$$= \angle PDB + \angle DBP = 180^{\circ} - \angle BPD.$$Thus, $XYZM$ is cyclic if and only if $\angle BPD = 2 \angle BCD$. Now, observe that $$\angle BAD = 180^{\circ} - \angle DBA - \angle BDA = 180^{\circ} - \angle PBC - \angle PDC$$$$= \angle BCD + \angle DBP + \angle BDP = \angle BCD + (180^{\circ} - \angle BPD).$$It follows that $ABCD$ is cyclic if and only if $\angle BPD = 2 \angle BCD$, which finishes. $\blacksquare$ Remark: Solving USA TST 2010/7 and reading this article by Evan Chen helped me with this question.

We shall start with proving $AP = CP \Rightarrow ABCD$ is cyclic, as this is the harder portion. The key is to focus on $PBD$ as the reference triangle. Then, we see that $BA$ and $BC$ are isogonal, as well as $DC$ and $DA$. Thus, $A$ and $C$ are isogonal conjugates of each other (with respect to $\triangle PBD$). In the diagram below, I've extended PC and marked the equal angles red, in case it is confusing what it means for $PA$ and $PC$ to be isogonal in this context. A simple angle chase demonstrates that the perpendicular bisector of $AC$ (shown in dashed orange) is the external angle bisector of $\angle BPD$. Thus, this perpendicular bisector of $AC$ must pass through $I_B$ and $I_D$, the $B-$ and $D-$ excenters, respectively. Now comes the key claim: Claim: $BCI_BAI_D$ and $DCI_DAI_B$ are both cyclic. Proof: We shall show $BCI_BA$ is cyclic, as the others are analogous. Observe that $BI_B$ is the angle bisector of $\angle PBD$, but since $\angle PBC = \angle DBA$, we get that $BI_B$ is also the angle bisector of $\angle ABC$! Thus, $I_B$ is the intersection of the perpendicular bisector of $AC$ and the angle bisector of $\angle ABC$, which is well known to be the arc midpoint of $\widehat{AC}$ on $(ABC)$, unless $BC = BA$. But we know $BC = BA$ is impossible, as that would imply $DB$ bisects $\angle CDA$. A similar argument can be done to get $I_D$ lies on $(ABC)$: $BI_D$ is the external angle bisector of $\angle ABC$, and $I_DA = I_DC$, so $I_D$ must be the arc midpoint of $\widehat{ABC}$. $\Box$ Using the claim, both $B$ and $D$ lie on circumcircle $(ACI_BI_D)$, so $ABCD$ is cyclic as desired. Now we prove $ABCD$ cyclic $\Rightarrow AP = CP$. Let $X = BP \cap (ABCD)$ and $Y = DP \cap (ABCD)$. Because $\angle DBA = \angle XBC$, we have $ACDX$ is an isoceles trapezoid, so $DX || AC$. Similarly, we have $BY || AC$. Putting these together, we have $BY || DX$, so $BYXD$ is an isoceles trapezoid as well. This means that $BY, AC,$ and $DX$ share a perpendicular bisector. Furthermore, we know that $P$, the intersection of diagonals of isoceles trapezoid $BYXD$, must lie on the perpendicular bisector of $BY$ and $XD$. Thus, $P$ is on the perpendicular bisector of $AC$ as well, proving $PA = PC$.

Solved with huashiliao2020.

Very conceptual problem demonstrating how one deals with isogonal conjugates. For one direction, let $ABCD$ be cyclic and denote $X = \overline{BP} \cap (ABCD)$, $Y = \overline{DP} \cap (ABCD)$. By the angle conditions, $ADXC$ and $ABYC$ are isosceles trapezoids. Thus it follows that $\overline{BY}, \overline{DX}, \overline{AC}$ share a perpendicular bisector, and the result follows. For the other direction, notice that $A$ and $C$ are isogonal conjugates with respect to triangle $BPD$. Construct $X$ on $\overline{BP}$ such that $XP=DP$ and $Y$ similarly. The triangles $APB$ and $CPY$ are congruent as $\angle APB = \angle CPY$, thus $\overline{BY} \parallel \overline{AC} \parallel \overline{DX}$. On the other hand, $\angle CBX = \angle DBA = \angle CYX$, hence $BYCX$ is cyclic, and similarly $DYCX$ is cyclic. It follows that all the points $A, B, Y, C, X, D$ are concyclic, as needed.

whoops I first prove that if $ABCD$ is cyclic then $AP=CP$. The condition implies that $\overline{BD}$ and $\overline{BP}$ are isogonal in $\triangle BAC$, and $\overline{DB}$ and $\overline{DP}$ are isogonal in $\triangle DAC$. Let $B'$ and $D'$ be points such that $BB'AC$ and $DD'AC$ are isosceles trapezoids (with $\overline{BB'} \parallel \overline{DD'} \parallel \overline{AC}$), which also lie on $(ABCD)$. Then $B,P,D'$ are collinear, as are $D,P,B'$, so in fact $P=\overline{BD'} \cap \overline{DB'}$ which lies on the perpendicular bisector of $\overline{AC}$ by symmetry. I will now prove that we don't need to consider the other direction! We use the following claim. Claim: Let $X,Y,Z$ be points and $\ell$ be a line. Then either there are at most two points $R \in \ell$ such that $\ell$ and $\overline{RZ}$ are isogonal in $\triangle RXY$, or (algebraically) the isogonality always holds. Proof: We use complex numbers with $\ell$ being the real axis (WLOG), denoting $X=x$ etc. The isogonality condition is equivalent to $$\frac{r-x}{r-0} \div \frac{r-z}{r-y} \in \mathbb{R} \iff \frac{(r-x)(r-y)}{r-z} \in \mathbb{R} \iff (r-x)(r-y)(r-\overline{z})=(r-\overline{x})(r-\overline{y})(r-z).$$Upon expansion and simplification this means that any $r$ is the root of a fixed quadratic (in terms of $x,y,z$). If the quadratic is identically zero then the isogonality is true for all $P \in \ell$, otherwise it has at most two solutions. $\blacksquare$ Fix a choice of $B,A,C$, as well as line $\overline{BD}$; note the convex condition implies $\overline{BD}$ intersects segment $\overline{AC}$. Then if $AP=CP$, $P$ is fixed as well, since $\overline{BP}$ is a fixed line as $D$ varies. Now apply the claim with $\ell=\overline{BD}$ and $X=A,Y=C,Z=P$. Obviously we can't have $\overline{AP} \parallel \overline{CP}$, so at least one of them is not parallel to $\ell$; WLOG $\overline{CP}$. Then if $R$ is placed at $\overline{CP} \cap \ell$, then the isogonality does not hold, since $A \not \in \overline{BD}$ so $\angle (\overline{AR},\ell) \neq 0$ but $\angle (\overline{CR},\ell)=0$. Hence the "at most two" part of the claim applies. Since $B$ and $\ell \cap (ABC) \neq B$ are valid positions for $R$ in the language of the claim, the first being tautological and the second being true due to the first part of this solution, it follows that $D$ must be placed at the second intersection of $\ell$ with $(ABC)$, whence $ABCD$ is cyclic. $\blacksquare$

This took soo long. The problem is quite easy but i don't know why. The case where $ABCD$ is cyclic is easy. IF $AP=PC$ , then construct $D'$ such that $ADD'C$ is cyclic. Let $D'P$ intersect again at$ B'$. Then by angle chasing B=B' We are done

Why consider the other direction when you can just not !! By shifting $D$ along $BD$, it follows that at most one value of $D$ on a fixed line works as the other line is fixed. As such, showing the result for cyclic $ABCD$ finishes. Let $F$ be the midpoint of $\widehat{ADC}$ and $E$ the midpoint of $\widehat{ABC}$. It remains to show that $P, E, F$ are collinear. Note that $\angle BPF = \angle BFP + \angle PBF = \angle BFE + \angle FBD$ $\angle EPD = \angle PED + \angle EDP = \angle FED + \angle BDE$. As such, it follows that $\angle BPF + \angle EPD = 2 \angle BDE + 2 \angle FBD = \angle BDP + PBD = BPD$ which finishes.

Claim $$\angle APD+\angle CPD=180^{\circ}$$ Proof Let $X$ be a point on $BD$ such that $\angle BAX=\angle CAD$. Now we can see that $P$ and $Q$ are isogonaly conjugate with respect to three of the four angle, so $X$ and $P$ are isogonaly conjugate with respect to $ABCD$. \end{proof} Now let we start with the angle chase. $o$ means orange, $g$ means green and $x$ and $y$ are like the diagram that is (hopefully) attached. $$\angle APB = 180^{\circ}-x-o-g=180^{\circ}-CPD=PDC+PCD=ADB+ACD-y=180^{\circ}-BAD-y=180^{\circ}-g-o-x$$ So this means that $x=y$, or $PA=PC$. The opposite direction is similar.