There are two key steps to solving this problem. Show that if we move $D$ on the perpendicular bisector of $EC$ with unit velocity, the quantity \[[BCD]+[ADE] - [ABD] - [ABP]\]varies linearly with time, and furthermore, the rate of change is $0$ only if $P\in CE$. Show that the area condition is satisfied if $\angle EDC=2\angle ADB$. It's fairly straightforward to see that these together imply the problem. We begin with the first point. Indeed, the rate of change of the area sum is \[|\phi(BC)| + |\phi(AE)| - |\phi(AB)|,\]where $\phi$ is the projection map onto line $CE$. We're given that this is zero, and it is easy to see that this implies that \[\omega_A\cap CE = \omega_B\cap CE\]where $\omega_A$ is the circle with center $A$ and radius $AE$ and $\omega_B$ is the circle with center $B$ and radius $BC$, so $P\in CE$, as $P$ is the intersection of $\omega_A$ and $\omega_B$ that lies within the pentagon. This proves the first part. For the second part, construct $Q$ to be the reflection of $E$ over $DA$. By the angle condition, $Q$ is also the reflection of $C$ over $BD$. Now, given $[BCD]=[DQB]$ and $[ADE]=[AQD]$, it is easy to see that the area condition is equivalent to $[ABP]=[BAQ]$. But this is obvious as $Q$ is the reflection of $P$ over $AB$, as it is the other intersection of $\omega_A$ and $\omega_B$, so we're done. This completes the proof. Remark: Noticing the two part split is key, and the motivation for noticing it is trying to understand what the angle condition is saying. Indeed, $D$ is the only point that we can move and have the quantity change linearly, so it is natural to try that. One would expect that if the rate of change is nonzero, that there is only one solution for $D$, but the problem seems to suggest that there are many (basically all values of $D$). Furthermore, if the rate of change is nonzero, we expect exactly one solution to exist, no matter where $P$ is, so by reverse logic, that point must be the point satisfying the angle condition. At this point, the two part split is forced. To solve the second of the two part split (the first part is pretty trivial), the key is to notice that we just fold in $BCD$ and $ADE$ into $ABD$, and $Q$ just formalizes this.

Here are two approaches. First solution using linearity Let $Q$ be the reflection of $P$ across $\overline{AB}$, so $[ABD] + [ABP] = [ABD] + [AQB] = [AQBD]$. Proceeding forwards, all areas are signed. Now $D$ varies on the perpendicular bisector of $\overline{EC}$. Claim: Define \[ f(X) = [BCX] + [AXE] - [AQBX] \]for $X$ on the perpendicular bisector of $\overline{EC}$. Then $f$ is linear. Proof. Obvious. $\blacksquare$ Assume for contradiction $P$, $C$, $E$ are not collinear. We now compute some special cases. $f(D) = 0$ by assumption. Let $O_1$ denote the circumcenter of $\triangle QCE$. Since $\triangle AEO_1 \cong \triangle AQO_1$ and $\triangle BCO_1 \cong \triangle BQO_1$, it follows $f(O_1) = 0$. Let $O_2$ denote the circumcenter of $\triangle PCE$. Since $\triangle AEO_2 \cong \triangle APO_2$ and $\triangle BCO_2 \cong \triangle BPO_2$, it follows $f(O_2) = [APBQ] = 2[ABP] \ne 0$. So $f$ is a nonconstant linear function, hence injective, and thus $D$ must actually coincide with $O_1$. (The figure is thus somewhat misleading.) [asy][asy] size(11cm); pair C = dir(160); pair E = dir(20); pair Q = dir(250); pair B = 0.65*(C+Q); pair A = 0.8*(E+Q); pair O_1 = origin; pair P = -Q+2*foot(Q, B, A); draw(Q--P, lightred); pair D = 1.3*(C+E); filldraw(A--B--C--D--E--cycle, invisible, black+dotted); draw(B--P--A--Q--cycle, dotted); draw(E--Q--C, deepgreen); draw(E--P--C, blue); draw(C--E); filldraw(CP(B, P), invisible, orange); filldraw(CP(A, P), invisible, orange); draw(A--B, red); pair O_2 = circumcenter(C, P, E); draw(B--O_1--A, deepgreen+dashed); draw(B--O_2--A, blue+dashed); draw(O_1--O_2, dashed); dot("$C$", C, dir(C)); dot("$E$", E, dir(E)); dot("$Q$", Q, dir(Q)); dot("$B$", B, dir(B)); dot("$A$", A, dir(A)); dot("$O_1$", O_1, dir(45)); dot("$P$", P, dir(90)); dot("$D$", D, dir(45)); dot("$O_2$", O_2, dir(O_2)); /* TSQ Source: !size(11cm); C = dir 160 E = dir 20 Q = dir 250 B = 0.65*(C+Q) A = 0.8*(E+Q) O_1 = origin R45 P = -Q+2*foot Q B A R90 Q--P lightred D = 1.3*(C+E) R45 A--B--C--D--E--cycle 0.1 lightcyan / black dotted B--P--A--Q--cycle dotted E--Q--C deepgreen E--P--C blue C--E CP B P 0.05 yellow / orange CP A P 0.05 yellow / orange A--B red O_2 = circumcenter C P E B--O_1--A deepgreen dashed B--O_2--A blue dashed O_1--O_2 dashed */ [/asy][/asy] However, in that case we then get \[ \angle EDC = \angle EDQ + \angle QDC = 2\angle ADQ + 2 \angle QDB = 2 \angle ADB \]which contradicts the given hypothesis. Second solution using Bretschneider (Ankan Bhattacharya) Let $Q$ be the reflection of $P$ over $\overline{AB}$, and let $F$ be the point such that $\triangle DFC \stackrel{+}{\cong} \triangle DAE$. Then the area condition reads \begin{align*} [DCB] + [DAE] & = [DBA] + [PBA]\\ \iff [DCB] + [DFC] & = [DBA] + [BQA]\\ \iff [DFCB] & = [DBQA]. \end{align*}Now, quadrilaterals $DBQA$ and $DBCF$ have the same side lengths in the same order and the same area. If they were congruent, then $\angle CDE = \angle FDA = 2 \angle BDA$, contradiction. [asy][asy] pair A, B, C, D, E, F, P, Q; A = dir(0); B = dir(180); P = dir(110); C = B + abs(B-P) * dir(140); E = 2*foot(A, P, C) - P; D = (C+E)/2 + rotate(90) * (E-C)/2 * 0.6; Q = reflect(A, B) * P; F = D + (C-D)*(A-D)/(E-D); fill(D--B--Q--A--cycle^^D--B--C--F--cycle, lightyellow); draw(A--P^^A--E, cyan); draw(A--Q^^C--F, cyan+2pt); draw(B--P, magenta); draw(B--Q^^B--C, magenta+2pt); draw(C--D^^D--E, brown); draw(D--A^^D--F, green+2pt); draw(D--B, orange+2pt); draw(C--E, dashed); dot("$A$", A, dir(300)); dot("$B$", B, dir(220)); dot("$C$", C, dir(220)); dot("$D$", D, dir(80)); dot("$E$", E, dir(30)); dot("$F$", F, dir(140)); dot("$P$", P, dir(50)); dot("$Q$", Q, dir(260)); [/asy][/asy] Thus, $DBQA$ and $DBCF$ are not congruent. From Bretschneider's formula (!) we obtain \begin{align*} \angle DFC + \angle DBC + \angle DBQ + \angle QAD & = 360^{\circ}\\ \iff \angle CBQ + \angle EAQ & = 360^{\circ} \end{align*}(interpreting $\angle CBQ$ as a reflex angle). Finally, \[ \angle CPE = \angle CPQ + \angle QPE = \frac{1}{2} (\angle CBQ + \angle QAE) = \frac{1}{2} \cdot 360^{\circ} = 180^{\circ} \]as desired.

I solved a G5 synthetically Rephrase the problem in terms of triangle $ABD$ to make the innate symmetry more clear. Let $\omega_A, \omega_B, \omega_D$ be the circles centered at $A, B, D$ respectively and with radii $AP = AE$, $BP = BC$, and $DC = DE$ respectively. Note that the area condition rewrites as $[ABD] = [PBA] + [CDB] + [EAD]$ with signed areas. Let $P', C', E'$ be the reflections of $P, C, E$ over sides $AB, BD, DA$ respectively. Now $[ABD] = [ABP'] + [BDC'] + [DAE']$ (again with signed areas), so the parallels to $AB, BD, DA$ through $P', C', E'$ respectively concur at some point $X$ (indeed, they concur at the point with homogenized barycentric coordinates $([ABP'] : [BDC'] : [DAE'])$). Let $Q$ be the radical center of $\omega_A, \omega_B, \omega_D$. Note $PP'$ is the radical axis of $\omega_A$ and $\omega_B$, et cetera, so $PP', CC', EE'$ concur at $Q$. Now visibly $P', C', E'$ are all on the circle with diameter $QX$. Now inverting at $Q$ with power $QP \times QP' = QC \times QC' = QE \times QE'$ shows that $P, C, E$ are collinear as desired. Remark: The condition $\angle EDC \neq 2\angle ADB$ was tacitly used to prevent $C'$ and $E'$ from being the same point.

The "only if" part can be shown by easy sine rule bash on the equation $CP+PE=CE$

This seems similar to spartacle's solution, although just in case it's new I'll keep it here for its cleanliness.

Let $O$ be the circumcenter of $\triangle CEP$ with circumradius $R$, whose angles are $\angle A, \angle B, \angle C$. Let $x, y, z$ be the signed positive lengths $OA, OB, OD$. Then $RHS \Longleftrightarrow 2[ABD] = [AEP]+[BPC]+[DCE]+[CEP] = [AEOP]+[BPOC]+[DCOE]$. This is equivalent to $\sum xR\sin A = \sum yz \sin A$. Assume none of $A, B, C, = 0$. Note that since $P$ lies on the same side of $AB$ as $D$ from convexity, $[DBP]+[PAD] = [DBPA] = [DBA] - [BAP] < [DBA]$ so $xR\sin A+yR\sin B < xy \sin C$. This means $xy\sin C + yz \sin A + xz \sin B = xR\sin A + yR \sin B + zR\sin C < xy \sin C + zR \sin C$ so $x \sin B + y \sin A < R \sin C$. However, since convexity, we have that $x > R \cos A, y > R \cos B$, so $R \sin C < R \sin C$ contradiction. Therefore $CEP$ is collinear, finishing the proof. (If $CEP$ is already collinear it's just make all the inequalities equalities and go backwards.)

Denote by $Q,R$ reflections of $P$ over $AB$ and it's midpoint respectively, $S=AE\cap BC, O$ is the circumcenter of $QCE.$ For arbitrary point $X$ we consider function $f(X)=\text{area} (AXE)+\text{area} (BCX)-\text{area} (AQBX),$ where areas are directed. Since there exist composition of reflections $E\stackrel{OA}{\mapsto} Q\stackrel{OB}{\mapsto} C$ it follows $f(O)=0$ and $\angle EOC=2\angle AOB$ so $D\neq O.$ Direct part. By assumption $f(D)=0=f(O).$ Suppose $P\notin CE$ and $O'$ is the circumcenter of $PCE,$ so there exist composition of reflections $E\stackrel{O'A}{\mapsto} P\stackrel{O'B}{\mapsto} C,$ therefore $f(O')=2\cdot \text{area} (APB)\neq 0$ and by linearity of $f$ it follows $D=O,$ contradiction $\Box$ Converse part. From $P\in CE$ we deduce $\angle ARB=\angle AQB=\angle APB=\pi-\angle BPC-\angle APE=\angle ASB,$ so $\angle RAE=\angle RBC\stackrel{\text{SAS}}{\implies} |RC|=|RE|.$ Now it's clear $f(R)=0,R\neq O$ and by linearity $f(D)=0$ for $D\in OR\text{ } \Box$

Let $P'$, $C'$, and $E'$ be reflections of $P$, $C$, $E$ over $AB$, $BD$, $DA$, respectively. There are three induced circles in the diagram, the circle centered at $D$ passing through $C$, $C'$, $E$, and $E'$; the circle centered at $B$ passing through $C$, $C'$, $P$, and $P'$; and the circle centered at $A$ passing through $P$, $P'$, $E$, and $E'$. By Radical Center, $CC'$, $EE'$, and $PP'$ concur, say at $X$. Let the parallel through $C'$ parallel to $DB$ and the parallel through $E'$ parallel $DA$. We have \[[BCD]+[ADE]=[BC'D]+[ADE']=[BYD]+[AYD]=[ABD]+[ABY]\]so it suffices to show that $PY\perp PP'$ if and only if $P$ is on $CE$. Evidently, $\angle YE'X=\angle YC'X=90^\circ$ so the first one is equivalent to $P'E'XC'$ being cyclic. But this is obvious: $\angle XCP=\angle XP'C'$ and $\angle XCE=\angle XC'E'$ so $P'E'XC'$ cyclic if and only if $P$ is on $CE$ as desired.

Same as others. I think I totally overcomplicated this problem when I first did it last May (?). Hopefully my geometry skills are not getting worse. Draw the circles centered at $A$ with radius $AE$, $B$ with radius $BC$, and $D$ with radius $DE=DC$. Let the second intersections of these circles be $E'$, $C'$, and $P'$, and note that the radical axes are perpendicular at their midpoints to $\overline{AD}$, $\overline{BD}$, and $\overline{AB}$, respectively. By radical axis theorem on these three circles, $EE'$, $CC'$, and $PP'$ concur at some point $Q$. By taking reflections, we rewrite the condition as \[ [AP'BD] = [ADE']+[BDC']. \]Thus, it suffices to show that $P$ lies on $\overline{CE}$ if and only if the line through $E'$ parallel to $AD$, the line through $C'$ parallel to $BD$, and the line through $P'$ parallel to $AB$ concur. Let the intersection of the first two lines be $R$. Then $RC'QE'$ is cyclic since $\angle RC'Q = \angle RE'Q = 90^{\circ}$, and furthermore $R$ is the antipode of $Q$ on that circle. On the other hand, $P'C'QE'$ is cyclic if and only if $P$ lies on $\overline{CE}$ since $P$ lying on $\overline{CE}$ is equivalent to \[ \angle P'C'Q + \angle P'E'Q = \angle P'PC + \angle P'PE = 180^{\circ}. \]Thus, $\angle RP'Q = 90^{\circ}$, or $\overline{P'R} \parallel \overline{AB}$, and we are done.