First, note that the condition is equivalent to $c+b=3a$, so we have $s=2a$. We proceed with barycentric coordinates on $\triangle ABC$. Now we have $$B-\text{excircle } (\omega): -a^2yz-b^2zx-c^2xy+(x+y+z)((2a-c)^2x+(2a)^2y+a^2z)=0$$which gives us $E=(a:0:b-a)$. Similarly, $D=(0,-1,2)$. Next, $$C-\text{excircle }(\Omega): -a^2yz-b^2zx-c^2xy+(x+y+z)((2a-b)^2x+a^2y+4a^2z)=0$$which gives us $F=(a:2a-b:0)$. Note that $P$ can be parameterized as $(a:2a-b:t)$. Since $DE$ has equation $DE: ax+2ay+az=bx$, we find $t=3b-5a$ and so $P=(a:2a-b:3b-5a)$. We can now compute displacement vectors $\overrightarrow{BC}=(0,1,-1)$ and $$\overrightarrow{PB}=\left(\frac{a}{2b-2a}, \frac{4a-3b}{2b-2a}, \frac{3b-5a}{2b-2a}\right).$$Thus, by EFFT, it remains to show $$a^2\left(\frac{3b-4a}{2b-2a}+\frac{3b-5a}{2b-2a}\right)+b^2\left(\frac{-a}{2b-2a}\right)+c^2\left(\frac{a}{2b-2a}\right)=0,$$which is obvious as the LHS simplifies as $$\frac{a(3a-b-c)(3a-b+c)}{2(a-b)},$$which is $0$ as $b+c=3a$.

Beautiful problem! [asy][asy] unitsize(3cm); pair A, B, C, D, E, F, G, IB, IC; B = (0, 0); C = (1, 0); path EE = shift((B+C)/2)*xscale(3/2)*yscale(sqrt(2))*unitcircle; A = IP(EE, (0.3,0)--(0.3,1434)); D = 2 * C - B; E = C + abs(B - C) * unit(A - C); F = B + abs(B - C) * unit(A - B); G = 2 * B - C; IC = extension(G, G + rotate(90) * (B - G), F, F + rotate(270) * (B - F)); IB = extension(D, D + rotate(90) * (C - D), E, E + rotate(270) * (C - E)); draw(A--B--C--cycle, blue+linewidth(0.8)); draw(D--G, blue+linewidth(0.8)); draw(circle(IC, abs(IC-G)), blue+linewidth(0.4)); draw(IC--G, blue+linewidth(0.4)); draw(circle(IB, abs(IB-D)), blue+linewidth(0.4)); draw(B--E, blue+linewidth(0.4)); draw(G--F, blue+linewidth(0.4)); draw(E--D, red); draw(C--IC, red); draw(C--F, heavygreen); draw(B--IC, heavygreen); clip(box(G - (.5, .25), (D.x, A.y) + (.25, .25))); dot("$A$", A, N); dot("$B$", B, S); dot("$C$", C, S); dot("$D$", D, S); dot("$E$", E, NE); dot("$F$", F, NW); dot("$G$", G, S); //dot("$I_C$", IC, NW); dot(IC); [/asy][/asy] Let the $C$-excircle touch line $BC$ at $G$. It is well known that $CG = \tfrac12 (AB + BC + CA)$, so the length condition implies that $CG = 2BC$. It follows that $BG = BC$. Similarly, we can prove that $CD = BC$. Finally, we have $BG = BF$ by equal tangents. But this means that $B$ is the circumcenter of $\triangle CFG$ and also lies on $CG$, so $\triangle CFG$ must be a right triangle with right angle at $F$, i.e. $CF \perp FG$. Now observe that by symmetry, $FG$ is perpendicular to the $B$-external angle bisector, so $CF$ is parallel to the $B$-external angle bisector. Similarly, $ED$ is parallel to the $C$-internal angle bisector since both are perpendicular to the $C$-external angle bisector. It follows that if we translate the conclusion of the problem left by $BC$ (i.e. with the displacement vector $\overrightarrow{CB}$), we wish to show that the $B$-external angle bisector, the $C$-external angle bisector, and the line through $G$ perpendicular to $BC$ concur. But they do at the $C$-excenter, so we are done. $\blacksquare$

cool problem sketch: note that BED's circumcenter is C let BI_B intersect CF at X, now angle chase since you can get CF, DE angles to BC and also EBX to get that P,E,X,B concyclic therefore PEC and BEI_B are spirally similar since PEB+BEC=BEC+CEI_B therefore EPB=I_BCE=EBD and we're done

Quite a beautiful problem actually diagram credits to @cosmicgenius lol [asy][asy] unitsize(3cm); pair A, B, C, D, E, F, G, IB, IC, I, P, M, X, N; B = (0, 0); C = (1, 0); path EE = shift((B+C)/2)*xscale(3/2)*yscale(sqrt(2))*unitcircle; A = IP(EE, (0.3,0)--(0.3,1434)); D = 2 * C - B; E = C + abs(B - C) * unit(A - C); F = B + abs(B - C) * unit(A - B); G = 2 * B - C; IC = extension(G, G + rotate(90) * (B - G), F, F + rotate(270) * (B - F)); IB = extension(D, D + rotate(90) * (C - D), E, E + rotate(270) * (C - E)); I = incenter(A,B,C); P = intersectionpoint(C--C+100*(F-C), D--D+100*(E-D)); M = (B+P)/2; X = foot(I, B, C); N = (X+C)/2; draw(A--B--C--cycle, blue+linewidth(0.8)); draw(D--G, blue+linewidth(0.8)); draw(circle(IC, abs(IC-G)), blue+linewidth(0.4)); draw(circle(IB, abs(IB-D)), blue+linewidth(0.4)); draw(E--D, red); draw(C--M, red); draw(C--F, heavygreen); draw(B--I, blue); draw(B--M, blue); draw(I--N, blue); draw(I--X, dashed+blue); clip(box(G - (.5, .25), (D.x, A.y) + (.25, .25))); dot("$A$", A, N); dot("$B$", B, S); dot("$C$", C, S); dot("$D$", D, S); dot("$E$", E, NE); dot("$F$", F, NW); dot(G); dot("$I$", I, N); dot("$M$",M,N); dot("$X$",X,S); dot("$N$",N,S); [/asy][/asy] First note that $BC = CD$ by length chasing. Let $DE$ meet the line through $B$ perpendicular to $BC$ at $P'$; then the midpoint $M$ of $BP'$ is the intersection of $CI$ and the perpendicular to $BC$ at $B$, and by a homothety of factor 0.5 at $B$, it suffices to show that the line through $M$ and the midpoint of $BC$ is perpendicular to $BI$, since $BF = BC$. Note that this is equivalent to $BI \perp IN$ if $N$ is the midpoint of $CX$ and $X$ is the foot from $I$ to $BC$. So, it suffices to show that $XI^2 = XB \cdot XN$, or $XB \cdot XC = 2 \cdot XI^2$. By the Incenter-Excenter Lemma, the $A$-excenter $I_A$ is the antipode of $I$ wrt $(IBC)$, so by Power of a Point from $X$ to $(BIC)$ it suffices to show that the $A$-exradius is twice the inradius, which is true since $\frac{AB + AC + BC}{AB + AC - BC} = \frac{4 \cdot BC}{2 \cdot BC} = 2$. $\square$

let's actually not bash Let $D'$ be the foot of the perpendicular from $I_C$ to line $BC$. Note that $3a=b+c\rightarrow s=2a$. Claim: $D'B=BC=CD$. We have that $D'C=DB=s=2a$, and $BC=a$, so we have $D'B=a$ and $CD=a$. Note that $\angle D'FC=\angle BED=90$ due to $D'B=BF=BC=CE=CD.$ Note that $BI_C$ is perpendicular to $D'F$, since $D'B=BF$ and $BI_C$ bisects $\angle D'BF$, and since $CP$ is also perpendicular to $D'F$, we have that $$BI_C\parallel CP$$. On the other hand, we already have that $ED$ is perpendicular to $BE$. $CI_C$ is also perpendicular to $BE$ since $BC=EC$. Therefore, $$DP\parallel CI_C$$. Since $CD=BC$, line $CI_C$ is just line $DP$ translated left by $BC$, and $I_CB$ is just line $CP$ translated left by $BC$ as well. Therefore, point $I_C$ is just point $P$ translated parallel to $BC$ by a length of $BC$. Therefore, $\triangle I_CD'B$ and $\triangle PBC$ are congruent, so we are done.

[asy][asy] import geometry; unitsize(2.5cm); pair C = (0, 0), B = (1, 0); path A_ellipse = shift((B + C) / 2) * xscale(3 / 2) * yscale(sqrt(2)) * unitcircle; pair A = intersectionpoint((-0.3, 0)--(-0.3, 3), A_ellipse), O_1 = excenter(A, C, B), O_2 = excenter(A, B, C), D = foot(O_1, B, C), Q = foot(O_2, B, C), Y = foot(O_1, A, B), X = foot(O_2, A, C), EE = foot(O_1, A, C), F = foot(O_2, A, B), P = extension(D, EE, C, F); draw(A--B--C--cycle, linewidth(1.2)); draw(D--C, linewidth(1.2)); draw(B--Q); draw(A--Y); draw(A--X, linewidth(1.2)); draw(B--P, heavygreen); draw(Q--O_2, heavygreen); draw(B--Q, linewidth(1.2)); draw(P--O_2); draw(D--P, blue); draw(C--P, red); draw(O_1--C); draw(C--O_2, blue); draw(B--O_2, red); draw(excircle(A, B, C)); draw(excircle(A, C, B)); dot("$A$", A, dir(200)); dot("$B$", B, S); dot("$C$", C, S); dot("$D$", D, S); dot("$O_1$", O_1, N); dot("$O_2$", O_2, N); dot("$P'$", P, N); dot("$Q$", Q, S); dot("$X$", X, W); dot("$F$", F, dir(200)); dot("$E$", EE, dir(-30)); clip(box(D - (.5, .5), (Q.x, O_2.y) + (.5, .5))); [/asy][/asy] Let the $C$-excircle touch line $BC$ at $Q$ and line $AC$ at $X$. Also let $O_1$ be the $B$-excenter and $O_2$ be the $C$-excenter. Let $P'$ be the point so that $BQO_2P'$ is a rectangle. We show that $P=P'$. Note that $AB+BC+CA=AC+AX+BC+BQ=CX+CQ=2CQ$ and $AB+BC+CA=4BC$. Thus $CQ=2BC$, so $BC=BQ$. Similarly, $DC=CB$. Then, as both $FBQO_2$ and $BQO_2P'$ are cyclic, we know $FBQO_2P'$ is cyclic. Hence \[\angle{FP'B}=\angle{FQB}=\frac{\angle{FO_2Q}}{2}=\angle{BO_2Q}=90^\circ-\angle{P'O_2B}.\]But as $CB=BQ=P'O_2$ and $CB\parallel P'O_2$, $P'O_2BC$ is a parallelogram, so this is equal to $90^\circ-\angle{P'CB}=\angle{CP'B}$. Thus, $\angle{FP'B}=\angle{CP'B}$, so $C$, $F$, and $P'$ are collinear. Now, we also have \[\angle{EDC}=\frac{\angle{ECB}}{2}=\frac{\angle{XCQ}}{2}=\angle{O_2CQ}=\angle{P'O_2C}.\]Again, as $DC=CB=P'O_2$ and $DC\parallel P'O_2$, $P'O_2CD$ is a parallelogram, so this is equal to $\angle{P'DC}$. Hence, $D$, $E$, and $P'$ are collinear. Thus, $P'$ is the intersection of lines $CF$ and $DE$, so $P'=P$, and $BQO_2P$ is a rectangle. Thus $\angle{PBC}=90^\circ$, as desired.

Let the $C$-excircle touch $AC$ and $BC$ at $X$ and $G$, respectively, and let the $B$-excircle touch $AB$ at $Y$. Since $s=2a$, we have that $a=BG=BF=CD=YF=EX$. Claim. $XYFPE$ is cyclic. Proof. Using the isosceles triangles, note $\angle CPD = \angle PCB - \angle PDB = \pi/2 - \angle B/2 - \angle C/2 = \angle A/2$. Then, it's immediate from $\angle FXA = \angle AYE = \angle A/2$. Since $BG=BF=BC$, $\angle GFC = 90^\circ$. By Power of a Point, $CP\cdot CF = CE\cdot CX = CB\cdot CG$, so $PFGB$ is cyclic, and we're done.

The condition implies that $a = b + c \Rightarrow BF = BC = CE$ and $AF + AE = BC$. Note also that $CD = CE = BC$. Let $Q$ be the projection of $B$ onto $CF$, $I_B$ be the $B$-excentre of $\triangle ABC$, and $G$ the contact point of the $B$-excircle with $AB$. Note that since $AG = AE$, $FG = BC = CD$ and so $BG = 2 \cdot BC = BD$. Claim 1: $PBQE$ cyclic. Proof: Let $\theta_A = 1/2 \hat{A}$, and definte $\theta_B, \theta_C$ analogously. A homothety scale factor 2 from $B$ maps $A \to G, C \to D$ hence $FC \parallel GD$. Now $\measuredangle EPQ = \measuredangle DPC = \measuredangle PDG = \theta_A$ by alternate segment theorem. Furthermore, $\measuredangle EBQ = \measuredangle EBC - \measuredangle QBC = 90 - \theta_C - \theta_B = \theta_A$. Hence $\measuredangle EPQ = \measuredangle EBQ$. $\square$ Claim 2: $B - Q - I_B$. Proof: $\triangle FBC$ isosceles $\Rightarrow BQ$ is the internal bisector of $\hat{B}$. $I_B$ also lies on this bisector. $\square$ Claim 3: $(PBQE)$ tangent to $BC$. Proof: Note that since $CB = CE = CD$, $BD$ is the diameter of $(EBD)$. By Claim 2, $\measuredangle I_BQC = 90 = \measuredangle I_BEC = \measuredangle I_BDC$ hence $I_BEQCD$ cyclic. But note that $C$ is the circumcentre of $\triangle EBD$, and since $I_B$ is the second point of intersection of the perpendicular bisector of $ED$ with $(ECD)$, $BI_B$ is the $B$-symmedian. Thus $Q$ is the $B$-dumpty point, and in particular $\measuredangle EBQ = \measuredangle BDQ$. Consider $\triangle PBC$. $Q$ lies on the median $PC$ and also satisfies $\measuredangle CDQ = \measuredangle EBQ = \measuredangle DPQ$. Hence $Q$ is the $P$-humpty point, and $(PBQ)$ is tangent to $BC$. $\square$ By alternate segment theorem, $\measuredangle PBC = \measuredangle BQP = 90$. $\blacksquare$

Nice problem! Let the $C$-excircle touch $BC$ at $G$. Since $a={b+c}/{3} \implies s=2a \implies CF=CD=CB=BF=BG$ and hence $\angle BED=\angle GFC=90$ Redefine $P$ to be $G$-antipode in $(GBF)$. Hence, $\angle PFG=90\implies P-F-C$. $\angle DBE=90-\angle BDE = 90-C/2$. Similarly $\angle GCP=90-B/2$. Hence $\angle CPD=\angle PCB- \angle PDC=A/2$. Also $\angle BPC=\angle BPF=\angle BGF=B/2$. Hence $\angle BPE = \angle BPC+\angle CPD = B/2 + A/2 = 90-C/2 = \angle DBE \implies BD$ tangent $(BPE)$. As $\angle GBP=90 \implies BEP = 90 \implies P-E-D$. Hence $ P = CF \cap DE$, and as $\angle GBP=90$, we are done!

Point circle strikes again… Let $B$-excircle be $w_B$ with center $I_B$. Like many other solutions above, note that $BF=BC=CD=CE$ by length chasing. Since $BF=BC$ and $BI_B$ bisects $\angle ABC$, we have $BI_B\perp FC$. However, we know that $BC^2=DC^2$ so $C$ lies on the radical axis of $(B,0)$ and $w_B$. Thus, $FC$ is the radical axis of $(B,0)$ and $w_B$ but $P-F-C$ is collinear. We have $$PB^2=PE\cdot PD.$$It’s easy to see that we’re left to prove that $BE\perp PD$ which is true since $CB=CE=CD$ andddd we’re done!

OO cute problem! I am not sure if this is correct or not, so can someone check? If $BC=a$. Note that $CE=CD=\frac{AB+BC+CA}{2}-C=a$. So $C$ is the centre of $(BDE)$ and hence $\angle BDE=90$. So $C$ is midpoint of $BE$. Define $I$ as the incentre of $ABC$. Let the incircle touch $CA$ at $M$, $BC$ at $L$. Let $M$ antipode be $N$ and $L$ antipode be $J$. Claim: $B-N-J-D$ collinear ( By symmetry we also get $C-J-F$) Proof: By homothety, $B-N-D$. Note that if $\angle BCA=2\theta$, we get $\angle BED=\theta\implies CI||EA$. We have $\angle EBD=90-\theta$ and $\angle ILM =\theta\implies \Delta LMJ \text{ is similar to } \Delta EDB$. Note that $\angle MJN=90$. Since $JL\perp BC\implies \angle LJB=90-\angle LBJ=\theta$. And $\angle LJN=\angle JLM=\theta\implies J-N-B$. Define $P'$ as the intersection of the perpendicular at $B$ to $BC$ and $CE$. Define $C'=CI\cap BC$. Since $LJ||BP'$ and $I$ is midpoint of $NJ\implies C'$ is midpoint of $BP'$. Now define $Y$ as midpoint of $NJ$. Note that $C-I-Y-C'$ ( as $CI$ is perpendicular bisector of $LM$ and $LMNJ$ rectangle). Note that $\Delta BCY$ is similar to $\Delta BED$. So $E-D-P'$. So $P'=P$. And we are done!

Here is an "honors-geometry-only" solution not using either cyclic quadrilaterals or power of a point, just good old congruent triangles... found by a contestant. Defining $a$, $b$, $c$, $s$ in the usual way, the condition requires $s-a=a$, ergo \[ a = s-a = BK = BF = CE = CD \]where $K$ is the contact point of the $C$-excircle on $\overline{BC}$. We have $\overline{I_CB} \parallel \overline{CFP}$ because they're both perpendicular to the $\angle B$-bisector, and $\overline{I_CC} \parallel \overline{PED}$ because they're both parallel to the $\angle C$-bisector. Since $BC = CD$, it follows by ASA congruence that \[ \triangle I_CBC \cong \triangle PCD. \] [asy][asy] size(12cm); pair B = dir(232); pair C = dir(308); pair A = dir(110); pair I = incenter(A, B, C); pair D = 2*C-B; pair E = -B+2*foot(B, C, I); pair F = -C+2*foot(C, B, I); pair I_B = extension(C, midpoint(D--E), B, I); pair I_C = extension(A, I_B, C, I); draw(B--D); pair K = 2*B-C; pair P = extension(D, E, C, F); filldraw(B--I_C--C--cycle, invisible, deepgreen+1.5); filldraw(C--P--D--cycle, invisible, deepgreen+1.5); filldraw(A--B--C--cycle, invisible, red+1.5); draw(K--D, red); pair T_B = foot(I_B, A, B); pair T_C = foot(I_C, A, C); draw(T_B--A--T_C, red); filldraw(CP(I_B, D), invisible, grey); filldraw(CP(I_C, F), invisible, grey); draw(P--B, blue); draw(I_C--K, blue); clip(box((-2.3,-1),(2.3,2))); dot("$B$", B, dir(225)); dot("$C$", C, dir(315)); dot("$A$", A, dir(A)); dot("$D$", D, dir(315)); dot("$E$", E, dir(75)); dot("$F$", F, dir(200)); dot("$I_C$", I_C, dir(I_C)); dot("$K$", K, dir(225)); dot("$P$", P, dir(90)); [/asy][/asy] The foot from $I_C$ to line $BC$ is $K$, so since $KB = BC = CD$, it follows the from $P$ to line $BC$ is $B$.

sorry not sorry First, the length condition implies that $s=2a$, so $s-a=a$, and we thus have $BC=BF=CE=CD$. We employ cartesian coordinates. WLOG let $BC=1$, and set $A=(0,a)$, $B=(c-1, 0)$ and $C=(c,0)$, so $D=(c+1,0)$, ignoring the "sidelength definition" of $a$ and other variables. The $AB+AC=3BC$ condition translates into $$\sqrt{a^2+c^2}+\sqrt{a^2+(c-1)^2}=3\qquad(1).$$Because $CE=1$, we can easily compute $$E=\left(c\left(1-\frac{1}{\sqrt{a^2+c^2}}\right),\frac{a}{\sqrt{a^2+c^2}}\right),$$and similarly $$F=\left((c-1)\left(1-\frac{1}{\sqrt{a^2+(c-1)^2}}\right),\frac{a}{\sqrt{a^2+(c-1)^2}}\right).$$Evidently it is sufficient to show that the slope of $\overline{CF}$ is twice the slope of $\overline{DE}$. We have \begin{align*} \frac{\frac{a}{\sqrt{a^2+(c-1)^2}}}{(c-1)\left(1-\frac{1}{\sqrt{a^2+(c-1)^2}}\right)-c}&=2\cdot \frac{\frac{a}{\sqrt{a^2+c^2}}}{c\left(1-\frac{1}{\sqrt{a^2+c^2}}\right)-(c+1)} &\iff\\ c\left(\sqrt{a^2+c^2}-1\right)-(c+1)\sqrt{a^2+c^2}&=2\left((c-1)\left(\sqrt{a^2+(c-1)^2}-1\right)-c\sqrt{a^2+(c-1)^2}\right) &\iff\\ -c-\sqrt{a^2+c^2}&=2\left(-c-\sqrt{a^2+(c-1)^2}+1\right)&\iff\\ c&=2+\sqrt{a^2+c^2}-2\sqrt{a^2+(c-1)^2}&\iff\\ c+4&=3\sqrt{a^2+c^2}, \end{align*}where we use $(1)$ to obtain the last equality. Noting that $\left(\sqrt{a^2+c^2}\right)^2-\left(\sqrt{a^2+(c-1)^2}\right)^2=2c-1$, the last equation is then equivalent to $$\frac{\left(\sqrt{a^2+c^2}\right)^2-\left(\sqrt{a^2+(c-1)^2}\right)^2+1}{2}+4=3\sqrt{a^2+c^2}.$$By the difference of squares factorization and $(1)$, this becomes $$\frac{3\left(\sqrt{a^2+c^2}-\sqrt{a^2+(c-1)^2}\right)+1}{2}+4=3\sqrt{a^2+c^2} \iff 9=3\sqrt{a^2+c^2}+3\sqrt{a^2+(c-1)^2},$$which is just $(1)$. Thus we are done. $\blacksquare$

Ok so here is an actual synthetic-ish solution with significant help from CT17 As above we have $BC=BF=CE=CD$. Furthermore it is evident that $\angle CDE=\frac{\angle D}{2}$. We also have $\angle BCF=90^\circ-\frac{\angle B}{2}$, since the $B$-angle bisector is perpendicular to $\overline{CF}$ by symmetry (since $BC=BF$). The problem is equivalent to proving that $\tan \angle BCF=2\tan \angle BDE$, or $$\frac{r}{s-b}=\frac{2r}{s-c}$$by considering the triangles formed by the incenter $I$ and the $A$-intouch point, as well as $B$ and $C$. To this end, we will prove the following key claim. Claim: If $Q$ is the antipode of the $A$-intouch point with respect to the incircle, then $I$ is the orthocenter of $\triangle QBC$. Proof: Clearly $I$ lies on the $Q$-altitude of $\triangle QBC$, so we just have to prove that it also lies on the $C$-altitude. Consider the point $Q'$, which is the antipode of the $B$-intouch point. It is well-known that $Q'$ lies on $\overline{BE}$ (the proof is by homothety). But because $CB=CE$, $Q$ also lies on $\overline{BE}$ by symmetry. Again, because $CB=CE$, $\overline{CI} \perp \overline{BE}$, so we are done. $\blacksquare$ To finish, from our above claim we obtain $\triangle BDI \sim \triangle QDC$, which implies the desired ratio above. $\blacksquare$