I think $T$ must be between $E$ and $D$, here I found a nice solution, but if it is not, then I am getting a contradiciton!

Actually, I think this point T is not well defined, as I was trying to draw the figure here...

Pascual2005 wrote: I think $T$ must be between $E$ and $D$, here I found a nice solution, but if it is not, then I am getting a contradiciton! Actually, I found half of it to be true when $T$ is outside the segment $DE$, but in my drawing $D$ lies on the segment $AB$, while $E$ liues on the ray $(CA$, with $A$ between $E$ and $C$. I think it might depend on this, but I'm too tired to analyze the cases. I'll just go ahead and write what my drawing is showing me . Suppose first that $P$ is the second intersection of $AT$ with the circle $(ADE)$, and let $Q$ be the midpoint of the arc $DE$ of $(ADE)$ which does not contain $A$. By Ptolemy's Theorem applied to the quadrilateral $PDQE$, we get $PD\cdot QE+PE\cdot QD=QP\cdot DE$, and, since $QD=QE$ and $DE=AB=AC$, we get $(PD+PE)\cdot QD=QP\cdot AB\ (*)$. However, the triangles $ABC,QDE$ are similar, and $\angle DQP=\angle BAT$, while $P,T$ lie on the circumcircles of $QDE,ABC$ respectively, so the figures $QDEP,ABCT$ are similar. This gives $AT\cdot QD=QP\cdot AB$. If we compare this with $(*)$ we get exactly what we want: $PD+PE=AT$. The converse is not true though, I believe. In the general case, there will be a second point $P'$ on $AT$ satisfying $P'D+P'E=AT$. This can be seen as follows: the locus of the points $X$ with $XD+XE=AT$ is an ellipse with foci $D,E$, passing through $P$, the second intersection point between $AT$ and $(ADE)$, as was shown above. In order to prove that $P$ is the only intersection point between this ellipse and $AT$, we would have to show that $AT$ is tangent to the ellipse in $P$ or, in other words, that it is the external angle bisector of $\angle PDE$. However, it's clear from the construction above that $PQ$ is the angle bisector of $\angle PDE$, so the converse of what we have shown in the paragraph above holds iff $AT\perp QP$, which is equivalent to $DE\perp BC$, for example.

According to the official statement, $T$ is the intersection of segment $DE$ and $(ABC)$, therefore $T$ is between $D$ and $E$.

Does this work China TST 2000 P1 wrote: Let $ABC$ be a triangle such that $AB = AC$. Let $D,E$ be points on $AB,AC$ respectively such that $DE = AC$. Let $DE$ meet the circumcircle of triangle $ABC$ at point $T$. Let $P$ be a point on $AT$. Prove that $PD + PE = AT$ if and only if $P$ lies on the circumcircle of triangle $ADE$. Solution: Suppose $P \equiv AT \cap \odot (ADE)$. Let $F$ be a point on $PD$ such $PF=PE$ $\implies$ $\Delta FDE \cong \Delta TAC$ which follows due to angle chasing $\angle CTA=\angle ABC=\angle DFE$ ($\Delta ABC \sim \Delta PFE$) Now, for the Converse, Suppose $PD+PE=AT$. Let $P' \equiv AT \cap \odot (ADE)$. A similar procedure shows $P'D+P'E$ $=$ $PD+PE$ $\implies$ $P' \equiv P$ $\qquad \blacksquare$ [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 = -6.388586639866752, xmax = -0.5608141615290031, ymin = -4.488179703036392, ymax = -1.024954254843885; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); draw((-4.161474257432977,-1.309540257225541)--(-4.98,-3.27)--(-3.4899584286477756,-3.32509310594351)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-4.161474257432977,-1.309540257225541)--(-4.98,-3.27), linewidth(0.8)); draw((-4.98,-3.27)--(-3.4899584286477756,-3.32509310594351), linewidth(0.8) + rvwvcq); draw((-3.4899584286477756,-3.32509310594351)--(-4.161474257432977,-1.309540257225541), linewidth(0.8)); draw(circle((-4.203388378118096,-2.443144565385998), 1.1343789142049292), linewidth(0.4)); draw((-4.676513856607254,-2.5431170986199776)--(-3.231600218980625,-4.100554536877056), linewidth(0.4) + dtsfsf); draw((-4.676513856607254,-2.5431170986199776)--(-3.708747275481901,-3.9741111444537656), linewidth(0.4) + dtsfsf); draw((-3.708747275481901,-3.9741111444537656)--(-3.231600218980625,-4.100554536877056), linewidth(0.4) + sexdts); draw((-4.676513856607254,-2.5431170986199776)--(-3.4899584286477756,-3.32509310594351), linewidth(0.4) + dtsfsf); draw((-3.231600218980625,-4.100554536877056)--(-3.432220064335965,-4.382999816968901), linewidth(0.4)); draw(circle((-3.0610097105440572,-2.493310555357611), 1.6162717400546005), linewidth(0.4) + linetype("4 4")); draw((-4.161474257432977,-1.309540257225541)--(-3.708747275481901,-3.9741111444537656), linewidth(0.4) + dtsfsf); draw((-3.4899584286477756,-3.32509310594351)--(-3.231600218980625,-4.100554536877056), linewidth(0.4)); draw((-3.432220064335965,-4.382999816968901)--(-3.708747275481901,-3.9741111444537656), linewidth(0.4) + linetype("2 2") + sexdts); /* dots and labels */ dot((-4.161474257432977,-1.309540257225541),dotstyle); label("$A$", (-4.310651370951243,-1.2303593503918546), NE * labelscalefactor); dot((-4.98,-3.27),dotstyle); label("$B$", (-5.180040380014743,-3.308294619307359), NE * labelscalefactor); dot((-3.4899584286477756,-3.32509310594351),dotstyle); label("$C$", (-3.4078243230776084,-3.3990550103634383), NE * labelscalefactor); dot((-4.676513856607254,-2.5431170986199776),dotstyle); label("$D$", (-4.855213717287722,-2.4962279624898054), NE * labelscalefactor); dot((-3.231600218980625,-4.100554536877056),dotstyle); label("$E$", (-3.173758051406666,-4.235005980616802), NE * labelscalefactor); dot((-3.789421973810941,-3.499291968496834),dotstyle); label("$T$", (-3.770865887301927,-3.4516004999222214), NE * labelscalefactor); dot((-3.708747275481901,-3.9741111444537656),dotstyle); label("$P$", (-3.6896592216201713,-3.9245099059512674), NE * labelscalefactor); dot((-3.432220064335965,-4.382999816968901),dotstyle); label("$F$", (-3.670551770871523,-4.36875813585734), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

i have a nicer solution by length chasing BUT failed to contradict many cases