Excellent problem for practicing Involution. Let $PA, PB$ intersects the incircle at $U,V$ and let the tangent to incircle at $P$ intersects $DE$ at $T$. Let the incircle touches $AB$ at $F$ and let $Q = PF\cap DE$. Then by isogonality, there exists involution $\Psi$ which swaps $(D,E), (K,L), (T,\infty)$. Let $\Psi(Q)=R$, we find $$-1 = (PU;EF) = (TK;EQ) = (\infty L: DR)$$or $L$ is the midpoint of $DR$. Similarly $K$ is the midpoint of $ER$, implying the conclusion.

Construct a point $Q$ on line $DE$ such that $\measuredangle{DQP}=\measuredangle{FEP}$ so that $\triangle{DQP}\sim \triangle{FEP}$, where $F$ is the intersection of the incircle and segment $\overline{AB}$. Since $PA$ is the $P$-symmedian of $\triangle{FEP},$ from our similarity we obtain $LQ=DL$ as $\angle{KPE}=\angle{LPD}$. Similarly since $PB$ is the $P$-symmedian of $\triangle{FDP}$, we obtain $KQ=EK$ as $\triangle{EQP}\sim \triangle{FDP}$ and $\angle{KPE}=\angle{LPD}$. Thus we have $KL=KQ+LQ=\dfrac{1}{2}DE$, as desired.

Very nice! [asy][asy] unitsize(150); pair C, A, B, I, D, E, F, G, P, K, L, X; C = dir(130); A = dir(210); B = dir(330); I = incenter(A, B, C); D = foot(I, B, C); E = foot(I, C, A); F = foot(I, A, B); G = extension(A, D, B, E); P = point(incircle(A, B, C), intersections(incircle(A, B, C), G, (D+E)/2)[0]); K = extension(A, P, D, E); L = extension(B, P, D, E); X = K + L - (D+E)/2; draw(P--F^^P--X, gray(0.5)); draw(D--F--E, gray(0.5)); //draw(C--D^^C--E, gray(0.7)); draw(E--A--B--D); draw(A--P--B); draw(D--P--E); draw(D--E); draw(incircle(A, B, C)); dot("$A$", A, dir(A-I)); dot("$B$", B, dir(B-I)); //dot("$C$", C, dir(C-I)); dot("$D$", D, dir(D-I)); dot("$E$", E, dir(E-I)); dot("$F$", F, dir(F-I)); dot("$P$", P, dir(P-I)); dot("$K$", K, dir(130)); dot("$L$", L, dir(80)); dot("$X$", X, dir(320)); [/asy][/asy] I think this is essentially the same as the above solution, but more roundabout (essentially reproving the well-known symmedian property). Let the incircle touch $\overline{AB}$ at $F$. Construct the point $X$ on $\overline{DE}$ with $\triangle PDX \sim \triangle PFE$ and $\triangle PEX \sim \triangle PFD$. Since $\angle APF = \angle XPL$ and $\angle AFP + \angle PXD = \angle AFP + \angle PEF = 180^{\circ}$, \[\frac{PA}{AF} = \frac{PL}{XL}.\]Since $\angle EPA = \angle LPD$ and $\angle AEP + \angle PDL = \angle AEP + \angle PDE = 180^{\circ}$, \[\frac{PA}{AE} = \frac{PL}{DL}.\]Since $AE = AF$ it follows $XL = DL$. Similarly $XK = EK$ so $DE = 2KL$ as desired.

Let $PA,PB,AB, FK, FL$ meet the incircle again at $X,Y,F,K',L'$. By Pascal on hexagon $FK'YPXL'$, we know $KL, K'Y, L'X$ meet at some point $Z$. Now since $DD\cap FF =B\in PY$, we know $(P,Y;D,F)$ is harmonic. Projecting through $L$ yields $(Y,P;E,L')$ is harmonic. From $\angle EPA=\angle DPB$ we deduce $XY||DE$, hence projecting the harmonic pencil $(XY, XP; XE, XL')$ onto line $DE$ yields $(\infty_{DE}, K; E,Z)$ is harmonic, so $K$ is the midpoint of $EZ$. Similarly, $L$ is the midpoint of $DZ$, so $KL=\frac{1}{2}DE$ as desired.

Denote by $X,Y$ the second intersections of $PA, PB$ with the incircle, by $M,N$ the midpoints of $PE,PD,$ respectively, by $U,W$ the midpoints of $PX,PY,$ respectively. Since $\angle EPX= \angle EPA=\angle BPD=\angle DPY$, we get that $EXYD$ is an isosceles trapezoid, also $\Delta PEK \sim \Delta PYD, \Delta PEX \sim \Delta PLD$. Clearly $PEXF$ and $PFYD$ are the harmonic quadrilaterals, so $\Delta XEU \sim \Delta FXU$, so $ \angle XEU= \angle UXF= \angle PXF$ and $\angle NLD=\angle UEX=\angle PXF $ (because in the similarity $\Delta PEX \sim \Delta PLD, U,N$ fit to each other). Analogiously, $\angle MKE=\angle WDY=\angle WYF=\angle PYF$, so $ \pi = \angle PXF+\angle PYF=\angle NLD+\angle MKE$, so $MK||NL$. Also $MN||KL$ as the midline in $\Delta PED$, so $MKLN$ is a parallelogram and $\frac{DE}{2}=MN=KL$, which we had to prove.

What is IOM ...

International Olympiad of Metropolises, held in Moscow...

invert with center $ P, $ power $ PD \cdot PE, $ followed by reflection on the bisector of $ \angle DPE, $ denoting inverse points with $ ^{*}. $ Clearly, $ A^* \in PB, B^* \in PA $ and $ DE $ is tangent to $ \odot (PDA^*), \odot (PEB^*), \odot (PA^*B^*) $ at $ D, E, T, $ respectively, so $$ \left\{\begin{array}{cc} LD^2 = LP \cdot LA^* = LT^2 \\\\ KE^2 = KP \cdot KB^* = KT^2 \end{array}\right\| \Longrightarrow 2KL = DE. $$

My diagram has the $A$-labelling oops. Let $M, N$ be midpoints of $\overline{DE}, \overline{DF}$ respectively. Observe that $\triangle PFK \sim \triangle PDM$ and $\triangle PLE \sim \triangle PND$. Then $$FK+EL=\frac{PF\cdot DM+PE\cdot DN}{PD}=\tfrac{1}{2}EF$$by Ptolemy’s theorem on cyclic quadrilateral $DEPF$. $\blacksquare$

Same as above but posting it because it's really nice! CantonMathGuy wrote: The incircle of a triangle $ABC$ touches the sides $BC$ and $AC$ at points $D$ and $E$, respectively. Suppose $P$ is the point on the shorter arc $DE$ of the incircle such that $\angle APE = \angle DPB$. The segments $AP$ and $BP$ meet the segment $DE$ at points $K$ and $L$, respectively. Prove that $2KL = DE$. Let $M, N$ be the midpoints of $FE, FD$ respectively. Then since $PA, PB$ are the $P$ symmedians of $\triangle PEF, \triangle PDF,$ hence $\measuredangle FPM=\measuredangle KPE=\measuredangle DPL=\measuredangle NPF.$ Thus $\triangle PEK \sim \triangle PFN$ and $\triangle PLD \sim \triangle PMF$ yield \begin{align*} EK+LD &=\frac{PE}{PF} \cdot FN+\frac{PD}{PF} \cdot MF \\ &= \frac{1}{2PF} \left( PE \cdot FD+PD \cdot EF \right) \\ &\overset{\text{Ptolemy}}{=} \frac{1}{2PF} \cdot PF \cdot ED=\frac{1}{2} ED \end{align*}and so $2KL=ED,$ as desired. $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); 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 = -27.27583773109999, xmax = 72.74139959930677, ymin = -28.2254391648684, ymax = 46.29165267053318; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); 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Define $X := PA \cap \omega_{DPE}, Y := PB \cap \omega_{DPE}$, $N$ to the touchpoint of incircle with $AB$. Firstly note the following well known lemma, which Lemma Let $A,B,C,D$ be four con-cyclic harmonic points with $(A, B; C, D) = -1$. Then after an inversion $\Psi$ centered at $A$ with arbitrary radius, $\Psi(B)$ is the midpoint $\overline{\Psi(C)\Psi(D)}$. (This can be proved very easily with length chasing/inversion distance formula; I'm omitting the proof). Perform an arbitrary inversion $\Psi$ centered at $P$. Observe that $\angle \Psi(E)P\Psi(X) = \angle EPX = \angle EPA = \angle DPB = \angle DPY = \angle \Psi(D)P\Psi(Y)$ $\Psi(X), \Psi(E), \Psi(F), \Psi(Y)$ are colinear. Observe that $-1 = (P, X; E,N) = (P,Y; N,D)$. By lemma, $\Psi(X)$ is the midpoint of $\Psi(E)\Psi(N)$, and $\Psi(Y)$ is the midpoint of $\Psi(D)\Psi(N)$. Writing them as vectors, $\Psi(N) = 2 \Psi(X) - \Psi(E)$ and $\Psi(N) = 2 \Psi(L) - \Psi(D)$. Equating them, $2(\Psi(K)-\Psi(L)) = \Psi(E)-\Psi(D)$, which means $2\overline{\Psi(K)\Psi(L)} = \overline{\Psi(E)\Psi(D)}$ By inversion distance fomula, $2KL = DE \Leftrightarrow \frac{1}{2} = \frac{\Psi(K)\Psi(L) \cdot P\Psi(E) \cdot P\Psi(D)}{\Psi(E)\Psi(D) \cdot P\Psi(K) \cdot P\Psi(L)}$ Writing $\Psi(X) = K, \Psi(Y) = L, \Psi(K) = X, \Psi(L) = Y, \Psi(E) = E, \Psi(D) = D$ (btw this is what you would actually get after $\sqrt{PD \cdot PE}$ inversion after reflection along $P$ angle bisector), the new problem reads: IOM 2018/P6, inverted wrote: Let $\Delta PED$ be a triangle. Points $K, L \in \overline{ED}$ such that $\angle EPK = \angle LPD$ and $2 \overline{KL} = \overline{EF}$. Let $X = PK \cap \omega_{PED}, Y = PL \cap \omega_{PED}$. Prove that $\frac{\overline{XY} \cdot \overline{PE} \cdot \overline{PD}}{\overline{ED} \cdot \overline{PX} \cdot \overline{PY}} = \frac{1}{2}$. By $\angle PXE = \angle PDE = \angle PDL$, and $\angle EPK = \angle LPD$, so $\Delta PEX \sim \Delta PLD$. So $\frac{\overline{PE}}{\overline{PX}} = \frac{\overline{PL}}{\overline{PD}}$. Also, by very easy angle chasing, $XY || ED$, so $\frac{\overline{PL}}{\overline{PX}} = \frac{KL}{XY}$. Putting 'em together, $$ \frac{\overline{XY} \cdot \overline{PE} \cdot \overline{PD}}{\overline{ED} \cdot \overline{PX} \cdot \overline{PY}} = \frac{\overline{XY} \cdot \overline{PL} \cdot \overline{PD}}{\overline{ED} \cdot \overline{PD} \cdot \overline{PY}} $$$$ = \frac{\overline{XY} \cdot \overline{PL}}{\overline{ED} \cdot \overline{PY}} = \frac{\overline{XY} \cdot \overline{KL}}{\overline{ED} \cdot \overline{XY}} = \frac{KL}{ED} = \frac{1}{2}$$, as desired.

CantonMathGuy wrote: The incircle of a triangle $ABC$ touches the sides $BC$ and $AC$ at points $D$ and $E$, respectively. Suppose $P$ is the point on the shorter arc $DE$ of the incircle such that $\angle APE = \angle DPB$. The segments $AP$ and $BP$ meet the segment $DE$ at points $K$ and $L$, respectively. Prove that $2KL = DE$. Dušan Djukić Solution. We change the labellings as shown in the diagram below. Let $\Omega$ be the incircle. Let $X=\Omega\cap PB,~Y=\Omega\cap PC.$ Reflect everything upon the angle bisector of $\angle BAC.$ For a point $Z,$ let $Z'$ denote the image of the point $Z$ after reflection over the angle bisector of $\angle BAC.$ Note that $X$ and $Y$ swap their places with each other in this transformation since $\angle FPX=\angle EPY$ and the angle bisector of $\angle BAC$ is the perpendicular bisector of $EF.$ It is therefore obvious that $(P',X,C'),~(P', Y,B'),~(B',D',C')$ are triplets of collinear points. Define $J$ as the intersection of $\overline{PD'}$ and $\overline{EF}.$ Let $P_{\infty}$ be the point at infinity along the line $\overline{EF}.$ It is easy to see that $PP'\cap EF=P_{\infty}.$ [asy][asy] import graph; size(13cm); real labelscalefactor = 0.5; pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); pen dotstyle = black; real xmin = -9.436557641281185, xmax = 6.667007004924652, ymin = -4.031437069534323, ymax = 7.168620002740333; pen qqffff = rgb(0,1,1); pen wqwqwq = rgb(0.3764705882352941,0.3764705882352941,0.3764705882352941); /* draw figures */ draw((-4.482453181325971,6.478700055319052)--(xmin, 3.2302325581395372*xmin + 20.95806626197435), linewidth(0.4)); /* ray */ draw((-4.482453181325971,6.478700055319052)--(xmax, -0.95*xmax + 2.2203695330593765), linewidth(0.4)); /* ray */ draw((-6.823916595960116,-1.0847713003247377)--(3.2498213507216693,-0.8669607501262124), linewidth(0.4)); draw(circle((-3.2176138586912058,1.6346187905993497), 2.640798770333854), linewidth(0.8) + qqffff); 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label("$X$", (-5.647483606879812,0.8442096659086242), NE * labelscalefactor); dot((-0.6115082807676208,2.0612707120908293),linewidth(4pt) + dotstyle); label("$Y$", (-0.5489501635309063,2.167599494100281), NE * labelscalefactor); dot((-4.443101949640309,2.754288821332742),linewidth(4pt) + dotstyle); label("$K$", (-4.39374587490877,2.8641204563064164), NE * labelscalefactor); dot((-2.2726715253727092,3.3210104755813257),linewidth(4pt) + dotstyle); label("$L$", (-2.22060047282563,3.4352676453154474), NE * labelscalefactor); dot((-3.1487923044337536,3.0922463504200324),linewidth(4pt) + dotstyle); label("$J$", (-3.0982168852053595,3.1984505181653615), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Note that $P'FXD'$ is a harmonic quadrilateral, therefore, \begin{align*} (F,J;K,P_{\infty})\overset{P}=(F,D';X,P')=-1 \end{align*}So, $KJ=FK.$ Similarly, $JL=LE.$ Thus, $EF=2\cdot KL.$ And we are done.$\blacksquare$

Nothing new here. Consider an inversion at $P$ with radius $\sqrt{ PD \cdot PE}$ followed by a reflection over the angle-bisector of $\angle DPE$. This clearly swaps $D$ and $E$ Let $A', B', K'. L'$ denote the images of $A,B,K,L$ respectively. Since $\angle APE = \angle DPB$, it is clear that $P,A,K,B', L'$ are collinear, as are $P,B,L,A',K'$. Moreover, $(PA'B'), (PB'E)$, and $(PA'D)$ are all tangent to the line $DE$. Let $T$ be the point of tangency of of $(PA'B')$ to $DE$. Then by power of a point \begin{align*} KE^2 = KP \cdot KB' = KT^2 \\ LD^2 = LP \cdot LA' = LT^2 . \end{align*}Hence \[ KE + LD = KT + LT = KL \implies 2KL = DE, \]as desired.

Probably a better way to prove collinearity but I'm bad so it is what it is. Some definitions: Let $X,Y$ denote intersections of $AP, BP$ with incircle $\omega,$ respectively. Note the angle conditions imply $XY\parallel DE.$ Let $XD\cap EY = J$ and let $F$ denote the $C$ intouch point, $P'$ be the intersection of $\omega$ with line through $P$ parallel to $XY,$ $Q=P'J\cap \omega, M=QP\cap DE.$ Claim: $F,J,P$ are collinear. Proof. Since $FXEP, FYDP$ are harmonic and $XYDE$ is isosceles, $$\frac{XF}{XE} = \frac{PF}{PE} \qquad \frac{YD}{YF}=\frac{PD}{PF} \iff \frac{XF}{YF} = \frac{PD}{PE}$$$$\frac{\sin \angle FPX}{\sin \angle FPY} = \frac{\sin \angle FYX}{\sin \angle FXY} =\frac{XF}{YF} = \frac{PD}{PE} = \frac{\sin \angle DEP}{\sin \angle EDP} = \frac{\sin \angle DXP}{\sin \angle EYP}.$$Hence, applying Trig Ceva on $\triangle XYP$ implies $FP, XD, YE$ are concurrent. $\blacksquare$ Therefore, $QP'EX$ is a reflection of $FPDY$ and hence harmonic. Thus, $$(Q,E; X, P') \stackrel{P}{=} (M,E; L, \infty) = -1 \qquad \text{and} \qquad (Q,E; X, P') \stackrel{J}{=} (P', Y; D, Q) \stackrel{P}{=} (\infty, K; D, M) = -1$$Each implying $L$ is the midpoint of $M,E$ and $K$ is the midpoint of $M,D,$ respectively, so we're done.

CantonMathGuy wrote: The incircle of a triangle $ABC$ touches the sides $BC$ and $AC$ at points $D$ and $E$, respectively. Suppose $P$ is the point on the shorter arc $DE$ of the incircle such that $\angle APE = \angle DPB$. The segments $AP$ and $BP$ meet the segment $DE$ at points $K$ and $L$, respectively. Prove that $2KL = DE$. Dušan Djukić Took wayyy longer than I expected. Great problem! Let $D'$ and $E'$ be the midpoints of $DF$ and $EF$ respectively and $M \in DE$ such that $(PM,PF)$ are isogonal wrt $\angle EPD$. Claim 01. $\triangle PMD \sim \triangle PEF$ and similarly, $\triangle PME \sim \triangle PDF$. Proof. We have $\measuredangle PDM \equiv \measuredangle PDE = \measuredangle PFE$ and by definition, $\measuredangle EPF = \measuredangle MPD$. Then we are done. Now, notice that $PA$ is the symmedian of $\triangle PEF$. Similarly, $PB$ is the symmedian of $\triangle PDF$. Hence, we have \[ \measuredangle E'PF = \measuredangle EPA = \measuredangle BPD \]Claim 02. $\triangle LPD \sim \triangle E'PF$. Proof. To prove this, notice that $\measuredangle PDL \equiv \measuredangle PDE = \measuredangle PFE \equiv \measuredangle PFE'$ and $\measuredangle LPD = \measuredangle BPD = \measuredangle E'PF$. Hence, we are done. To finish this, notice that from the above two claims, we have \[ \frac{PD}{MD} = \frac{PF}{EF} \ \text{and} \ \frac{PD}{LD} = \frac{PF}{E'F} \]This is enough to conclude that $MD = 2 LD$. Similarly, $ME = 2KM$. Therefore, \[ DE = MD + ME = 2ML + 2KM = 2KL. \]

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); 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 = -12.28263718487124, xmax = 14.465298350412207, ymin = -9.127117983570622, ymax = 7.3920569277905965; /* image dimensions */ pen qqzzcc = rgb(0,0.6,0.8); pen qqttzz = rgb(0,0.2,0.6); pen qqzzff = rgb(0,0.6,1); pen ffqqtt = rgb(1,0,0.2); pen yqqqyq = rgb(0.5019607843137255,0,0.5019607843137255); pen qqttcc = rgb(0,0.2,0.8); draw((5.1559507895739305,6.104814200068148)--(-10.729057784076815,-7.499505718748189)--(11.115247550456308,-7.250395505123642)--cycle, linewidth(0.4) + qqzzcc); draw(arc((3.072441999264836,2.8758461726351774),0.8204888201007191,-158.54956644956755,-143.06578754745865)--(3.072441999264836,2.8758461726351774)--cycle, linewidth(0.4) + linetype("4 4") + qqzzff); draw(arc((3.072441999264836,2.8758461726351774),0.8204888201007191,-51.541508367939535,-31.81477987859267)--(3.072441999264836,2.8758461726351774)--cycle, linewidth(0.4) + linetype("4 4") + qqzzcc); /* draw figures */ draw((5.1559507895739305,6.104814200068148)--(-10.729057784076815,-7.499505718748189), linewidth(0.4) + qqzzcc); draw((-10.729057784076815,-7.499505718748189)--(11.115247550456308,-7.250395505123642), linewidth(0.4) + qqzzcc); draw((11.115247550456308,-7.250395505123642)--(5.1559507895739305,6.104814200068148), linewidth(0.4) + qqzzcc); draw(circle((3.2795798186087652,-2.229687021977993), 5.109733386146436), linewidth(0.4) + qqttzz); draw((-0.044184569637912494,1.6512878000123163)--(3.072441999264836,2.8758461726351774), linewidth(0.4) + ffqqtt); draw((3.072441999264836,2.8758461726351774)--(-10.729057784076815,-7.499505718748189), linewidth(0.4) + ffqqtt); draw((11.115247550456308,-7.250395505123642)--(3.072441999264836,2.8758461726351774), linewidth(0.4) + ffqqtt); draw((3.072441999264836,2.8758461726351774)--(7.945841541788876,-0.1475302535564318), linewidth(0.4) + ffqqtt); draw((-0.044184569637912494,1.6512878000123163)--(7.945841541788876,-0.1475302535564318), linewidth(0.4) + yqqqyq); draw(circle((3.072441999264836,2.8758461726351774), 4.382260917729214), linewidth(0.4) + qqttcc); draw(circle((2.892836567534781,1.9533789973135698), 0.9397892320370378), linewidth(0.4) + green); /* dots and labels */ dot((5.1559507895739305,6.104814200068148),linewidth(4pt) + dotstyle); label("$C$", (5.275823565284151,6.325421461659657), NE * labelscalefactor); dot((-10.729057784076815,-7.499505718748189),linewidth(4pt) + dotstyle); label("$B$", (-11.133952836730233,-7.349392206685724), NE * labelscalefactor); dot((11.115247550456308,-7.250395505123642),linewidth(4pt) + dotstyle); label("$A$", (11.238042324682711,-7.021196678645434), NE * labelscalefactor); dot((7.945841541788876,-0.1475302535564318),linewidth(4pt) + dotstyle); label("$E$", (8.065485553626596,0.0623568015574727), NE * labelscalefactor); dot((-0.044184569637912494,1.6512878000123163),linewidth(4pt) + dotstyle); label("$D$", (-0.24880115672735778,1.9221314604524444), NE * labelscalefactor); dot((3.072441999264836,2.8758461726351774),linewidth(4pt) + dotstyle); label("$P$", (3.169902260358972,3.098165435930147), NE * labelscalefactor); dot((1.1006570899967043,1.3935462335757312),linewidth(4pt) + dotstyle); label("$L$", (1.200729092117246,1.621285559748846), NE * labelscalefactor); dot((4.935479118947397,0.5302014832958309),linewidth(4pt) + dotstyle); label("$K$", (5.057026546590626,0.7460974849747417), NE * labelscalefactor); dot((3.8267312405510463,2.0584720054759034),linewidth(4pt) + dotstyle); label("$B'$", (4.154488844479835,2.086229224472589), NE * labelscalefactor); dot((1.9546744367412656,1.8981013617116638),linewidth(4pt) + dotstyle); label("$A'$", (2.267364558248181,1.6759848144222276), NE * labelscalefactor); dot((2.6864253618079252,1.036537563475007),linewidth(4pt) + dotstyle); label("$M$", (2.787007477645303,1.265740404371866), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Ez. Here is a shabbily presented but well detailed proof. We perform an inversion $\Gamma$ which is inverting the configuration about a circle centered at $P$ with radius $\sqrt{PD \cdot PE}$ and reflecting about angle bisector of $\angle PDE$ where $\Gamma$ takes a point $X$ to a point $X'$. We see that points $D$ and $E$ swap under $\Gamma$ and $\angle APE = \angle DPB$ implies that lines $\overline{PA}, \overline{PB}$ are isogonal with respect to $\triangle PDE$ and so $A' \in \overline{PB}$ and $B' \in \overline{PA}$. Now, $\Gamma$ takes line $\overline{DE}$ to $\odot (\triangle PDE)$ and therefore lines $\overline{AB}, \overline{BD}, \overline{AE}$ are all tangent to $\odot (\triangle PDE)$ implies that the circumcircles of $\triangle PDA', \triangle PEB', \triangle PA'B'$ are all tangent to $\overline{DE}$. Let $M = \odot (\triangle PA'B') \cap \overline{DE}$. Then since $K \in \overline{PA}$ belongs to radical axis of circumcircles of $\triangle PEB'$ and $\triangle PA'B'$ (which means that $K$ has equal power with respect to both these circles) and is also belonging to common tangent of these two circles, we have that $KM = KE$ and using symmetrical arguments we get that $L \in \overline{PB}$ belongs to radical axis of circumcircles of $\triangle PDA'$ and $\triangle PA'B'$ and is also belonging to common tangent of these two circles, we have that $LM = LD$. Therefore, $KL = KM + ML = \dfrac{KM + KE}{2} + \dfrac{LM + LD}{2} = \dfrac{EM + DM}{2} = \dfrac{DE}{2}$ or $2KL = DE$ as desired. Remark : This was my first solution. I later on also found involution solution by MarkBcC, the similarity solution and the Ptolemy solution too, however I think in terms of process involved in the proof, the inversion proof is the most easy-going proof and is rather straightforward! Also I am interested if someone found a proof using a different inversion like inverting about circumcircle of $\triangle PDE$ or something like that. Remark : This proof is not new nice job HKIS and TelvCohl

Let lines $AP$ and $BP$ meet the incircle at $X$ and $Y$, respectively. Let $M$ and $N$ be the midpoints of $\overline{EF}$ and $\overline{DF}$, respectively. Claim: We have $\triangle PEF \sim \triangle PKN$. Proof: It suffices to show that $\triangle PEK \sim \triangle PFN$. Since quadrilateral $FPDY$ is harmonic, it follows that $\angle KPE = \angle BPD = \angle NPF$. Moreover, $\angle PEK = \angle PFN$ so the similarity follows. Claim: Line $FP$ bisects $\overline{KN}$. Proof: Let $J$ be the midpoint of $\overline{KN}$. We have \[\angle JPN = \angle MPF = \angle EPA = \angle DPB = \angle FPN,\]so $J$ lies on $\overline{FP}$. Similarly, line $FP$ bisects $\overline{ML}$. Since $\overline{MN} \parallel \overline{KL}$, lines $FP$, $KN$ and $ML$ in fact concur. Therefore, it follows that $MNLK$ is a parallelogram.

Might be a fakesolve?? First, $A$ center the problem and let $BP$, $CP$ meet the incircle again at $X,Y$. Let $P'$ and $D'$ be on the incircle with $DD'\parallel PP' \parallel EF$. We are given that $XY \parallel EF$ as well and $DEXP$, $DFYP$ are harmonic. Reflect each of these across the perpendicular bisector of $EF$. Then, $D'FXP'$ and $D'EYP'$ are harmonic. Projecting these through $P$ onto $EF$ yields that $PX$ bisects $FT$ and $PY$ bisects $ET$, where $T=PD'\cap EF$. Thus $2KL=2TK+2TL=TE+TF=EF$.