We $\sqrt{bc}$-invert, denoting images with $\bullet'$. Then we can rearrange \begin{align*} \angle DPA+\angle AQD &= \angle QIP \\ \iff 360^\circ - \angle PAQ - \angle PDQ &= \angle AQ'I_A+\angle AP'I_A \\ &= 360^\circ - \angle P'AQ' - \angle P'I_AQ'. \end{align*}Since $\angle PAQ = \angle P'AQ'$, it suffices to show that $\angle PDQ = \angle P'I_AQ'$. [asy][asy] defaultpen(fontsize(10pt)); size(300); pair A, B, C, D, E, F, I, M, IA, EE, FF, P, Q, PP, QQ, X, IA; A = dir(130); B = dir(210); C = dir(330); I = incenter(A, B, C); D = foot(I, B, C); E = foot(I, C, A); F = foot(I, A, B); X = extension(E, F, B, C); P = intersectionpoints(Line(E, F, 20), circumcircle(A, B, C))[1]; Q = intersectionpoints(Line(E, F, 20), circumcircle(A, B, C))[0]; IA = 2*dir(270) - I; EE = extension(A, B, IA, rotate(90, IA)*A); FF = extension(A, C, IA, rotate(90, IA)*A); PP = intersectionpoints(Line(B, C, 20), circumcircle(A, EE, FF))[1]; QQ = intersectionpoints(Line(B, C, 20), circumcircle(A, EE, FF))[0]; M = midpoint(B--C); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(incircle(A, B, C), heavyred); draw(Q--X--B^^C--PP, orange); draw(B--EE--FF--C, orange); draw(circumcircle(A, EE, FF), lightblue); draw(arc(circumcenter(P, D, Q), circumradius(P, D, Q), 170, 380), heavygreen+dashed); dot("$A$", A, dir(A)); dot("$B$", B, dir(215)); dot("$C$", C, dir(270)); dot("$D$", D, dir(270)); dot("$E$", E, dir(80)); dot("$F$", F, dir(130)); dot("$P$", P, dir(150)); dot("$Q$", Q, dir(20)); dot("$X$", X, dir(200)); dot("$E'$", EE, dir(EE)); dot("$F'$", FF, dir(FF)); dot("$P'$", PP, dir(10)); dot("$Q'$", QQ, dir(300)); dot("$I_A$", IA, dir(300)); dot("$I$", I, dir(240)); dot("$M$", M, dir(270)); [/asy][/asy] Claim: We have $PQCB\sim P'Q'E'F'$. Proof. Both quadrilaterals are cyclic, so it suffices to show that the angles are equal. We have \begin{align*} \angle PBC &= 180^\circ - \angle PAC = \angle P'F'E' \\ \angle PCB &= \angle PAB = \angle P'AF' = \angle P'Q'F' \end{align*}and analogously $\angle QCB = \angle Q'E'F'$ and $\angle Q'P'E' = \angle QBC$. $\blacksquare$ Claim:Let $M$ be the midpoint of $\overline{BC}$. Then $PQMD$ is cyclic. Proof. Let $X = \overline{EF}\cap \overline{BC}$. Then $(\overline{XD}; \overline{BC}) = -1$, so $XD\cdot XM = XB\cdot XC = XP\cdot XQ$, which is enough. $\blacksquare$ Combining these two claims solves the problem, as now we have $\angle P'I_AQ' = \angle PMQ = \angle PDQ$. Remark: The main motivation for $\sqrt{bc}$-inversion is to deal with the weird angle condition; after inversion at $A$ or $D$ it reduces to a single angle equality. However, if you're not careful with the manipulations, you might have to deal with $D'$, the mixtilinear ex-touch point.

parmenides51 wrote: Let $I$ be the incentre of acute-angled triangle $ABC$. Let the incircle meet $BC, CA$, and $AB$ at $D, E$, and $F,$ respectively. Let line $EF$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\angle DPA= \angle AQD =\angle QIP$. (Slovakia) slovakia? whats that??

We let lines $QD$ and $PD$ meet $(ABC)$ again at $X$ and $Y$, respectively. [asy][asy] size(10cm); pair A = dir(130); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair D = foot(I, B, C); pair E = foot(I, C, A); pair F = foot(I, A, B); pair T = extension(B, C, E, F); pair P = IP(T--F, unitcircle); pair Q = -P+2*foot(origin, T, P); draw(A--B--C--cycle, lightblue); filldraw(unitcircle, invisible, mediumcyan); filldraw(incircle(A, B, C), invisible, red); draw(B--T--Q, lightblue); pair X = -Q+2*foot(origin, Q, D); pair Y = -P+2*foot(origin, P, D); draw(X--A--Y, red); draw(P--Y, deepgreen); draw(Q--X, deepgreen); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(I)); dot("$D$", D, dir(270)); dot("$E$", E, dir(70)); dot("$F$", F, dir(130)); dot("$T$", T, dir(T)); dot("$P$", P, dir(135)); dot("$Q$", Q, dir(Q)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); /* TSQ Source: A = dir 130 B = dir 210 C = dir 330 I = incenter A B C D = foot I B C R270 E = foot I C A R70 F = foot I A B R130 T = extension B C E F P = IP T--F unitcircle R135 Q = -P+2*foot origin T P A--B--C--cycle lightblue unitcircle 0.1 lightcyan / mediumcyan incircle A B C 0.1 lightred / red B--T--Q lightblue X = -Q+2*foot origin Q D Y = -P+2*foot origin P D X--A--Y red P--Y deepgreen Q--X deepgreen */ [/asy][/asy] Claim: Line $AX$ is the polar of $P$ with respect to the incircle. Similarly, line $AY$ is the polar of $Q$ with respect to the incircle. Proof. Let $\overline{PQEF} \cap \overline{BC} = T$, so \[ -1 = (DT;BC) \overset{Q}{=} (XP;BC) \overset{A}{=} (P, \overline{AX} \cap \overline{FE}, F, E). \]In addition, $A$ is on the polar of $P$ already, so this proves the claim. $\blacksquare$ Now $\overline{IP} \perp \overline{AX}$ and $\overline{IQ} \perp \overline{AY}$, so \[ \measuredangle PIQ = \measuredangle XAY = \measuredangle APY + \measuredangle XQA = \measuredangle DQA + \measuredangle APD \]as needed.

Draw $T$, the center of spiral similarity $\phi$ sending $FE$ to $BC$. Clearly, $T = (AFE) \cap (ABC)$. It suffices to show that\[\angle TPD + \angle TQD = \angle PIQ.\]Let $P', Q'$ be on line $BC$ such that $\phi(P) = P'$ and $\phi(Q) = Q'$. Let $M$ be the midpoint of arc $BC$. It is easy to see that $\phi(I) = M$. Note that it remains to show that $T$ is the center of spiral similarity sending $DQ \to P'M$ and $DP \to Q'M$ and the result just follows since then\[\angle TPD + \angle TQD = \angle P'MQ' = \angle PIQ.\]Note that proving $T$ is the spiral center sending $DQ \to P'M$ is the same as proving that $T$ is the spiral center sending $DP' \to QM$. Indeed, note that\[\angle TMQ = \angle TPQ = \angle TP'Q' = \angle TP'D\]and\[\angle TQM = \overarc{TM} = \overarc{TD} = \angle TDP'\]where we use $\overarc{TD}$ on the circle $\gamma$ formed by $(ABC)$ shrunk homothety at $T$ to be tangent to $BC$ at $D$. Such a circle clearly exists by shooting lemma $MD \times MT = MI^2$. So indeed, $T$ is the spiral center sending $DQ \to P'M$ and thus similarly $DP \to Q'M$ and we are done. Remark: Defining $T$ was nonobvious.

yay multiple inversions! Invert about the incircle to get this diagram, where $A'$, $B'$, and $C'$ are midpoints of sides, $I$ is the circumcenter of $DEF$, and $P',Q'$ are the intersections of $(IEF)$ and $(A'B'C')$. We end up wanting to prove $\angle P'IQ'=\angle P'A'Q'+\angle P'DQ'$. [asy][asy] // Generated By AsyPadv1.2 import olympiad; import markers; import math; import graph; // change the unit size to fit your needs unitsize(1cm); // colored pens pen c000000 = rgb("000000"); // dependency level 0 /* You can change the coordinates of these points of dependency level 0. The drawing will retain the same relationships and qualities. Please be aware that as a result of this some of the image may be clipped off. */ pair D = (2.11, 19.01); dot(D, c000000); label("$D$", D, dir(102.17145820858599)); pair E = (0.56, 14.28); dot(E, c000000); label("$E$", E, dir(272.9254564999678)); pair F = (6.27, 14.18); dot(F, c000000); label("$F$", F, dir(264.981927106768)); // dependency level 1 // Do not change anything below, unless you are experienced in Asymptote. path segD_E = D--E; draw(segD_E, c000000); path segE_F = E--F; draw(segE_F, c000000); path segF_D = F--D; draw(segF_D, c000000); pair Ap = (E + F)/2; dot(Ap, c000000); label("$A'$", Ap, dir(291.037511025423)); pair Bp = (F + D)/2; dot(Bp, c000000); label("$B'$", Bp, dir(36.251331177569114)); pair Cp = (D + E)/2; dot(Cp, c000000); label("$C'$", Cp, dir(135.70731936854415)); path ccD_E_F = circumcircle(D, E, F); pair ccD_E_Fcenter = circumcenter(D, E, F); real ccD_E_Frad = circumradius(D, E, F); draw(ccD_E_F, c000000); path pbE_F = ((E+F)/2-20.0*unit(((F-E).y, -(F-E).x)))--((E+F)/2+20.0*unit(((F-E).y, -(F-E).x))); path pbF_D = ((F+D)/2-20.0*unit(((D-F).y, -(D-F).x)))--((F+D)/2+20.0*unit(((D-F).y, -(D-F).x))); // dependency level 2 pair I = intersectionpoint(pbE_F, pbF_D); dot(I, c000000); label("$I$", I, dir(88.86864852014583)); path ccAp_Cp_Bp = circumcircle(Ap, Cp, Bp); pair ccAp_Cp_Bpcenter = circumcenter(Ap, Cp, Bp); real ccAp_Cp_Bprad = circumradius(Ap, Cp, Bp); draw(ccAp_Cp_Bp, c000000); // dependency level 3 path ccE_I_F = circumcircle(E, I, F); pair ccE_I_Fcenter = circumcenter(E, I, F); real ccE_I_Frad = circumradius(E, I, F); draw(ccE_I_F, c000000); pair T = intersectionpoints(segE_F, ccAp_Cp_Bp)[0]; dot(T, c000000); label("$T$", T, dir(247.86801327873428)); // dependency level 4 pair Pp = intersectionpoints(ccE_I_F, ccAp_Cp_Bp)[0]; dot(Pp, c000000); label("$P'$", Pp, SW); pair Qp = intersectionpoints(ccE_I_F, ccAp_Cp_Bp)[1]; dot(Qp, c000000); label("$Q'$", Qp, S); // clip the drawing view clip((0.0, 13.34)--(0.0, 20.0)--(7.11, 20.0)--(7.11, 13.34)--cycle); [/asy][/asy] Now, consider the inversion and reflection centered at $D$ swapping $(E,B')$ and $(C',F)$. This also swaps $(I,T)$, so it swaps $(IEF)$ and $(A'B'C')$. Thus, it swaps $P'$ and $Q'$, giving $$\angle P'IQ'=360^\circ - \angle DIQ' - \angle DIP'= 360^\circ - \angle DQ'T-\angle DP'T$$$$= \angle P'DQ' +\angle P'TQ' = \angle P'DQ'+\angle P'A'Q'$$as desired.

v_Enhance wrote: We let lines $QD$ and $PD$ meet $(ABC)$ again at $X$ and $Y$, respectively. [asy][asy] size(10cm); pair A = dir(130); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair D = foot(I, B, C); pair E = foot(I, C, A); pair F = foot(I, A, B); pair T = extension(B, C, E, F); pair P = IP(T--F, unitcircle); pair Q = -P+2*foot(origin, T, P); draw(A--B--C--cycle, lightblue); filldraw(unitcircle, invisible, mediumcyan); filldraw(incircle(A, B, C), invisible, red); draw(B--T--Q, lightblue); pair X = -Q+2*foot(origin, Q, D); pair Y = -P+2*foot(origin, P, D); draw(X--A--Y, red); draw(P--Y, deepgreen); draw(Q--X, deepgreen); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(I)); dot("$D$", D, dir(270)); dot("$E$", E, dir(70)); dot("$F$", F, dir(130)); dot("$T$", T, dir(T)); dot("$P$", P, dir(135)); dot("$Q$", Q, dir(Q)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); /* TSQ Source: A = dir 130 B = dir 210 C = dir 330 I = incenter A B C D = foot I B C R270 E = foot I C A R70 F = foot I A B R130 T = extension B C E F P = IP T--F unitcircle R135 Q = -P+2*foot origin T P A--B--C--cycle lightblue unitcircle 0.1 lightcyan / mediumcyan incircle A B C 0.1 lightred / red B--T--Q lightblue X = -Q+2*foot origin Q D Y = -P+2*foot origin P D X--A--Y red P--Y deepgreen Q--X deepgreen */ [/asy][/asy] Claim: Line $AX$ is the polar of $P$ with respect to the incircle. Similarly, line $AY$ is the polar of $Q$ with respect to the incircle. Proof. Let $\overline{PQEF} \cap \overline{BC} = T$, so \[ -1 = (DT;BC) \overset{Q}{=} (XP;BC) \overset{A}{=} (P, \overline{AX} \cap \overline{FE}, F, E). \]In addition, $A$ is on the polar of $P$ already, so this proves the claim. $\blacksquare$ Now $\overline{IP} \perp \overline{AX}$ and $\overline{IQ} \perp \overline{AY}$, so \[ \measuredangle PIQ = \measuredangle XAY = \measuredangle APY + \measuredangle XQA = \measuredangle DQA + \measuredangle APD \]as needed. It is exactly just like my solution in our TST

Here's a rather convoluted solution using isogonal conjugates let $X=MY\cap (PMDQ)$. $Y$ is the foot of $M$ to $PQ$. It is well-known that $L, Y, I$ are collinear when $L$ is the midpoint of arc $BAC$ (See 2019Taiwan TST R2) [asy][asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9cm); 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 = -4.3, xmax = 13.42, ymin = -8.18, ymax = 6.3; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((4.56,-1.34), 4.650892387488664), linewidth(1.2) + wrwrwr); draw((6.38,2.94)--(0.517025203358866,-3.6389464529920206), linewidth(1.2) + wrwrwr); draw((0.517025203358866,-3.6389464529920206)--(8.607790360471837,-3.6304569844621124), linewidth(1.2) + wrwrwr); draw((8.607790360471837,-3.6304569844621124)--(6.38,2.94), linewidth(1.2) + wrwrwr); draw(circle((5.497295589807671,-1.403148764669088), 2.230570768167277), linewidth(1.2) + wrwrwr); draw((0.4177326542797788,0.7748099764706899)--(9.199165515092487,-1.0099343647444248), linewidth(1.2) + wrwrwr); draw(circle((5.026207754837738,0.9538705002486265), 4.611952452447697), linewidth(1.2) + wrwrwr); draw((4.5624077819153515,-3.6347017187270665)--(6.390385127239218,5.359449408890044), linewidth(1.2) + wrwrwr); draw((6.390385127239218,5.359449408890044)--(6.3525153422243745,-3.4632568023405748), linewidth(1.2) + wrwrwr); draw((0.4177326542797788,0.7748099764706899)--(6.390385127239218,5.359449408890044), linewidth(1.2) + wrwrwr); draw((6.390385127239218,5.359449408890044)--(9.199165515092487,-1.0099343647444248), linewidth(1.2) + wrwrwr); draw((0.4177326542797788,0.7748099764706899)--(4.5648800801972875,-5.990889827206969), linewidth(1.2) + wrwrwr); draw((4.5648800801972875,-5.990889827206969)--(9.199165515092487,-1.0099343647444248), linewidth(1.2) + wrwrwr); draw((4.5648800801972875,-5.990889827206969)--(4.555119919802712,3.3108898272069687), linewidth(1.2) + wrwrwr); draw((4.555119919802712,3.3108898272069687)--(5.497295589807671,-1.403148764669088), linewidth(1.2) + wrwrwr); draw((6.38,2.94)--(4.5648800801972875,-5.990889827206969), linewidth(1.2) + wrwrwr); draw((0.4177326542797788,0.7748099764706899)--(-0.07916551509248704,-1.6700656352555752), linewidth(1.2) + wrwrwr); /* dots and labels */ dot((4.56,-1.34),dotstyle); label("$O$", (4.64,-1.14), NE * labelscalefactor); dot((6.38,2.94),dotstyle); label("$A$", (6.46,3.14), NE * labelscalefactor); dot((0.517025203358866,-3.6389464529920206),dotstyle); label("B", (0.18,-3.98), NE * labelscalefactor); dot((8.607790360471837,-3.6304569844621124),dotstyle); label("C", (8.8,-3.86), NE * labelscalefactor); dot((4.5624077819153515,-3.6347017187270665),linewidth(4pt) + dotstyle); label("$M$", (4.14,-4.04), NE * labelscalefactor); dot((5.497295589807671,-1.403148764669088),linewidth(4pt) + dotstyle); label("$I$", (5.66,-1.32), NE * labelscalefactor); dot((5.499636079186368,-3.6337183049237343),linewidth(4pt) + dotstyle); label("$D$", (5.58,-3.48), NE * labelscalefactor); dot((7.609742555550423,-0.6868989697465355),linewidth(4pt) + dotstyle); label("$E$", (7.7,-0.38), NE * labelscalefactor); dot((3.8320356836860325,0.08088472446318594),linewidth(4pt) + dotstyle); label("$F$", (3.54,0.32), NE * labelscalefactor); dot((0.4177326542797788,0.7748099764706899),linewidth(4pt) + dotstyle); label("$P$", (0.08,0.96), NE * labelscalefactor); dot((9.199165515092487,-1.0099343647444248),linewidth(4pt) + dotstyle); label("$Q$", (9.38,-0.96), NE * labelscalefactor); dot((6.390385127239218,5.359449408890044),linewidth(4pt) + dotstyle); label("$X$", (6.48,5.52), NE * labelscalefactor); dot((6.3525153422243745,-3.4632568023405748),linewidth(4pt) + dotstyle); label("$H$", (6.44,-3.3), NE * labelscalefactor); dot((4.5648800801972875,-5.990889827206969),linewidth(4pt) + dotstyle); label("$N$", (4.54,-6.56), NE * labelscalefactor); dot((4.555119919802712,3.3108898272069687),linewidth(4pt) + dotstyle); label("$L$", (4.64,3.48), NE * labelscalefactor); dot((-0.07916551509248704,-1.6700656352555752),linewidth(4pt) + dotstyle); label("$W$", (-0.62,-2.04), NE * labelscalefactor); dot((5.258637752000502,-0.2090588427138896),linewidth(4pt) + dotstyle); label("$Y$", (5.34,-0.04), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] $\angle PIQ=\angle DPA+\angle DQA=180-(\angle XPA+\angle XQA)=180-(\angle PAQ-\angle PXQ) =180-\angle PAQ+\angle PXQ=\angle PNQ+180-\angle PMQ$ thus, it suffices to prove that $\angle PIQ+\angle PMQ=180+\angle PNQ$ note that $LA\perp AN, PQ\perp AN$, then $LA\parallel PQ$, $NM, NI$ are isogonal. Therefore, let's prove that $M, I$ are isogonal conjugates in $\triangle PNQ$. It remains to prove that $PM, PI$ are isogonal. By using the isogonality lemma in $\triangle PNM, \triangle PIY$, note that $NM\cap IY=L, PW\parallel NI\parallel MY$. It suffices to prove that $PW, PL$ are isogonal,which is true since $WN=LQ=PA$ and $WP\parallel MY\parallel AN \blacksquare$.

Comment: Nice problem with short statement (given that it is not known). The first solution is configuration-heavy, but I think it's somewhat routine to come up during the contest. The second solution is found after the TST. Solution 1. Configuration medley First we deal with the $AB=AC$ case. In this case, $A,I,D$ are colinear thus $AB\cdot AF = AI\cdot AD$. Moreover, $$\angle APF = \angle APQ = \angle AQP = \angle ABP\implies AB\cdot AF = AP^2$$thus $AI\cdot AD = AP^2$ which means $\angle APD = \angle AIP$. Similarly, $\angle AQD = \angle AIQ$ which gives the conclusion. Now we assume that $AB\ne AC$. We first define a bunch of points. Let $O$ be the circumcenter of $\triangle ABC$, $M_A, M_B, M_C$ be midpoints of minor arcs $BC, CA, AB$, and $N$ be the midpoint of arc $BAC$. The idea is rather than dealing with asymmetric $P,Q$, we investigate two circles; $\odot(PDQ)$, centered at $O_D$ and $\odot(PIQ)$, centered at $O_I$ respectively. Now we state two main claims about locations of $O_I$ and $O_D$. Claim: [Location of $O_I$] $M_BO_I$ and $M_CO_I$ are tangents to $\odot(ABC)$. Proof 1: Extend $IF$ to meet $\odot(AIB)$ again at $C_1$ and extend $IE$ to meet $\odot(AIC)$ again at $B_1$. Notice that $$FB_1\cdot FI = FA\cdot FB = FP\cdot FQ\implies B_1\in\odot(PIQ)$$and similarly $C_1\in\odot(PIQ)$. Now notice that $M_BO_I$ is perpendicular bisector of $IB_1$, thus it must be parallel to $AC$. Similarly $M_CO_I\parallel AB$ so they must be tangents to $\odot(ABC)$. $\blacksquare$ Proof 2: Let $X=EF\cap BC$, the radical center of $\odot(PIQ), \odot(BIC), \odot(BCPQ)$. Since polar of $X$ w.r.t. $\odot(DEF)$ is $AD$, we get that the radical axis $IX$ of $\odot(PIQ)$ and $\odot(BIC)$ is perpendicular to $AD$. Thus, $M_AO_I\parallel AD$. Hence combining with $OO_I\parallel AI$, we find that systems $\triangle M_AM_BM_C\cup O\cup O_I$ and $\triangle DEF\cup I\cup A$ are homothetic. Hence $O_IM_B$ and $O_IM_C$ are tangents as desired. $\blacksquare$ Claim: [Location of $O_D$] $O_D$ is midpoint of $M_BM_C$. Proof: Let $M$ be the midpoint of $BC$. This proof is divided into two steps. Step 1: $M\in\odot(PDQ)$ This is rather standard. Let $X=EF\cap BC$. Since $(BC;DX)=-1$, we get $$XP\cdot XQ = XB\cdot XC = XD\cdot XM$$which complete this step. Step 2: Completion. Notice that $IM_BNM_C$ is parallelogram. Thus, if $O_D'$ is the midpoint of $M_BM_C$, then note that line through $O_D'$ perpendicular to $DM$ is equidistant to $ID$ and $MN$. Thus, $O_D'D=O_D'M$. Moreover, $OO_D'\parallel AI$ thus $O_D'P=O_D'Q$ hence $O_D'=O$ as desired. $\blacksquare$ The two claims implies $O_I,O_D$ are inverses w.r.t. $\odot(ABC)$. Thus we can angle chase. $$\measuredangle OO_IP = \measuredangle OPO_D = \measuredangle POO_D + \measuredangle OO_DP$$however, $$\measuredangle OO_IP = \measuredangle(\perp PQ, \perp PP) = \measuredangle QIP$$where the tangent is w.r.t. $\odot(PIQ)$. Analogously, $$\measuredangle POO_D = \measuredangle PAQ \text{ and }\measuredangle OO_DP =\measuredangle QDP $$hence $$\measuredangle QIP = \measuredangle PAQ + \measuredangle QDP = \measuredangle APD + \measuredangle DQA$$which gives the result after unsigning angles and noticing that $D$ lies inside $\odot(APQ)$ $\implies \angle APD+\angle AQD < 180^{\circ}$. Solution 2: A short inversion solution Perform inversion around $A$ with radius $\sqrt{AP\cdot AQ}$, followed by reflection of angle bisector of $\angle PAQ$. Denote the image by $^*$. It's easy to deduce that $BF^*\parallel CE^*\perp AI$. $I^*$ is midpoint of $E^*F^*$. The key claim is Claim: $I^*$ lie on $\odot(PDQ)$. Thus $\odot(PDQ)$ gets mapped to $\odot(PIQ)$. Proof: Let $M$ be the midpoint of $BC$. First, we prove that $M\in\odot(PDQ)$. Let $T=EF\cap BC$. Notice that $$(B,C;D,T)=-1\implies TD\cdot TM = TB\cdot TC = TP\cdot TQ$$so $M$ indeed lie on $\odot(PDQ)$. Finally, note that $\{B,F^*\}$, $\{C,E^*\}$, $\{P,Q\}$, $\{M,I^*\}$ are symmetric across some line. Hence $PQMI^*$ is isosceles trapezoid, thus it's indeed cyclic so we are done. $\blacksquare$ Finally, we get $D^*\in\odot(PIQ)$ thus $$\angle APD + \angle AQD = \angle AD^*P + \angle AD^*Q = \angle PD^*Q = \angle PIQ$$so we are done.

That's nice problem. I would like introduce my generalization and proof.

Attachments:

Let $BI\cap O=T, CI\cap O=S$. Let's use several lemmas. Lemma 1. The circumcenter of $\triangle PDQ=O_2$ is midpoint of segment $ST$ Lemma 2. If the circumcenter of $\triangle PIQ=O_1$, $O_1S$ and $O_1T$ is tangent line of $O$. By angle chasing and using circumcenters, $$\angle DPA+\angle DQA=\angle PIQ$$$$\iff 360-\angle PAQ-\angle PDQ=\angle PIQ$$$$\iff 360-180+\angle PO_1O -180+\angle PO_2Q=180-\angle PO_1O$$$$\iff \angle O_2PO =\angle PO_1O$$ Because $O_1S$ is tangent to $O$, we can get $$OP^2=OS^2=OO_2\times OO_1$$,which ends the problem

Attachments:

Sharky-Devil Lemma OP!! parmenides51 wrote: Let $I$ be the incentre of acute-angled triangle $ABC$. Let the incircle meet $BC, CA$, and $AB$ at $D, E$, and $F,$ respectively. Let line $EF$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\angle DPA + \angle AQD =\angle QIP$. (Slovakia) Solution: $$S \overset{\text{Def}}{=}\odot(AEF) \cap \odot(ABC)$$By Fact 5, we know that $\overline{AI}$ passes through the midpoint of minor $\overarc{BC}$, let's say $M_A.$ Lemma (Miquel point in Intouch picture). $S$ is the Miquel point of $BCEF.$ $\overline{SD}$ is bisector of $\angle BSC.$ If $\overline{SI} \cap \overline{PQEF} = K$, then $\overline{DK} \perp \overline{PQEF}.$ $\overline{SI} \cap \odot{(ABC)} = R$ is the $A$-antipode w.r.t $\odot(ABC).$ $S, D, M_A$ are collinear. (Proof). Well known. (Incenter-Related Configurations) Since, $\overline {AI} \perp \overline{EF} \implies \overline{AIM_A} \perp \overline{PQEF}.$ Thus, $\overline{DK} \parallel \overline{AIM_A}.$ Motivation wrote: After observing the angle conditions: $\angle DPA + \angle AQD =\angle QIP$ (which needs to be proved), we claim the following. Claim 1: $\triangle SPI \sim \triangle SDQ.$ Proof. $$\angle AM_AR = \angle ASR = \angle ASI = 90^{\circ} \implies \overline{M_AR} \parallel \overline{PQEF} \implies PR = QM_A \implies \angle PSR = \angle DSQ.$$ Since, $\angle QPS = \angle QM_A S$, hence $\triangle SPK \sim \triangle SM_AQ.$ Also, $\triangle SKD \sim \triangle SIM_A$, and thus we're done. $\square$ Claim 2: $\triangle SPD \sim \triangle SIQ.$ Proof. Follows from Claim 1. To prove $\tfrac{SP}{SD} = \tfrac{SI}{SQ}$; from Claim 1 we have $\tfrac{SD}{SQ} = \tfrac{SP}{SI}.\quad \square$ So, \begin{align*} \measuredangle DPA + \measuredangle AQD &= (\measuredangle DPS + \measuredangle SPA) + \measuredangle AQD \\&= \measuredangle DPS + (\measuredangle SQA + \measuredangle AQD) \\&= \measuredangle DPS + \measuredangle SQD \\&= \measuredangle QIS + \measuredangle SIP \\&= \measuredangle QIP. \quad \blacksquare \end{align*}

How is everyone so good at synthetic? [asy][asy] size(9cm); defaultpen(fontsize(10pt)); pair A = dir(130), B = dir(210), C = dir(330), O = origin, I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,C,A), F = foot(I,A,B), P = intersectionpoint(unitcircle,E--E+dir(E--F)*69), Q = intersectionpoint(unitcircle,F--F+dir(F--E)*69), Y = foot(I,E,F), M = foot(O,B,C), N = dir(270), L = dir(90), X = foot(O,E,F), R = extension(O,X,L,I); draw(A--B--C--A); draw(unitcircle); draw(A--L^^P--Q, purple); draw(A--N--L--I); draw(incircle(A,B,C), gray); draw(P--R--Q--O--P^^O--R, blue); draw(P--I--Q, heavygreen); draw(arc(circumcenter(P,D,Q),circumradius(P,D,Q),-170,30), orange); dot("$A$", A, dir(130)); dot("$B$", B, dir(210)); dot("$C$", C, dir(330)); dot("$I$", I, dir(230)); dot("$D$", D, dir(265)); dot("$E$", E, dir(80)); dot("$F$", F, dir(135)); dot("$P$", P, dir(210)); dot("$Q$", Q, dir(20)); dot("$L$", L, dir(90)); dot("$N$", N, dir(270)); dot("$R$", R, dir(135)); dot("$M$", M, dir(315)); dot("$X$", X, dir(240)); dot("$Y$", Y, dir(240)); dot("$O$", O, dir(325)); [/asy][/asy] Let $O$ be the circumcenter, $M$ be the midpoint of $\overline{BC}$, $N = \overline{AI} \cap (ABC)$, and $L = \overline{NO} \cap (ABC)$. Claim: The midpoint $R$ of $\overline{IL}$ is the circumcenter of $\triangle PDQ$. Proof. Let $f(Z) = ZB/ZC$ for any point $Z$. Check by Ratio Lemma that $$f(P)f(Q) = f(\overline{PQ} \, \cap \, \overline{BC}) = f(\overline{EF} \, \cap \, \overline{BC}) = f(D) = f(D)f(M),$$so $PDMQ$ is cyclic. Then from $\overline{AL} \parallel \overline{PQ}$ we can see that $R$ is the intersection of perpendicular bisectors of $\overline{PQ}$ and $\overline{DM}$, which implies the conclusion. $\square$ Note that \begin{align*}\angle RPO &= 180^\circ - \angle PRO - \angle POR \\ &= \angle PDQ + \angle PAQ - 180^\circ \\ &= 180^\circ - \angle APD - \angle AQD\end{align*}and $$\angle IPQ + \angle IQP = 180^\circ - \angle PIQ,$$so it suffices to prove $\angle RPO = \angle IPQ + \angle IQP$. We will show that the tangents of these angles are equal. Suppose that $X = \overline{OR} \cap \overline{PQ}$ and $Y = \overline{AI} \cap \overline{PQ}$. We compute \begin{align*}\tan \angle RPO &= \tan (\angle RPX + \angle OPX) \\ &= \frac{\tfrac{RX}{PX} + \tfrac{OX}{PX}}{1-\tfrac{RX}{PX} \tfrac{OX}{PX}} \\ &= \frac{PX \cdot OR}{PX^2 - RX \cdot OX} \\ &= \frac{PQ \cdot IN}{PQ^2 - 4RX \cdot OX}\end{align*}and \begin{align*} \tan (\angle IPQ + \angle IQP) &= \frac{\tfrac{IY}{PY} + \tfrac{IY}{QY}}{1-\tfrac{IY}{PY} \tfrac{IY}{QY}} \\ &= \frac{IY(PY+QY)}{PY \cdot QY - IY^2} \\ &= \frac{IY \cdot PQ}{AY \cdot NY-IY^2}\end{align*}so it remains to show $$\frac{IN}{PQ^2 - 4RX \cdot OX} = \frac{IY}{AY \cdot NY-IY^2}.$$To deal with the ugly denominator on the LHS, we first note that by Pythagorean's Theorem $$\left(\frac{PQ}{2}\right)^2 = R^2 - OX^2 \implies PQ^2 = 4R^2 - 4OX^2,$$so $$PQ^2 - 4RX \cdot OX = 4R^2 - 4OX(OX+RX) = 4R^2 - 2OX \cdot IN.$$Using the fact that $R$ is the midpoint of $\overline{IL}$, we have $$2OX = 2OR - 2RX = IN - (AY-IY) = NY - AY.$$Thus we want \begin{align*}\frac{IN}{4R^2-IN(NY-AY)} &= \frac{IY}{AY \cdot NY-IY^2} \\ \iff 4R^2 \cdot IY &= IN(AY \cdot NY -IY^2) + IN\cdot IY(NY - AY) \\ &= IN(AY+IY)(NY-IY) \\ &= IN \cdot AI \cdot IN \\ &= IN \cdot 2Rr \\ \iff \frac{2R}{IN} &= \frac{r}{IY} = \frac{AI}{r},\end{align*}which is true as desired. $\blacksquare$

Nice problem. My solution is very similar to v_Enhance. Extend $\overline{PD}$ and $\overline{QD}$ to meet $(ABC)$ at $X$ and $Y,$ respectively. Note that $A$ is on the polar of $P$. Moreover, let $T = \overline{EF} \cap \overline{BC}$ and let $M = \overline{EF} \cap \overline{AY},$ and note that $(PM;FE) \stackrel{A}{=} (PY;BC) \stackrel{Q}{=} (TD;BC) = -1,$ so $Y$ lies on the polar of $P$ too. Thus the polar of $P$ is $\overline{AY},$ and so $\overline{IP} \perp \overline{AY}.$ Similarly $\overline{IQ} \perp \overline{AX}.$ Now let $\overline{AX}$ and $\overline{AY}$ meet $\overline{IQ}$ and $\overline{IP}$ at $X'$ and $Y'$; note that $X'$ and $Y'$ lie on $(AI).$ Now we are done by an angle chase; note that $$\angle APD + \angle AQD = \angle APX + \angle APY = 180 - \angle YAX = 180 - \angle X'AY' = \angle X'IY' = \angle PIQ.$$ Remarks: I think the easiest solution is probably by $\sqrt{bc}$ inversion, since the angle condition is most naturally interpreted like that. However, when trying it I couldn't figure out how to get rid of $D'$ from the picture. I think the idea of drawing $X$ and $Y$ is rather natural, since we want a way to deal with $\angle APD$ and $\angle AQD,$ noting that $\overline{AX}$ and $\overline{IQ}$ concurred on $(AEF)$ was a nice surprise.

$\textbf{Notation:}$ $M$ the mid point of BC $L, K$ midpoints of arcs $BC$ and $BAC$ in $(ABC)$ $S= (AEF)\cap (ABC)$ $T = KS\cap BC$ $R = KD\cap (ABC)$ $P', Q' = KP, KQ\cap BC$ 1- From the typical inversion at $L$ with radius $LI$ we know that $L, D, S$ and $L, R, T$ are collinear. By Brocard we know that the polar of $T$ wrt (ABC) crosses $D$ then $(DT;BC)= -1$, this fact shows that $T, P, F, E, Q$ are collinear. 2- Inversion at $T$ with radius $\sqrt{TR\cdot TC}$, shows that $PQMD$ is cyclic and $\angle DQM = \angle DPM$. 3- Inversion at $K$ with radius $KB$ shows that $PMLP'$ and $QMLQ'$ are cyclic. 4- Finally, realize that $PAQI\sim P'KQ'L$, then $\angle PIQ = P'LQ'=\angle P'LM +\angle Q'LM = \angle KPM +\angle KQM = (\angle APD -\angle APK -\angle DPM) + (\angle AQD +\angle AQK +\angle DQM)= \angle APD +\angle AQD.$

In a group solve long ago ... [asy][asy] size(10cm); pair A = dir(120); pair B = dir(210); pair C = dir(270); pair I = incenter(A, B, C); pair D = foot(I, B, C); pair E = foot(I, C, A); pair F = foot(I, A, B); pair E_x = extension(B, C, E, F); pair P = IP(E_x--F, unitcircle); pair Q = -P+2*foot(origin, E_x, P); draw(A--B--C--cycle, red); filldraw(unitcircle, invisible, lightred); filldraw(incircle(A, B, C), invisible, magenta); draw(B--E_x--Q, magenta); pair T_1 = -Q+2*foot(origin, Q, D); pair T_2 = -P+2*foot(origin, P, D); draw(T_1--A--T_2, mediumgreen); draw(P--T_2, blue); draw(Q--T_1, blue); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(I)); dot("$D$", D, dir(270)); dot("$E$", E, dir(70)); dot("$F$", F, dir(130)); dot("$E_x$", E_x, dir(E_x)); dot("$P$", P, dir(135)); dot("$Q$", Q, dir(Q)); dot("$T_1$", T_1, dir(T_1)); dot("$T_2$", T_2, dir(T_2)); /* TSQ Source: A = dir 120 B = dir 210 C = dir 270 I = incenter A B C D = foot I B C R270 E = foot I C A R70 F = foot I A B R130 E_x = extension B C E F P = IP E_x--F unitcircle R135 Q = -P+2*foot origin E_x P A--B--C--cycle lightblue unitcircle 0.1 lightcyan / mediumcyan incircle A B C 0.1 lightred / red B--E_x--Q magenta T_1 = -Q+2*foot origin Q D T_2 = -P+2*foot origin P D T_1--A--T_2 mediumgreen P--T_2 blue Q--T_1 blue */ [/asy][/asy] Solution 1 (using projective Geo) We worked projecting backwards to get this solution. We claim that $AT_1$ is Polar of $P$ wrt incircle ($\omega$), and similarly $AT_2$ is Polar of $Q$ wrt incircle. We see that $P \in \overline{EF} =$ Polar of $A$ wrt $\omega$. So by La Hire's Theorem, $A \in$ Polar of $P$ wrt incircle We know that $(B, C, D, E_x) = -1$ (Cevians induce harmonicity), and so projecting onto circumcircle ($\Omega$), we get that $(B, C; D, E_x) \overset{Q}{=} (B, C; T_1, P) = -1$ and projecting from $A$ onto line $PQ$ yields that $(B, C; T_1, P) = (F, E; P, AT_1 \in PQ) = - 1$ which proves our claim. Therefore, we have that $IP \perp AT_1$ and $IQ \perp AT_2$ ($\diamondsuit$) because of properties of polars and poles, and so $\angle PIQ = 180 - \angle XAY$ (due to $\diamondsuit$) and $180 -\angle XAY = \angle APY + \angle AXQ = \angle APD + \angle AQD$ which proves our problem. I'll add other solution later.

As v_Enhance, we denote the second intersection of $QD$ and $PD$ as $X$ and $Y$, respectively, and the intersection of $PQ$ and $BC$ as $T$ (possibly point of infinity). We first claim that $XY$ also passes through $T$. Since $AD, BE, CF$ are concurrent (well-known), the points $(T, B; D, C)$ are harmonic. By Brokard's theorem, if $\ell$ is the pole of $D$ w.r.t. to circle $ABC$, then $PQ$ and $XY$ intersect on $\ell$. If $D'$ is the projection from $OD$ to $\ell$, then $OD\cdot OD' = r^2 = OB^2=OC^2$. Thus $O, B, D', C$ are in fact cyclic and so $OD'$ is an internal angle bisector of angle $BD'C$. Now, $\ell$ and $OD'$ are perpendicular, so if $\ell$ intersects $BC$ at $T'$ then based on the two angle bisectors we have $(DT', DB; DD', DC)$ are harmonic pencil, so $(T', B; D, C)$ are harmonic. This gives $T=T'$ and therefore $PQ, BC, XY$ concur at $T$. Now, by Brokard's theorem again, $PX$ and $YQ$ intersect at the polar of $TD$ which is $BC$. This point, namely $U$, is the intersection of tangents to of $B$ and $C$ to the circumcircle. This means $BXCP$ and $BYCQ$ are both harmonic quadrilaterals. Also using the identities on the cevians of triangle (with some trigonometric flavour) we have $$ \frac{\sin\angle PIF}{\sin\angle PIE}\cdot \frac{IF}{IE} = \frac{PF}{PE} = \frac{\sin\angle PAF}{\sin\angle PAE}\cdot \frac{AF}{AE} $$but since $AF=AE$ and $IF=IE$ we have $$\frac{\sin\angle PIF}{\sin\angle PIE}=\frac{\sin\angle PAF}{\sin\angle PAE}=\frac{\sin\angle PAB}{\sin\angle PAC}= \frac{PB}{PC} = \frac{XB}{XC} = \frac{\sin\angle BAX}{\sin\angle XAC}$$and given that $\angle PIE = \angle PIF + \angle FIE = \angle PIF + 180^{\circ} - \angle BAC$ we have $\sin\angle PIE = \sin(\angle BAC - \angle PIF$, and $\sin \angle XAC = \sin(\angle BAC - \angle BAX)$, we have $$ \frac{\sin\angle PIF}{ \sin(\angle BAC - \angle PIF)} = \frac{\sin\angle BAX}{\sin(\angle BAC - \angle BAX)} $$which then gives $\angle PIF = \angle BAX$ after expanding. Similarly, we have $\angle QIE = \angle CAY$. Therefore, $$ \angle DPA + \angle AQD =\angle YPA + \angle AQX =\angle CPA + \angle CAY + \angle AQB + \angle BAX =\angle CBA + \angle QIE + \angle ACB + \angle PIF $$$$ =180^{\circ} - \angle BAC+ \angle QIE + \angle PIF =\angle FIE + \angle QIE + \angle PIF =\angle QIP $$as desired.

See my solution to this problem on my Youtube channel in the link below: https://www.youtube.com/watch?v=Dve63ZswPrg

Solution. WLOG, suppose that $AC\geq AB$. Define $L$ and $M$ to be the midpoints of $\widehat{BAC}$ and $\overline{BC}$, respectively. Let $N=\overline{AI}\cap (ABC)$. First off, we start with the following facts. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(12.0032cm); 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 = -4.4024, xmax = 7.6008, ymin = -2.6608, ymax = 6.3174; /* image dimensions */ pen ffxfqq = rgb(1.,0.4980392156862745,0.); pen ffdxqq = rgb(1.,0.8431372549019608,0.); pen yqqqyq = rgb(0.5019607843137255,0.,0.5019607843137255); pen ffqqff = rgb(1.,0.,1.); draw((0.728,4.454)--(0.,0.)--(7.,0.)--cycle, green); draw((3.5,5.611764570334504)--(-0.39539690417994094,1.5913933380162135)--(3.5,0.)--cycle, ffdxqq); draw((-0.39539690417994094,1.5913933380162135)--(0.728,4.454)--(1.910251402350007,1.623371933391118)--cycle, ffdxqq); draw(arc((1.910251402350007,1.623371933391118),0.484,112.66853692491676,180.7946229570821)--(1.910251402350007,1.623371933391118)--cycle); 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label("$N$",(3.69,-1.93)); draw(arc((1.910251402350007,1.623371933391118),0.484,32.150000003274556,112.66853692491676)); draw(arc((1.910251402350007,1.623371933391118),0.363,32.150000003274556,112.66853692491676)); draw(arc((6.325716349153725,4.398560292531148),0.484,156.76402499769395,237.2825619193362)); draw(arc((6.325716349153725,4.398560292531148),0.363,156.76402499769395,237.2825619193362)); draw((1.910251402350007,1.623371933391118)--(2.850180460301333,2.946954118274531), yqqqyq); draw((1.910251402350007,1.623371933391118)--(0.30813898467245193,1.8852349419383252), yqqqyq); draw((0.728,4.454)--(6.325716349153725,4.398560292531148), ffqqff); draw((6.325716349153725,4.398560292531148)--(1.910251402350007,1.623371933391118), ffqqff); draw((1.910251402350007,1.623371933391118)--(0.728,4.454), ffqqff); draw((3.5,0.)--(6.325716349153725,4.398560292531148), ffqqff); draw((6.325716349153725,4.398560292531148)--(3.5,5.611764570334504), ffqqff); draw((3.5,5.611764570334504)--(3.5,0.), ffqqff); draw((-0.39539690417994094,1.5913933380162135)--(1.9102514023500072,0.), linetype("2 2")); draw((1.9102514023500072,0.)--(6.325716349153725,4.398560292531148), linetype("2 2")); label("$X$",(-1.934,-0.2)); draw((-1.9373213494671266,0.)--(-0.39539690417994094,1.5913933380162135), ffdxqq); /* dots and labels */ dot((0.728,4.454),linewidth(3.pt) + dotstyle); dot((0.,0.),linewidth(3.pt) + dotstyle); dot((7.,0.),linewidth(3.pt) + dotstyle); dot((1.910251402350007,1.623371933391118),linewidth(3.pt) + dotstyle); dot((2.850180460301333,2.946954118274531),linewidth(3.pt) + dotstyle); dot((0.30813898467245193,1.8852349419383252),linewidth(3.pt) + dotstyle); dot((-0.39539690417994094,1.5913933380162135),linewidth(3.pt) + dotstyle); dot((6.325716349153725,4.398560292531148),linewidth(3.pt) + dotstyle); dot((-4.205620887546788,0.),linewidth(3.pt) + dotstyle); dot((3.5,-2.1829140988481996),linewidth(3.pt) + dotstyle); dot((3.5,5.611764570334504),linewidth(3.pt) + dotstyle); dot((3.5,0.),linewidth(3.pt) + dotstyle); dot((1.9102514023500072,0.),linewidth(3.pt) + dotstyle); dot((-1.9373213494671266,0.),linewidth(3.pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Observation 1. Quadrilateral $PQMD$ is cyclic. Proof. Consider the intersection of $EF$ and $BC$, say $R$. Since $(R,D;B,C)=-1$, we infer that $RP\cdot RQ=RB\cdot RC=RD\cdot RM$, hence the result. $\square$ More importantly, we have Observation 2. Triangles $\bigtriangleup PAI$ and $\bigtriangleup MLP$ are similar, as well as $\bigtriangleup QAI$ and $\bigtriangleup MLQ$. Proof. Since $\angle PAI=\angle PLM$, it suffices to show that $AP:AI=LM:LP$ (1) for the first similarity to be true. Let $X=\overline{LP}\cap \overline{BC}$. Observe that $P$ and $X$ are corresponding points for the similar quadrilaterals $EAFI$ and $CLBN$, thus $AP:AI=LX:LN$ (2). Besides that, recall that $BC$ and $(ABC)$ are images of each other under the inversion centered at $L$ with radius $LB=LC$, so $LP\cdot LX=LM\cdot LN$ (3). Equality (1) follows straightforwardly from (2) and (3). We can prove that $\bigtriangleup QAI\sim \bigtriangleup MLQ$ in a completely analogous way. $\square$ We turn to the solution of the problem. According to the second claim, we can deduce that $$\angle PIQ= \angle AIP+\angle AIQ=\angle LPM+\angle LQM=\angle APD+\angle AQD+(\angle LQA+\angle MQD-\angle APL- \angle DPM)$$however, the terms in brackets cancel out on account of $DPQM$ being cyclic. We're done! $\blacksquare$

Nice pole/polar problem! (Similar to v_Enhance's solution). Solution: We let $PQ \cap BC = W$. Since $AF$ and $AE$ are both tangent to the incircle, $EF$ is the polar of $A$ w.r.t the incircle of $\triangle{ABC}$. Note that $P$ and $Q$ would have to lie on that polar. So, by La Hire's Theorem, if $P$ lies on the polar of $A$ w.r.t to incircle, then $A$ lies on the polar of $P$ w.r.t the incircle. Now, let the polar of $P$ w.r.t the incircle intersect $(ABC), PI, PQ$ at $X, Y, Z$, respectively (note that the $AX \perp IP$). Since $Z$ lies on the polar of $P$ w.r.t the incircle, $(FE;ZP) \overset{A}{=} (BC;XP) \overset{Q}{=} (B, C;QX \cap BC, W) = -1$. Now, let the polar of $Q$ w.r.t the incircle intersect $(ABC), QI, PQ$ at $X', Y', Z'$, respectively (note that $AX' \perp IQ$). Since $Z'$ lies on the polar of $Q$ w.r.t the incircle, $(FE;Z'Q) \overset{A}{=} (BC;X'Q) \overset{P}{=} (B, C; PX' \cap BC, W) = -1$. Thus, $PX' \cap BC$ and $QX \cap BC$ must coincide, so $PX'$, $QX$, $BC$ are concur at a point, which we will denote as $V$. Notice that $AV, BE, CF$ concur (since $(BC;VW) = 1$), and so we can get that $V$ is the $A$-intouch point, but $D$ was already defined as that point, so thus $V$ and $D$ coincide. Now, the rest is just simple angle chasing. Note that $\angle{AYI}+\angle{AY'I} = 90$ (since $AX \perp IP$ and $AX' \perp IQ$), so $\angle{QIP} = \angle{Y'IY} = 180 - \frac{\overarc{XX'}}{2}$. Similarly, $\angle{DPA}+\angle{AQD} = \angle{X'PA}+\angle{XQA} = \frac{\overarc{AX'}+\overarc{AX}}{2} = \frac{360-\overarc{XX'}}{2} = 180-\frac{\overarc{XX'}}{2}$, as claimed. $\square$ Remark: A spiral similarity centered at the Sharky Devil Point also very poggers.

It's incredible how this problem has a short statement, but at the same problem is very challenging! Firstly, we invert about $I$, preserving the incircle. $P,Q$ are sent to $(IEF) \cap \omega$, where $\omega$ is the $9$-point circle of $DEF$, and call the images of $P,Q$ by $P^*,Q^*$, respectively, and $A$ is sent to the midpoint $M$ of $EF$, by the way. Moreover, observe that $\angle QIP= \angle DPA + \angle DQA \iff \angle IPD +\angle IQD+ \angle PDQ = \angle IPD+ \angle IDP+ \angle IQD + \angle IDQ =360º- \angle PID - \angle QID= \angle DPA + \angle DQA \iff \angle PDQ= \angle IPA + \angle IQA$. Thus, after the inversion, by the last observation, it's enough to prove that $ \angle P^*MQ^*= \angle IMP^* + \angle IMQ^*= \angle PDQ \iff \angle P^*MQ^*= \angle P^*DQ^*- \angle P^*IQ^*= \angle IP^*A + \angle IQ^*A \iff \angle P^*MQ^*= \angle IP^*A+ \angle IQ^*A. \quad (\star)$ Now, we perform another inversion $\Phi$ centered at $D$ with radius $\sqrt{\frac{DE.DF}{2}}$ followed by a reflection WRT the internal angle bisector of $\angle EDF$. It's well known that $\Phi$ swaps $\omega$ and $(IEF)$, then since $P^*,Q*$ both lie on $\omega, (IEF)$, $\Phi(P^*)=Q^*$ and $\Phi(Q^*)=P^*$. Furthermore, by inversion's angle properties, we have that $\angle IP^*A= \angle AH_DQ^*$ and $\angle IQ^*A= \angle AH_DP^*$, where $H_D$ is the foot of the perpendicular through $D$ to $EF$. Thus, $\angle IP^*A+ \angle IQ^*A= \angle P^*H_DQ^*$, and since $H_D \in \omega$, $\angle P^*H_DQ^*= \angle P^*MQ^*$, so $\angle P^*MQ^*= \angle IP^*A+ \angle IQ^*A$, so $(\star)$ is true, and inverting back we are done! $\blacksquare$

Solution with LYC. Let $M_A$ denote the arc midpoint of $BC$ wrt $(ABC)$ not including $A$. Let $A\ne K = (AEF)\cap (ABC)$. Let $KI\cap EF=T$. It is well-known that a spiral similarity at $K$ maps $EIFT\to CM_ABD$. Observe that $K$ is the center of the spiral sim $QT\to M_AP$ because $\angle KQT=\angle KQP=\angle KM_AP$ and \[\measuredangle KPM_A=\measuredangle KAM_A=\measuredangle KAI=90^\circ-\measuredangle AIK=90^\circ-\measuredangle AIT=\measuredangle ITQ=\measuredangle KTQ.\]Thus since $KI\cdot KD = KT\cdot KM_A$, there is a spiral sim about $K$ mapping $QI\to DP$, which implies the desired, as \[\angle PIQ = \angle PIK + \angle KIQ = \angle DQK + \angle KPD = \angle DQA+\angle APD.\]

dame dame

Denote $P',Q',$ as the inverses of $P,Q$ with respect to the incircle, and also let $X=PD \cap (ABC), Y=QD \cap (ABC), Z=EF \cap BC$. All poles and polars are taken with respect to the incircle. Claim 1: $AFP'IQ'E$ is cyclic Observe that $P$ lies on the polar of $A,$ so by La-Hire's $A$ lies on the polar of $P,$ so $\angle{IP'A}=90,$ implying $P'\in (AEF).$ Similarly, $Q'\in (AEF).$ Claim 2: $A,P',Y$ are collinear It suffices to show $Y$ lies on the polar of $P,$ equivalently $(P,Y;E,F)=-1.$ But $$(P,Y;E,F)\overset{A}{=}(P,Y;B,C)\overset{Q}{=}(Z,D;B,C)=-1,$$so this claim is proven. It similarly follows that $A,Q',X$ are also collinear. Now, $$\angle{DPA}+\angle{AQD}=\angle{XPA}+\angle{YQA}=\angle{XYA}+\angle{YXA}$$$$=180-\angle{YAX}=180-\angle{P'AQ'}=\angle{P'IQ'}=\angle{PIQ},$$as desired $\blacksquare$

[asy][asy] size(10cm); defaultpen(fontsize(10pt)); pair A = dir(120); pair B = dir(210); pair C = dir(330); pair I = incenter(A, B, C); pair D = foot(I, B, C); pair E = foot(I, A, C); pair F = foot(I, A, B); path ABC = circumcircle(A, B, C); pair P = point(ABC, intersections(ABC, E, F)[1]); pair Q = point(ABC, intersections(ABC, E, F)[0]); pair G = point(ABC, intersections(ABC, D, Q)[1]); pair H = point(ABC, intersections(ABC, D, P)[1]); pair L = extension(E, F, B, C); draw(B--A--C); draw(L--C); draw(L--Q); draw(P--I--Q); draw(Q--G); draw(P--H); draw(G--A--H); filldraw(ABC, invisible, green); filldraw(circumcircle(D, E, F), invisible, green); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)*dir(15)); dot("$C$", C, dir(C)); dot("$I$", I, dir(I)*dir(90)); dot("$D$", D, dir(D)*dir(15)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)*dir(340)); dot("$P$", P, dir(P)*dir(345)); dot("$Q$", Q, dir(Q)); dot("$H$", G, dir(G)); dot("$G$", H, dir(H)); dot("$L$", L, dir(L)); [/asy][/asy] Define $G, H$ as the second intersections of $PD, QD$ with the circumcircle of $\triangle ABC$ respectively. Claim: $\overline{AG}$ is the pole of $P$ with respect to the incircle. Proof. We have that $Q$ is on $\overline{EF}$ which is the pole of $A,$ so then $A$ is on the pole of $P$ by La Hire's. Next, we have that $\overline{AD}, \overline{BE}, \overline{CF}$ concur at the Nagel point, so then $-1 = (LD;BC)$ by the complete quadrilateral harmonic configuration. Finally, we have \[-1 = (LD;BC) \stackrel{Q}{=} (PG;BC) \stackrel{A}{=} (P\overline{AG}\cap EF; FE),\]so $AG$ is indeed the pole. $\blacksquare$ Finally, we have that since $\overline{AG} \perp \overline{PI}, \overline{AH} \perp \overline{QI},$ then \[\angle PIQ = \angle HAG = \angle HQG = \angle HQA + \angle AQG = \angle DQA + \angle APD.\]

Dear Mathlinkers, here Problem 1 Sincerely Jean-Louis

Similar to #4; posting for storage. Let $R$ and $S$ be the intersections of $\overline{PD}$ and $\overline{QD}$ with $(ABC),$ let $S'=\overline{AS}\cap\overline{PQ},$ and let $X=\overline{EF}\cap\overline{BC}.$ Notice $$-1=(XD;BC)\stackrel{Q}=(PS;BC)\stackrel{A}=(PS';FE)$$so $S'$ lies on the polar of $P$ with respect to the incircle. In addition, $A$ lies on the polar of $P$ by La Hire. Hence, $\overline{AS}\perp\overline{IP}.$ Similarly, $\overline{AR}\perp\overline{IQ}.$ Therefore, $$\measuredangle RPA+\measuredangle AQS=\measuredangle RSA+\measuredangle ARS=\measuredangle RAS=\measuredangle (RA,AS)=\measuredangle (QI,IP)=\measuredangle QIP.$$$\square$ Remarks: (Motivation) I extended $\overline{PD}$ so I could get a cyclic quadrilateral including $\angle APR.$ Then, the angle chasing was pretty natural, leading me to conjecture that $\overline{AS}\perp\overline{IP}.$ From here, poles/polars seemed relevant from the diagram and also because $I$ was the center of a circle. Projective chase was not that hard after defining $X$ which I had found from guessing that $RS,EF,$ and $BC$ concurred.

Invert wrt incircle, images denote as $\bullet '.$ Since $\angle AQD=\angle AQI+\angle DQI=\angle IA'Q'+\angle IDQ',$ and analogous equality holds, it's suffice to prove $\angle Q'A'P'+\angle Q'DP'=\angle Q'IP'.$ Denote by $H$ projection of $D$ onto $EF.$ Composition of inversion wrt $D$ with power $\frac{1}{2}|DE|\cdot |DF|$ and of symmetry wrt bisector of angle $DEF$ swaps $I,H,$ and so swaps points $P',Q'=\odot (EIF)\cap \odot (A'B'C'H).$ Hence $$\angle Q'A'P'=\angle Q'HP'=\angle Q'HD+\angle P'HD=\angle OQ'D+\angle OP'D,$$and now the desired equality is obvious.

Let $T',T$ be midpoints of minor and major arcs of $BC$ and Let $M$ be midpoint of $BC$. Let $TP$ and $TQ$ meet $BC$ at $X,Y$. Claim $: XPMT'$ and $YQMT'$ are cyclic. Proof $:$ Note that $TM.TT' = TB^2$. $\angle PBC = \angle XPB + \angle BXP = \angle BXP + \angle TCB = \angle BXP + \angle TBC \implies \angle PBT = \angle BXP \implies TB^2 = TP.TX$ so $TM.TT' = TB^2 = TP.TX$. we prove the other one with same approach. Claim $:XPQY$ is cyclic. Proof $:$ $TP.TX = TB^2 = TC^2 = TQ.TY$. $\angle DPA + \angle DQA = \angle DPT + \angle DQT$ Claim $: PDMQ$ is cyclic. Proof $:$ Let $EF$ meet $BC$ at $S$ we have $SP.SQ = SB.SC = SD.SM$. $\angle DPT + \angle DQT = \angle MPT + \angle MQT = \angle MT'X + \angle MT'Y = \angle XT'Y$. Note that $AFIE$ and $TBT'C$ are similar and $XPQY$ is cyclic so $\frac{AP}{AI} = \frac{TX}{TT'}$ and $\angle PAI = \angle PAT' = \angle PTT' = \angle XTT'$ so $XTT'$ and $PAI$ are similar so $\angle AIP = \angle TT'X$. with same approach we have $AIQ$ and $TT'Y$ are similar so $\angle AIQ = \angle TT'Y$ so $\angle XT'Y = \angle PIQ$ and we proved that $\angle XT'Y = \angle APD + \angle AQD$. we're Done.

Let $R = IQ\cap AEF\ne I$, $S = IP\cap (AEF)\ne I$, $K = AR\cap (ABC)\ne A$, $L = AS\cap (ABC)\ne A$, $T = AQ\cap (AEF)\ne A$, $U=EF\cap AR$. Claim: P,D,K collinear. Proof. Note that $U$ is the orthocenter of $AIQ$ so $I,U,T$ are collinear. Since $AFIE$ is harmonic, by taking an involution on $(AEF)$ through $U$ we get that \[-1 = (T, R; E,F) \stackrel{A}{=} (Q,K;C,B).\]But by Ceva and Menelaus, \[-1 = (EF\cap BC,D;C,B) \stackrel{P}{=}(Q,PD\cap (ABC);C,B)\]So $PD\cap (ABC)= K$. $\ \ \ \square$ Similarly, $Q,D,L$ are collinear. Now we angle-chase. Let $M=AI\cap EF$. Since $AMRQ$ and $AMSP$ are cyclic, \[\angle PIQ = 180^\circ - \angle IPQ - \angle IQP = 180^\circ- \angle IAS - \angle IAR = 180^\circ - \angle LAK = \angle AQD+ \angle APD \ \ \ \blacksquare\]

dolly33 wrote: By using the isogonality lemma in $\triangle PNM, \triangle PIY$, note that $NM\cap IY=L, PW\parallel NI\parallel MY$. It suffices to prove that $PW, PL$ are isogonal,which is true since $WN=LQ=PA$ and $WP\parallel MY\parallel AN \blacksquare$. Please anyone, what is "isongonality lemma"?

Let $G$ be the second intersection of $(AEIF)$ and $(ABC)$, let $X$ and $Y$ be the intersection of $GD$ and $GI$ respectively with $(ABC)$. The spiral similarity taking $EF$ to $BC$ similarly takes $I$ to $X$. Indeed, $X$ is on $AI$ and therefore \[\angle AGI=\angle AFI=90^\circ\]and so $AX\perp XY\implies XY\parallel PQ$. Therefore, $\angle PGX=\angle QGY$. Let $H$ be $GI$ intersection with $EF$ then $\triangle GHQ\sim \triangle GPX$ so $GP\cdot GQ=GH\cdot GX$. However, by spiral similarity $GH\cdot GX=GD\cdot GI$ so $\triangle GPD\sim \triangle GIQ$. Therefore, $\angle GPD=\angle GIQ$. Similarly, $\angle GQD=\angle GIP$. We are done.

Let $\{ QG, QH \}$ be the tangents from $Q$ to the incircle. Clearly, $A, G, H$ are collinear. $J = GH \cap (ABC), J \neq A$. By DDIT, there exists an involution swapping $\{AF, AE\}, \{AG, AH\}, \{AQ, AQ\} $. Project this involution from $A$ on $(ABC)$ and we get that there is an involution swapping $\{B, C\}, \{J, J\}, \{Q, Q\}$. Thus, $BC, JJ, QQ$ are concurrent, and therefore quadrilateral $BJCQ$ is harmonic. Projecting this from $P$ on the line $BC$ implies that $PJ \cap BC = D$. Similarly, if $\{PK, PL\}$ are tangents from $P$ to the incircle and $M = KL \cap (ABC)$ then $M \in QD$. $\angle{PIQ} = 180 - \angle{MAJ} = \frac{360^{\circ} - 2\angle{MAJ}}{2} = \frac{2\angle{JPA} + 2\angle{MQA} }{2} = \angle{DPA} + \angle{DQA}$.

Wow you are using involution swapping, so cool!!!

similar to others on the thread, but this problem is good enough that i'm posting this anyway. Let $\overline{PD} \cap (ABC) = X$ and $\overline{QD} \cap (ABC) = Y$ - it suffices to show that $\angle AIP = \angle AQY$ and $\angle AIQ = \angle APX$. Let $M$ be the midpoint of arc $BC$. Claim: $\overline{AY} \perp \overline{IP}$ and similar. Proof. In fact, I claim that $\overline{AY}$ is the polar of $P$ with respect to the incircle. Clearly $A$ lies on the polar by La Hire's, and also \[ -1 = (\overline{EF} \cap \overline{BC}, D; B, C) \overset{Q}{=} (P, Y; B, C) \overset{A}{=} (P, \overline{AY} \cap \overline{EF}; E, F) \]so $\overline{EF} \cap \overline{AY}$ also lies on the polar. $\square$ To finish, let $K$ be the Miquel point of $BCEF$ - now $K$ is the center of the spiral similarity sending $\overline{AP} \cap \overline{XY}$ to $Y$ and $I$ to $M$ and similar, so we have \[ \measuredangle PIQ = \measuredangle PIK + \measuredangle KIQ = \measuredangle YMK + \measuredangle KMX = \measuredangle DQK + \measuredangle KQD = \measuredangle DQA + \measuredangle APD \]as desired.

A solution using spiral, inversion and angle chase ig.

Let $R$ be the Sharky-Devil point, $M$ be the midpoint of $BC$, let $G$ denote the midpoint of arc $BC$, let $H$ be the foot of altitude from $D$ to $EF$ Claim 1: $RG$ and $RI$ are isogonal in $\triangle RPQ$

Proof

Claim 2: There is a spiral similarity centered at $R$ taking $PD$ to $IQ$

Proof

Claim 3:$\measuredangle RPI + \measuredangle RQI = \measuredangle PDQ$

Proof

Claim 4 : $\measuredangle PIQ = \measuredangle APD + \measuredangle AQD$

Proof

$\sqrt{SP\cdot SQ}$ inversion Let $S$ be the $A$-sharkydevil point. Cosndier a force overlay inversion centered at $S$ in $\triangle PSQ$. Claim: Circle $(SAEIF)$ is sent to line $BC$. Proof. Note that $E$ is sent to a point on $(ABC)$ as $PQ$ and $(ABC)$ are swapped. Then, if $E'=SE\cap (ABC)\neq S$, by Reim we have that $BE'\parallel PQ$. Thus $\measuredangle ESQ=\measuredangle E'SQ=\measuredangle PAB$ so $E$ is sent to $B$, finishing by symmetry. $\blacksquare$ Claim: $I$ is sent to $D$. Proof. By above, $I$ is sent to a point on line $BC$. However, note that from above $SE$ and $SB$ are isogonal wrt $\angle PSQ$. However, it is well known that $SD$ bisects $\angle BSC$ so $\angle BSD=\angle A/2$. But we also have $\angle ISE=\angle IAE=\angle A/2$ finishing as then $SD$, $SI$ are isogonal. $\blacksquare$ Thus we have $\triangle SIQ\sim \triangle SPD$ so $\angle SIQ=\angle SPD$. Similarly, $\angle SIP=\angle SQD$ so by adding, $\angle SQD+\angle SPD=\angle PIQ$. This finishes as \[\angle PIQ=\angle SQD+\angle SPD=360^{\circ} -\angle QDP-\angle QSP =360^{\circ}-\angle QDP-\angle QAP=\angle DPA+\angle AQD\]

Let $\Omega$ be the circumcircle and let $R = PQ \cap BC$, $K = \Omega \cap PD, L = \Omega \cap QD$. Then since we have $-1 = (B, C; R, D) \overset{P}{=} (B, C; Q, K)$, so the quadrilateral $BQCK$ is harmonic. Similarly $BPCL$ is harmonic. Now let $X = AP \cap (AEF)$ and $Y = AQ \cap (AEF)$. Then consider the following claim: Claim: $AK, EF, IY$ are concurrent, $AL, EF, IX$ are concurrent. Proof. We only need to prove that $AK, EF, IY$ are concurrent, since the other one is exactly same. Let $T = AK \cap EF$. Then we have $-1 = (B, C; K, Q) \overset{A}{=} (F, E; T, Q)$, thus $(F, E; T, Q)$ is harmonic. Let $M = AI \cap EF$. Then $M$ is the midpoint of $EF$, so we get $QM \cdot QT = QE \cdot QF = QA \cdot QY$, so $AYTM$ is cyclic. Therefore we have $\angle AYT = 90^{\circ}$, which means $Y, T, I$ are collinear. $\blacksquare$ Now note that $-1 = (B, C; K, Q) \overset{A}{=} (F, E; AK \cap (AEF), Y)$ and $-1 = (B, C; K, Q) \overset{A}{=} (F, E; AK \cap EF, Q) \overset{I}{=} (F, E; Y, IQ \cap (AEF))$, so $AK \cap (AEF)$ and $IQ \cap (AEF)$ coincides. Thus $IQ \perp AK$, so $\angle KAI = \angle PQI$. Similarly we get $IP \perp AL$ and thus $\angle LAI = \angle QPI$. Therefore $\angle QIP = 180^{\circ} - \angle PQI - \angle QPI = 180^{\circ} - \angle KAL = \angle KLA + \angle LKA = \angle DQA + \angle DPA$, as required. $\blacksquare$

Hehe multiple inversion Do inversion around incircle of $\triangle ABC$. Note $P'$ and $Q'$ are intersection of nine point circle of $\triangle DEF$ and $(IEF)$. Let $K$ be feet of perpendicular from $D$ to $EF$ and $A'$ be midpoint of $EF$. $$\angle DPA + \angle AQD = \angle DPQ + \angle PQD + \angle QPA + \angle AQP = \angle P'DI + \angle Q'DI + \angle P'A'I + \angle IA'Q = \angle P'DQ + \angle P'A'Q'$$ Now do $\sqrt{\frac{(DE.DF)}{2}}$. Observe that $I \leftrightarrow K$ , $P' \leftrightarrow Q'$.If $\omega$ is Nine point circle of $\triangle DEF$ then $\omega \leftrightarrow (IEF)$. Hence $\angle IP'D = \angle DKP'$ and $\angle IQ'D = \angle DKQ'$. Which give us $\angle IP'D + \angle IQ'D = \angle P'KD' = \angle P'A'Q'$ which implies $$\angle P'DQ' + \angle P'A'Q' = \angle P'DQ' + \angle IP'D + \angle IQ'D = 360 - \angle P'IQ' = \angle P'IQ' = \angle PIQ$$hence $\angle DPA+ \angle AQD = \angle QIP$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13.164861282226145cm); 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 = 3.0852669203412946, xmax = 22.25012820256744, ymin = -6.161514486937255, ymax = 3.9382774329213337; /* image dimensions */ /* draw figures */ draw(circle((10.625412961994623,-1.6404715098849263), 2.790630996071502), linewidth(0.4)); draw(circle((11.935857461024495,-1.9743429844097993), 5.891656036727269), linewidth(0.4)); draw((xmin, 0.18927915980387497*xmin-2.124797329376852)--(xmax, 0.18927915980387497*xmax-2.124797329376852), linewidth(0.4)); /* line */ draw(circle((10.142706480997312,0.9097642450575364), 2.5955168950680627), linewidth(0.4) + blue); draw((8.034377129280319,-0.6040571767992041)--(10.562717981163273,-4.430398156880074), linewidth(0.4)); draw((10.562717981163273,-4.430398156880074)--(12.658441036814962,0.27118175449837606), linewidth(0.4)); draw((10.625412961994622,-1.6404715098849263)--(9.66,3.46), linewidth(0.4)); draw((8.069201041891475,2.4709442944375706)--(10.625412961994622,-1.6404715098849263), linewidth(0.4)); draw((6.131435736786073,-0.9642443247265311)--(10.625412961994622,-1.6404715098849263), linewidth(0.4)); draw((10.625412961994622,-1.6404715098849263)--(16.969603238554054,1.087194913821775), linewidth(0.4)); draw(circle((10.315061592631968,-1.5614010346479905), 1.3953154980357538), linewidth(0.4) + red); draw((9.776072356517004,-0.2743905675534256)--(8.930879529653678,-1.385488114179013), linewidth(0.4)); draw((10.34640908304764,-0.16643771115041434)--(11.661422112626775,-1.195042345303694), linewidth(0.4)); draw((10.34640908304764,-0.16643771115041434)--(8.930879529653678,-1.385488114179013), linewidth(0.4)); draw((8.930879529653678,-1.385488114179013)--(10.562717981163273,-4.430398156880074), linewidth(0.4)); draw((10.562717981163273,-4.430398156880074)--(11.661422112626775,-1.195042345303694), linewidth(0.4)); draw((9.776072356517004,-0.2743905675534256)--(11.661422112626775,-1.195042345303694), linewidth(0.4)); draw((9.66,3.46)--(6.54,-4.34), linewidth(0.4)); draw((6.54,-4.34)--(17.22,-4.58), linewidth(0.4)); draw((17.22,-4.58)--(9.66,3.46), linewidth(0.4)); /* dots and labels */ dot((9.66,3.46),dotstyle); label("$A$", (9.709740685148336,3.5671269503430114), NE * labelscalefactor); dot((6.54,-4.34),dotstyle); label("$B$", (6.583079044033977,-4.2270331838017565), NE * labelscalefactor); dot((17.22,-4.58),dotstyle); label("$C$", (17.267714148561463,-4.463219854533416), NE * labelscalefactor); dot((11.935857461024497,-1.9743429844098006),linewidth(4.pt) + dotstyle); label("$O$", (11.981631517900498,-1.8876604451262706), NE * labelscalefactor); dot((10.625412961994622,-1.6404715098849263),linewidth(4.pt) + dotstyle); label("$I$", (10.665734352395532,-1.550250915509614), NE * labelscalefactor); dot((10.562717981163273,-4.430398156880074),linewidth(4.pt) + dotstyle); label("$D$", (10.609499430792756,-4.339503027007309), NE * labelscalefactor); dot((12.658441036814962,0.27118175449837606),linewidth(4.pt) + dotstyle); label("$E$", (12.701438514416033,0.3617364189847735), NE * labelscalefactor); dot((8.034377129280319,-0.6040571767992041),linewidth(4.pt) + dotstyle); label("$F$", (8.078927958667826,-0.5155283580185337), NE * labelscalefactor); dot((6.131435736786073,-0.9642443247265311),linewidth(4.pt) + dotstyle); label("$P$", (5.840778078877331,-0.7742089973913039), NE * labelscalefactor); dot((16.969603238554054,1.087194913821775),linewidth(4.pt) + dotstyle); label("$Q$", (17.009033509188694,1.1827662743853047), NE * labelscalefactor); dot((8.069201041891475,2.4709442944375706),linewidth(4.pt) + dotstyle); label("$S$", (8.112668911629491,2.566145345813597), NE * labelscalefactor); dot((9.776072356517004,-0.2743905675534256),linewidth(4.pt) + dotstyle); label("$K$", (9.822210528353889,-0.18936581272243233), NE * labelscalefactor); dot((10.34640908304764,-0.16643771115041434),linewidth(4.pt) + dotstyle); label("$A'$", (10.395806728702206,-0.07689596951688013), NE * labelscalefactor); dot((8.930879529653678,-1.385488114179013),linewidth(4.pt) + dotstyle); label("$P'$", (8.978686704312246,-1.291570276136844), NE * labelscalefactor); dot((11.661422112626775,-1.195042345303694),linewidth(4.pt) + dotstyle); label("$Q'$", (11.71170389420717,-1.1003715426874052), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Here is a sketch of my overkill solution: Firstly, we invert across the incircle Then we $\sqrt{bc}$ invert in $D$ (the triangle $EFD$) and note that points $P$ and $Q$ almost switched. Actually, this "almost" says that triangles $DBP$ and $DEQ$ are similar, and $DCQ$ and $DFP$ are similar. Now angle chasing hands the desired equality