It was really hard to do a synthetic solution after 3 months of bashing , projective geo ,... Assume that $I=A$ and redefine $A$ to be a point st $I$ is the incenter of $\triangle ABC$ .Now $D$ is the midpoint of arc $(BAC)$ in $(ABC)$ and also , $EI \perp AI$ which means $A,F,D$ are collinear.Let $S = AI \cap (EIC)$ and $T = AI \cap (EIB)$.The main claim is the either $XP$ or $YP$ is perpendicular to $BC$.Let $G = AI \cap BC$.Note that since $AF || IE$ , we have $\frac{EF}{EG} = \frac{IA}{IG}$.Since $\angle FAS = 90 = \angle FCS$ , $FACS$ is cyclic and $\angle FSG =\angle C$ So $SGF \sim AGC$ and by $\frac{EF}{EG} = \frac{IA}{IG}$ , $SE$ bisects $\angle GSF$. Now, note that if $P' = IB \cap (EIC)$ , $\angle ESP' =\angle ECP' = \frac{\angle B}{2} - \frac{\angle B-\angle C}{2} = \frac{\angle C}{2}$. But $\angle ESF = \frac{\angle C}{2}$ so $S,P',F$ are collinear and $\angle EP'F=90$ and $P' \in \{P,Q\}$ ;Wlog $P$(the other case is similar) . Now by $SGF \sim AGC$ , $\angle PFE=\frac{\angle A}{2}$ and similarly $\angle XFE = \frac{\angle A}{2}$ and $XP \perp BC$.Now , note that $\angle XYE = \frac{\angle A}{2}$ and $\angle ISE = \frac{\angle C}{2}$ and $\angle YIE = 180 - \angle CIG = 180 - \frac{\angle A+\angle C}{2}$ So $\angle IYE+\angle ISE+\angle YIS = 180$ which means $S,I,Y$ are collinear and $SCAYF$ is cyclic so since $SY$ bisects $\angle ASF$ , we have $YA=YF$ . Similarly , $QA=QF$ and $QF$ is the perpendicular bisector of $AF$. So if $U,V,W = XP \cap EF , XP \cap QY , EI \cap YQ$ , since $EI || AF$ , $U,V,W,E$ lie on a circle and $\angle UVW = \angle UEV$ which means $\angle IEB = \angle XEY + \angle PEQ$ and we're done. Remark. In the other case where we have $YP \perp BC$ , we should define $V = XQ \cap PY$ which changes the final equation to $\angle IEB = \angle PEQ - \angle XEY$...btw I guess It's really hard to find a non-synthetic sol(other than inversion and DDIT) since it's gonna be really hard to put a difference between $X,Y$ , or $P,Q$. So any computational solution to this would be an instructive masterpiece:)

原来是老毛子的题 In this quite simple exercise, we can observe from the peculiar conclusion we want to prove that there always exist two points on the circle which are symmetrical about a diameter among the four intersections. Additionally, we will find many pairs of similar triangles, concurrent circles, and parallel lines. We adopt an approach similar to 同一法, using the properties we found to inversely construct corresponding points in the original problem. In this process, we utilize two sets of parallel lines. Without loss of generality, let $AB > AC$. We reconstruct the original figure from six points $A,B,C,D,E,F$. Draw a line through $E$ parallel to $DB$ intersecting $BA$ at $T$, we aim to demonstrate that $T$ lies on $\odot EF$. This is equivalent to proving that $\frac{ET}{EF}=\cos\angle DBC$. Let $\angle ABC=B$, $\angle ACB=C$. Hence, $\frac{BE}{CE}=\left(\frac{\sin C}{\sin B}\right)^2$ leading to $BE=\frac{\sin^2 C}{\sin^2 C-\sin^2 B} BC$ . Also, $\frac{BF}{CF}=\frac{\sin \angle BDF}{\sin \angle CDF}=\frac{\sin 2C}{\sin 2B}$ which implies $BF=\frac{\sin 2C}{\sin 2C-\sin 2B}\cdot BC$. Furthermore, since $\frac{ET}{\sin B}=\frac{BE}{\sin C}$, substituting and simplifying yields $\frac{ET}{EF}=\frac{\sin B\sin C(\sin 2C-\sin 2B)}{\sin 2B\sin^2C-\sin 2C\sin^2 B}=\cos(B+C)$, thereby completing the first step. Similarly, draw a line through $E$ parallel to $DC$ intersecting $AC$ at $S$, then $S$ also lies on $\odot EF$. Let $BT$, $CS$ intersect $\odot EF$ again at $T'$ and $S'$, respectively. Note that $ACET$ and $ABES$ are cyclic and so $AC \parallel T'F$, $AB \parallel S'F$. The directed angle $\measuredangle T'FT + \measuredangle S'FS = \measuredangle ACB + \measuredangle BFT + \measuredangle ABC + \measuredangle CFS = \measuredangle ACB + \measuredangle ABC = \measuredangle AEB$. Then in the original problem, $\{P,Q\}=\{T,T'\}$, $\{X,Y\}=\{S,S'\}$. Thus, there exists a pair among these three angles such that their sum equals the third. The specific relation is determined by the sign of the directed angle, namely, the positional relationship between $S$, $T$, $S'$, $T'$, which is essentially determined by whether $B+2C$ and $C+2B$ are greater than 90 degrees.

Attachments:

Complex bash with $ABC$ unit circle. We have $D=BB\cap CC$ and $E=AA\cap BC$, so $d=\frac{2bc}{b+c}$ and $e=\frac{a(ab+ac-2bc)}{a^2-bc}$. Since $F$ lies on $BC$ and $DF\parallel AE$, we have $f=b+c-bc\overline f$ and \begin{align*} (d-f)(\overline{a-e})&=(\overline{d-f})(a-e)\\ \frac{2bc}{b+c}-f&=-a^2\left(\frac2{b+c}-\frac{b+c-f}{bc}\right)\\ bc((b+c)f-2bc)&=a^2((b+c)f-b^2-c^2)\\ f&=\frac{a^2b^2+a^2c^2-2b^2c^2}{(b+c)(a^2-bc)} \end{align*}Now, $p$ and $q$ are solutions to \begin{align*} (z-e)(\overline{z-f})+(z-f)(\overline{z-e})&=0\\ \left(z-\frac{a(ab+ac-2bc)}{a^2-bc}\right)\left(\frac{a+b-z}{ab}-\frac{2a^2-b^2-c^2}{(b+c)(a^2-bc)}\right)+\left(z-\frac{a^2b^2+a^2c^2-2b^2c^2}{(b+c)(a^2-bc)}\right)\left(\frac{a+b-z}{ab}-\frac{2a-b-c}{a^2-bc}\right)&=0\\ ((a^2-bc)z-a(ab+ac-2bc))((b+c)(a^2-bc)(a+b-z)-ab(2a^2-b^2-c^2))+((b+c)(a^2-bc)z-(a^2b^2+a^2c^2-2b^2c^2))((a^2-bc)(a+b-z)-(ab)(2a-b-c))&=0\\ (-2a^4b-2a^4c+4a^2b^2c+4a^2bc^2-2b^3c^2-2b^2c^3)z^2+(\text{something})z+(-2a^5c^2-4a^4b^2c+2a^4bc^2-2a^3b^4+2a^3b^3c+4a^3b^2c^2+4a^2b^4c-2a^2b^3c^2+2a^2b^2c^3-2ab^3c^3-2b^4c^3)&=0\\ (b+c)(a^2-bc)^2z^2+(\text{something})z+(a^2c+ab^2-abc-b^2c)(a^3c+a^2b^2-ab^2c-b^2c^2)&=0 \end{align*}From this and the fact that each root of the has degree one, we can guess that the roots are $p=\frac{a^2c+ab^2-abc-b^2c}{a^2-bc}$ and $q=\frac{a^3c+a^2b^2-ab^2c-b^2c^2}{(b+c)(a^2-bc)}$. It suffices to check one root; the other follows by Vietas. Indeed, $\frac{p-e}{p-f}=\frac{\frac{a^2c+ab^2-abc-b^2c}{a^2-bc}-\frac{a(ab+ac-2bc)}{a^2-bc}}{\frac{a^2c+ab^2-abc-b^2c}{a^2-bc}-\frac{a^2b^2+a^2c^2-2b^2c^2}{(b+c)(a^2-bc)}}=\frac{(b+c)(-a^2b+ab^2+abc-b^2c)}{(b+c)(a^2c+ab^2-abc-b^2c)-a^2b^2-a^2c^2+2b^2c^2}=\frac{-b(b+c)(a-b)(a-c)}{-b(a-b)(a-c)(b-c)}=\frac{b+c}{b-c}$ which is obviously pure imaginary. Also, $\frac{q-e}{q-f}=\frac{\frac{a^3c+a^2b^2-ab^2c-b^2c^2}{(b+c)(a^2-bc)}-\frac{a(ab+ac-2bc)}{a^2-bc}}{\frac{a^3c+a^2b^2-ab^2c-b^2c^2}{(b+c)(a^2-bc)}-\frac{a^2b^2+a^2c^2-2b^2c^2}{(b+c)(a^2-bc)}}=\frac{a^3c+a^2b^2-ab^2c-b^2c^2-a(b+c)(ab+ac-2bc)}{a^3c+a^2b^2-ab^2c-b^2c^2-a^2b^2-a^2c^2+2b^2c^2}=\frac{c(a-b)^2(a-c)}{c(a^2-b^2)(a-c)}=\frac{a-b}{a+b}$, which is also pure imaginary. We now consider the spiral similarity that brings $e$ to $f$ and to $-1$ and $1$ (thus bringing $(EF)$ to the unit circle). We have $\frac{p'+1}{p'-1}=\frac{p-e}{p-f}=\frac{b+c}{b-c}$, so $p'=\frac bc$. We have $\frac{q'+1}{q'-1}=\frac{q-e}{q-f}=\frac{a-b}{a+b}$, so $q'=-\frac ab$. Similarly, $x'=\frac cb$ and $y'=-\frac ac$. We have $EB$ intersects $(EF)$ again at $F$ and $EA$ intersects $(EF)$ again at $G$. We have $f'=1$ and by angle chasing, $g'=\frac{(ab)(ac)}{(bc)^2}$ using the fact that $EG$ is tangent to $(ABC)$ at $A$. Since $\frac{q'}{p'}\cdot\frac{y'}{x'}=\frac{g'}{f'}$, taking the spiral similarity back from the unit circle to $(EF)$ and halving all angles by the inscribed angle theorem gives the angle identity.

Here is a synthetic solution. Let $DF$ intersect $AB$ and $AC$ at $H$ and $K$ respectively. Note that $D$ is the midpoint of $HK$ and the center of $(BCHK)$ (if you have trouble seeing this, try making $\Delta AHK$ the reference triangle and applying the "three tangents lemma"). Claim. Redefine $P$ as the second intersection of $AB$ and $(ACE)$. Then $EPF = 90^\circ$. Proof. Note that $\measuredangle HFC = \measuredangle AEC = \measuredangle APC$ so $F,H,C,P$ are concyclic. So \[ \measuredangle EPF = \measuredangle EPA + \measuredangle HPF = \measuredangle ECA + \measuredangle HCB = \measuredangle BCK + \measuredangle HCB = \measuredangle HCK = 90^\circ \quad \square\] Similarly, redefine $X$ as the second intersection of $AC$ and $(ABE)$, then $\angle EXF = 90^\circ$. Also, note that $\measuredangle PCE = PAE = \measuredangle BCA = \measuredangle ECX$, so $P$ and $X$ are reflections across $EF$, so $PX \perp EF$. Now \begin{align*} \measuredangle PEQ + \measuredangle XEY &= \measuredangle XEQ + \measuredangle PEY = \measuredangle XPQ + \measuredangle PXY \\ &= 90^\circ - \measuredangle PBE + 90^\circ - \measuredangle XCE \\ &= \measuredangle EBA + \measuredangle ECA = \measuredangle EBA + \measuredangle BAE = \measuredangle BEA \end{align*}hence proved.

The tangent to $A$ has equation $z+a^2\overline z=2a$ and $BC$ has equation $z+bc\overline z=b+c$, so $\overline e=\frac{2a-b-c}{a^2-bc}$. Then, as $d=\frac{2ef}{e+f}$, we get that the line through $D$ parallel to $AE$ has equation $z+a^2\overline z=\frac{2(bc+a^2)}{b+c}$, so $\overline f=\frac{2a^2-b^2-c^2}{(b+c)(a^2-bc)}$. This means $$\overline{e-f}=-\frac{2(a-b)(a-c)}{(b+c)(a^2-bc)}$$and $$\overline{e-c}=-\frac{(c-a)^2}{c(a^2-bc)}.$$ Let $O=\frac{E+F}2$ be the center of the circle with diameter $EF$. We have $\overline{o-c}=-\frac12\overline{(e-f)}+\overline{e-c}=\frac{(a-c)(c^2-ab)}{(b+c)c(a^2-bc)}$. Therefore, we get $$|\cos\angle XEY|=|\cos\frac12\angle XOY|=\frac{CO}{\frac12EF}\sin C=\frac{AC|c^2-ab|}{|b+c||a^2-bc|}\frac{|b+c||a^2-bc|}{AB\cdot AC}\sin C=\frac{\sin C|c^2-ab|}{AB}=\frac{|c-ab/c|}{2R}.$$ Since $ab/c$ is the intersection of the line through $C$ parallel to $AB$ and $(ABC)$, this means the distance from $C$ to $ab/c$ is $2R|\sin(A-B)|$, which means $$|\cos\angle XEY|=|\sin(A-B)|=|\cos(A-90-B)|.$$ Since $C>B$, and $A$ is obtuse, we get that $E$ and $F$ are both on ray $BC$ outside of segment $BC$. This means $\angle AEB=C-B$. Also, $AC$ must intersect $(EF)$ at two points on the opposite side of $BC$ as $A$. This means $\angle XEY<90^\circ$, so $\angle XEY=|A-90-B|$. Similarly, $\angle PEQ=|A-90-C|$. Since $A-90-C+C-B=A-90-B$, we get that $\angle AEB=\pm\angle PEQ\pm\angle XEY$ for some combination of signs. We are done unless all of the signs are negative, but this is clearly impossible as all three angles are greater than zero. Therefore, one of the angles is the sum of the other two.