I thought this was the easiest standard geometry (G1, G2, G3, G4, G5, G7) in this SL. Oops. [asy][asy] unitsize(35); pair A, B, C, D, I, Ib, Id, Oa, Oc, X; Ib = (0, 1); Id = (0, -2); A = (4, 0); C = (-2.5, 0); B = extension(A, reflect(A, Ib) * C, C, reflect(C, Ib) * A); D = extension(A, reflect(A, Id) * C, C, reflect(C, Id) * A); Oa = circumcenter(A, Ib, Id); Oc = circumcenter(C, Ib, Id); I = extension(A, Oa, C, Oc); X = extension(Oa, Oc, I, bisectorpoint(B, I, D)); draw(A--I^^C--I, gray); draw(A--Ib--C--Id--cycle, gray); draw(Ib--Id, purple); draw(Oa--X, lightred + dashed); draw(B--I--D, dashed + gray); draw(X--tangent(X, Oa, abs(A-Oa), 1), dotted); draw(X--tangent(X, Oa, abs(A-Oa), 2), dotted); draw(I--X, heavycyan); draw(circumcircle(A, Ib, Id)^^circumcircle(C, Ib, Id), lightblue); draw(circle(Ib, abs(Ib.y))^^circle(Id, abs(Id.y)), green); draw(A--B--C--D--cycle, red); draw(A--C, red); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$I$", I, dir(I)); dot("$I_B$", Ib, dir(Ib)); dot("$I_D$", Id, dir(Id)); dot("$O_A$", Oa, dir(Oa)); dot("$O_C$", Oc, dir(Oc)); dot("$X$", X, dir(X)); [/asy][/asy] Lemma. The incircles of $\triangle ABC$ and $\triangle ADC$ are tangent to each other. (Thus $\overline{I_BI_D} \perp \overline{AC}$.) Proof. Since $ABCD$ is circumscribed $AB + CD = AD + BC$. Now the tangent length from $A$ to incircle $\triangle ABC$ equals $\tfrac{1}{2}(AB + AC - BC)$ while the tangent to $\triangle ADC$ has length $\tfrac{1}{2}(AD + AC - DC)$, which are evidently equal. Now let $O_A$ be the circumcenter of $\triangle AI_BI_D$ and observe \[\angle BAO_A = \angle BAI_B + \angle I_BAO_A = \angle I_BAC + \angle CAI_D = \frac{1}{2} \angle BAD\]by isogonality and so $O_A \in \overline{AI}$; similarly if $O_C$ is the circumcenter of $\triangle CI_BI_D$ then $O_C \in \overline{CI}$. But note $\overline{AC} \perp \overline{I_BI_D} \perp \overline{O_AO_C}$ so $\overline{AC} \parallel \overline{O_AO_C}$ now. Consequently \[\frac{XO_A}{XO_C} = \frac{AO_A}{CO_C} = \frac{O_AI}{O_CI}\]and $\overline{IX}$ is the external bisector of $\angle AIC$. Since $\angle AIB + \angle CID = 180^{\circ}$, $\overline{IX}$ is also the internal bisector of $\angle BID$. Similarly $\overline{IY}$ is the external bisector of $\angle BID$ and the internal bisector of $\angle AIC$, so $\angle XIY = 90^{\circ}$ as claimed.

I though this was the hardest standard geo in this shortlist, but it's still really easy for a G7. Let $\Gamma_a, \Gamma_b, \Gamma_c, \Gamma_d$ and $O_a, O_b, O_c, O_d$ be the circumcircles and circumcenters of $AI_bI_d, BI_cI_a, CI_dI_b, DI_aI_c$. First, because $ABCD$ is inscribed we have $AB + CD = BC + AD$, which implies that $AC + BC - AB = AC + CD - AD$. Thus the incircles of $ABC$ and $ACD$ are tangent to $AC$ at the same point, implying that $I_bI_d$ is perpendicular to $AC$. Analogously, $I_aI_c$ is perpendicular to $BD$. Now notice that $$\angle I_bAO_a = 90^{\circ} - \angle AI_dI_b = \angle CAI_d = \frac{\angle CAD}{2} = \frac{\angle BAD}{2} - \frac{\angle BAC}{2}= \angle BAI - \angle BAI_b = \angle I_bAI$$ Implying that $O_a$ lies on line $AI$, with similar collinearities holding for the other three circumcenters. Because $O_aO_c$ is perpendicular to the common chord of $\Gamma_a$ and $\Gamma_c$, which is $I_bI_d$, and $AC$ is also perpendicular to $I_bI_d$ we find $$\frac{IO_a}{IO_c} = \frac{AO_a}{CO_c} = \frac{r_a}{r_c}$$ We now aim to show that $I_bII_dX$ is cyclic. Because circles $\Gamma_a$ and $\Gamma_c$ both pass through $I_b$ and $I_d$ it suffices to show that $$\frac{\text{Pow}(I, \Gamma_a)}{\text{Pow}(I, \Gamma_c)} = \frac{\text{Pow}(X, \Gamma_a)}{\text{Pow}(X, \Gamma_c)}$$ It is clear that the respective ratios are $\frac{O_aI^2 - r_a^2}{O_cI^2 - r_c^2}$ and $\frac{r_a^2}{r_c^2}$, which are equal due to the ratios we previously proved to be equal. Thus $I, I_b, X, I_d$ are indeed concyclic, and because $XI_b = XI_d$ it follows that $IX$ is the internal angle bisector of angle $I_bII_c$, which coincides with angle $BID$. Analogously, $IY$ is the internal angle bisector of $\angle CIA$. It now suffices to show that lines $IA$ and $IC$ are $I$-isogonal in triangle $BID$, which follows from simple angle chasing.

Due to our team leader, this is a candidate for IMO P3 but end up losing C5 by 1 vote. By a well known lemma (or simple side-chasing), incircles of $\Delta ABD$, $\Delta CBD$ are tangent. So $BD\perp I_AI_C$. Furthermore since $$\angle I_ABI_C = \frac{\angle B}{2} = \angle ABI$$we get that $\angle I_CBD = \angle I_ABA = \angle I_ABD$ so $\{BI, BD\}$ are isogonal w.r.t. $\angle I_ABI_C$. But $BD\perp I_AI_C$ so the center $O_B$ of $\odot(BI_AI_C)$ lies on $BI$. Similarly, the center $O_D$ of $\odot(DI_AI_C)$ lies on $DI$. Since $O_BO_D\perp I_AI_C\perp BD$ so $O_BO_D\parallel BD$. Hence $$\frac{O_BY}{O_DY} = \frac{O_BB}{O_DD} = \frac{O_BI}{O_DI}$$so $IY$ externally bisects $\angle O_BIO_D \equiv \angle BID$. Similarly $IX$ externally bisects $\angle AIC$. Now it suffices to prove that $IX$ internally bisects $\angle BID$ too, which is equivalent to $\{IA, IC\}$ are isogonal w.r.t. $\angle BID$. Finally, note by angle chasing that $$\angle AID + \angle BIC = \left(180^{\circ} - \frac{\angle A}{2}-\frac{\angle D}{2}\right) + \left(180^{\circ} - \frac{\angle B}{2} - \frac{\angle D}{2}\right) = 180^{\circ}$$implying the desired isogonality so we are done.

The proof begins with the following lemmas. Lemma: The incircles $(I_B)$ and $(I_D)$ touch $\overline{AC}$ at the same point $T$, with $\overline{I_B T I_D} \perp \overline{ATC}$. Proof. Since $ABCD$ is circumscribed, $AD+BC = AB+CD$, and it follows $AD+AC-DC = AB+AC-BC$, which implies the result. $\blacksquare$ Lemma: We have $\overline{O_A O_C} \parallel \overline{AC}$, and $O_A \in \overline{AI}$, $O_C \in \overline{CI}$. Proof. The first observation follows by noticing that $\overline{O_A O_C}$ is the perpendicular bisector of $\overline{I_D T I_B}$. For the second part, note that $\overline{AO_A}$ is isogonal to $\overline{AT}$ with respect to $\angle I_B A I_D$; from which it follows that $\overline{AO_A}$ bisects $\angle BAD$. $\blacksquare$ [asy][asy] size(14cm); pair A = dir(150); pair C = -conj(A); pair B = dir(240); pair I_B = incenter(A, B, C); pair T = foot(I_B, A, C); real r = 2.718; path wA = CR(A, r-abs(B-C)); path wC = CR(C, r-abs(B-A)); pair D = IP(wA, wC); pair I_D = incenter(A, D, C); draw(A--B--C--D--cycle, heavycyan); draw(incircle(A, B, C), heavycyan+dotted); draw(incircle(A, D, C), heavycyan+dotted); pair I = extension(D, I_D, B, I_B); draw(CP(I, foot(I, A, B)), dotted+red); draw(circumcircle(A, I_B, I_D), blue); draw(circumcircle(C, I_B, I_D), blue); pair O_A = circumcenter(A, I_B, I_D); pair O_C = circumcenter(C, I_B, I_D); draw(I_B--I_D, orange); draw(A--I--C, red); draw(A--C, orange); draw(O_A--O_C, orange); draw(A--I_B--C--I_D--cycle, lightblue); pair Z = IP(Line(O_A, O_C), circumcircle(I, I_B, I_D)); pair X = extension(I, I+dir(90)*(Z-I), O_A, O_C); draw(X--O_A, orange); pair H_1 = IP(CP(midpoint(X--O_C), X), CP(O_C, C)); pair H_2 = OP(CP(midpoint(X--O_C), X), CP(O_C, C)); draw(H_1--X--H_2, heavygreen); draw(circumcircle(X, Z, I), dashed+heavygreen); draw(B--I--D, red); dot("$A$", A, dir(A)); dot("$C$", C, dir(C)); dot("$B$", B, dir(B)); dot("$I_B$", I_B, dir(340)); dot("$T$", T, dir(135)); dot("$D$", D, dir(D)); dot("$I_D$", I_D, dir(20)); dot("$I$", I, dir(315)); dot("$O_A$", O_A, dir(270)); dot("$O_C$", O_C, dir(270)); dot("$X$", X, dir(X)); /* TSQ Source: !size(11cm); A = dir 150 C = -conj(A) B = dir 240 I_B = incenter A B C R340 T = foot I_B A C R135 !real r = 2.718; !path wA = CR(A, r-abs(B-C)); !path wC = CR(C, r-abs(B-A)); D = IP wA wC I_D = incenter A D C R20 A--B--C--D--cycle 0.1 lightcyan / heavycyan incircle A B C heavycyan dotted incircle A D C heavycyan dotted I = extension D I_D B I_B R315 CP I foot I A B dotted red circumcircle A I_B I_D 0.1 lightblue / blue circumcircle C I_B I_D 0.1 lightblue / blue O_A = circumcenter A I_B I_D R270 O_C = circumcenter C I_B I_D R270 I_B--I_D orange A--I--C red A--C orange O_A--O_C orange A--I_B--C--I_D--cycle lightblue Z := IP Line O_A O_C circumcircle I I_B I_D R45 X = extension I I+dir(90)*(Z-I) O_A O_C X--O_A orange H_1 := IP CP midpoint X--O_C X CP O_C C H_2 := OP CP midpoint X--O_C X CP O_C C H_1--X--H_2 heavygreen circumcircle X Z I 0.1 lightgreen / dashed heavygreen B--I--D red */ [/asy][/asy] Remark: [Luke, Ankan] The first two claims are essentially a generalization of RMM 2015/4. In the RMM problem the quadrilateral with an incircle is degenerated to a triangle. Now, the heart of the problem is the following claim. Lemma: Line $\overline{IX}$ is the external angle bisector of $\angle AIC$ and the internal angle bisector of $\angle BID$. Proof. As for the main proof, we let $\gamma$ be the Appolonian circle of $(O_A)$ and $(O_C)$; this coincides with the circumcircle of $\triangle X I_B I_D$ which is in fact isosceles. From \[ \frac{IO_A}{IO_C} = \frac{AO_A}{CO_C} \]we see $I$ lies on $\gamma$ too. Thus $\overline{IX}$ is the external angle bisector of $\angle O_A I O_C \equiv \angle AIC$. Moreover, $\overline{IX}$ is now the internal angle bisector of $\angle I_B I I_D \equiv \angle BID$. $\blacksquare$ Similarly, $\overline{IY}$ is the external angle bisector of $\angle BID$ and the internal angle bisector of $\angle AIC$. So $\angle XIY = 90^{\circ}$ follows.

Claim 1. $\overline{I_AI_C} \perp \overline{BD}$ Proof. Let $T$ be the $A$-intouch point of $\triangle ABD$ and $T'$ be the $C$-intouch point of $\triangle CBD$. By Pitot's Theorem, \[2BT = AB + BD - AD = CD + BD - CB = 2BT'.\]Thus, $T\equiv T'$ and the result follows. Similarly, $\overline{I_BI_D}\perp \overline{AC}$ Let $\omega_a$ be the circumcircle of $\triangle AI_bI_d$. Similarly define, $\omega_b$, $\omega_c$ and $\omega_d$. Let $O_a$ be the center of $\omega_a$. Similarly define $O_b$, $O_c$ and $O_d$. Claim 2. $O_a \in \overline{AI}$ Proof. Since, $CA\perp I_bI_d$, $\angle I_bAO_a = \angle CAI_d$ \[\angle BAO_a = \angle BAI_b + \angle I_bAO_a = \angle I_bAC + \angle CAI_d = \angle I_bAI_d = \tfrac12(\angle BAC + \angle DAC) = \tfrac12\angle BAD\]Thus, $O_a\in\overline{AI}$. Claim 3. $\overline{O_aO_c}\parallel \overline{AC}$. Proof 3. Since, $\overline{I_bI_d}$ is the the radical axis of $\omega_a$ and $\omega_c$. Also, $\overline{AC} \perp \overline{I_bI_d}$. Thus, $\overline{AC}\parallel \overline{O_aO_c}$. Claim 4. $\overline{IX}$ is the external angle bisector of $\angle AIC$. Proof. \[\frac{XO_a}{XO_c} = \frac{AO_a}{CO_c} = \frac{IO_a}{IO_c}\]Thus, $\odot(II_bI_cX)$ is the Apollonian circle of $\omega_a$ and $\omega_c$. Therefore, $IX$ is the external bisector of $\angle O_aIO_c \equiv \angle AIC$. Claim 5. $\overline{IY}$ is the internal angle bisector of $\angle AIC$. Proof. Note that by symmetry, $YI_a = YI_c$. Since, $Y, I, I_a, I_c$ are concylic, the result follows. Thus, we are done. All the solutions posted are very, very similar. Any other proofs?

Here's a solution by pure angle chasing. Lemma: Let 2 circles $\omega_1$ and $\omega_2$ with centers $O_1,O_2$ intersect at $A,B$, where $\omega_2$ is bigger than $\omega_1$. Their common external tangents meet at $X $. If $\angle AO_1B=x, \angle AO_2B=y $, then $\angle AXB= \frac{x-y}{2}. $

Back to problem: Let $\angle A=2a, \angle B=2b, \angle C=2c, \angle D=2d. $ We will show $X,I_b,I,I_d $ cyclic. Let $O_1,O_2$ be the circumcenters of $\triangle AI_bI_d, CI_bI_d $ respectively (here assuming quadrialteral $ABID$ convex). Then $\angle I_bO_1I_d=2a, \angle I_bO_2I_d=2c $, so by lemma $\angle I_bXI_d=a-c,$, furthermore $\angle I_bII_d=b+2c+d$, summing them we get $a+b+c+d $ which is $180^{\circ}$. Since $XI_b=XI_d$, we get that $IX $ is the angle bisector of $\angle BID $, similarly $IY $ is the angle bisector of $\angle AIC $. Finally (assume $AICB $ convex) $\angle XIY=\angle BIX +\angle AIY- \angle AIB= \frac{b+2c+d}{2} +\frac{a+2d+c}{2} - (180^{\circ}-a-b)= \frac{3}{2} (a+b+c+d) -180^{\circ}= 90^{\circ}$ and we are done.

Surprisingly, I've solved a geometry problem. First, let $(P,\omega)$ denote the square root of the power of a point from $P$ to a circle $\omega$. Let $\omega_{A,B,C,D}$ be the 4 incircles in the problem. Because $ABCD$ is circumscribed, that means $AB+CD=BC+AD$. But then \[ (A,\omega_B)+(C,\omega_B) = AC = (A,\omega_D)+(C,\omega_D)\]\[ (A,\omega_B)+(C,\omega_D) = AB + CD - (B,\omega_B)-(D,\omega_D) = (A,\omega_D)+(C,\omega_D)\]and the two equations imply $(A,\omega_B)=(A,\omega_D)$. Hence $\omega_B,\omega_D$ are both tangent to $AC$ at the same point. Let $J_a,J_b$ be the circumcenters of $AI_bI_D,CI_bI_d$. Then since $AC \perp I_bI_c$, we have that $AC$ and $AJ_a$ are symmetric around the perp. bisector of $\angle I_bAI_d$. But so are $AC$ and $AI$ by easy angle-chasing. Therefore, $A,J_a,I$ are collinear and similarly $C,J_c,I$ are collinear so that \[ I = AJ_a \cap CJ_c.\]Notice $AC\perp I_bI_d \perp J_aJ_c$ and so the parallelity gives us \[ \frac{J_aI}{IJ_c} = \frac{J_aA}{J_cC} = \frac{rad(AI_bI_d)}{rad(CI_bI_d)} = \frac{J_aX}{XJ_c},\]with the last equation because $X$ is the external center of similitude of the two circles. Therefore $IX$ is the external angle bisector of $\angle J_aIJ_b = \angle AIC$. Similarly, $YI$ is the external angle bisector of $\angle BID$. Since the external and internal angle bisectors of an angle are perpendicular, to finish the problem it suffices to prove the (internal) angle bisectors of $BID$ and $AIC$ are perpendicular. That is equivalent to proving $\angle AID + \angle BIC =180$, but this follows easily by angle-chasing from the condition that $I$ is the incenter of $ABCD$. $\square$

Doable for a G7, if one is familiar with the configuration (atleast doable at home, cause one has geogebra; otherwise the diagram's just a big fat mess!). My solution is quite similar to some of the above solutions, but still posting since it is a G7 (and also cuz this is my 200th post ). See the remark at the end for some motivation. Anyway, here's my solution: Let $\odot (AI_bI_d)=\omega_a,\odot (BI_aI_c)=\omega_b,\odot (CI_bI_d)=\omega_c,\odot (DI_aI_c)=\omega_d$, and let $O_a,O_b,O_c,O_d$ be their respective centers. Also, Let $R_a,R_b,R_c,R_d$ be their respective radii. We start off with a well known lemma. LEMMA $I_aI_c \perp BD$, and $O_bO_d \parallel BD$. PROOF: Let the incircles of $\triangle DAB$ and $\triangle BCD$ meet $BD$ at $P$ and $Q$ respectively. Then $BP=\frac{1}{2}(AB+BD-AD)$ and $BQ=\frac{1}{2}(BC+BD-CD)$. But, By Pitot's Theorem, we have $AB+CD=BC+AD \Rightarrow AB-AD=BC-CD \Rightarrow BP=BQ \Rightarrow P \equiv Q$. This gives that $I_aI_c \perp BD$. Also as $I_aI_c$ is the radical axis of $\omega_b$ and $\omega_d$, so we get that $O_bO_d \perp I_aI_c \Rightarrow O_bO_d \parallel BD$. $\Box$ Return to the problem at hand. Note that in $\triangle BI_aI_c$, we have $BD$ and $BO_b$ are isogonal in $\angle I_aBI_c$ $\Rightarrow \angle I_aBO_b=\angle I_cBD=\frac{1}{2} \angle CBD=\frac{1}{2}(\angle CBA-\angle DBA)=\angle ABI-\angle ABI_a=\angle I_aBI$ $\Rightarrow O_b$ lies on $BI$. All other such results hold cyclically. Thus, we have that $$\frac{YO_b}{YO_d}=\frac{R_b}{R_d}=\frac{BO_b}{DO_d}=\frac{IO_b}{IO_d}$$where the last equality follows from the fact that $\triangle IO_bO_d \sim \triangle IBD$. As $Y$ lies on $O_bO_d$, the above equality gives that $YI$ is the external angle bisector of $\angle BID$. In a similar fashion, we have that $XI$ is the external angle bisector of $\angle AIC$. So it suffices to show that the external angle bisector of $\angle AIC$ is the internal angle bisector of $\angle BID$, which is an easy angle chase. Hence done. $\blacksquare$ REMARK: My solution was inspired by just one thought, basically the involvement of so many incenters and the weird condition to be proved, all of which made me think that maybe $XI$ and $YI$ were internal and external angle bisector of some angle, which will help me prove the problem. $\angle AIC$ or $\angle BID$ seemed like the best choice, and noticing that the external angle bisector of $\angle AIC$ must be the internal angle bisector of $\angle BID$ was a kind of a confirmation that I was going on the right path. Some angle chase, along with the lemma (which was, by chance, known to me) and the rest is all history!! EDIT: 200th post on AOPS. Yipee!!

Can someone tell me what the shooting lemma is?

Suppose that the point at which the incircle of $ABC$ hits $AC$ is $K_B$, and likewise for $ADC$. Clearly, $K_BA=\frac{AB-BC+AC}{2}=\frac{AD_CD+AC}{2}=K_DA$ by Pitot, so $K_B=K_D$. Therefore, these two circles are tangent to each other, and $IBI_D\perp AC$. Likewise, $I_AI_C\perp BD$. Define the circumcenter of $AI_BI_D$ to be $O_A$ and similarly for other 3. It is easy to see that $BI$ and $BD$ are isogonal in $BI_AI_C$, so $O_B$ lies on $BI$. As $I_AI_C$ is the radical axis of $(O_B)$ and $(O_D)$, we have $O_BO_D\perp I_AI_C$, so $O_BO_D\parallel BD$. Now, extend $IY$ to intersect $BD$ at $Y'$. Due to $I_BI_D\parallel BD$, there exists a homothety which brings $YO_BO_D$ (which is a line) to $Y'BD$. Therefore, $\frac{Y'B}{Y'D}=\frac{YO_B}{YO_D}$. If $R_B$, $R_D$ are the radii of the two circles, this is equal to $\frac{R_B}{R_D}$ by similar triangles. Now, by extend law of sines, $2R_B\sin\frac{B}{2}=I_AI_C=2R_D\sin\frac{D}{2}$, so the ratio is $\frac{\sin\frac{D}{2}}{\sin\frac{B}{2}}=\frac{BI}{DI}$. We get a similar result for $X'$, and we wish to show $\angle X'IY'=90$. I will proceed with complex numbers. Suppose that the points at which the incircle touches $AB$, $BC$, $CD$, and $AD$ are $W$, $X$, $Y$, $Z$ (for convenience, we will assume the original points $X$, $Y$ don't exist anymore.) This implies that $a=\frac{2wz}{w+z}$ and similar expressions. Then, by the ratio we found earlier, we know that $y'=\frac{d|b|-|d|b}{|b|-|d|}$. We wish to show that $\frac{y'}{x'}$ is purely imaginary, so we can get rid of constant factors. $y'=C\left(\frac{d}{|d|}-\frac{b}{|b|}\right)=\sqrt{\frac{d}{\overline{d}}}-\sqrt{\frac{b}{\overline{b}}}=\sqrt{yz}-\sqrt{wx}$, where $\sqrt{yz}$ is defined as the midpoint of the arc $YZ$. There is a similar expression for $x'$, so we need to show $\frac{\sqrt{yz}-\sqrt{wx}}{\sqrt{wz}-\sqrt{xy}}$ is purely imaginary. Conjugating, we realize that this expression is either equal to, or the negation of its conjugate, so it is purely real or imaginary. However, it is obviously not real, since that would imply $X'$ and $Y'$ both lie on the intersection of $AC$ and $BD$, so it is imaginary, as desired.

ISL 2017 G7 wrote: A convex quadrilateral $ABCD$ has an inscribed circle with center $I$. Let $I_a, I_b, I_c$ and $I_d$ be the incenters of the triangles $DAB, ABC, BCD$ and $CDA$, respectively. Suppose that the common external tangents of the circles $AI_bI_d$ and $CI_bI_d$ meet at $X$, and the common external tangents of the circles $BI_aI_c$ and $DI_aI_c$ meet at $Y$. Prove that $\angle{XIY}=90^{\circ}$. Solution with Kayak : First consider the following lemma, Lemma 1. Let two circles of unequal radius $\Omega_1$ and $\Omega_2$ with centers $A$ and $B$ respectively, intersect at $X$ and $Y,$ and the intersection of both external common tangents be $T,$ then it follows that \[\angle XTY=\tfrac12|\angle XAY-\angle XBY|.\]Proof. Let the tangency points $P,Q,R$ and $S$ as shown in the diagram. Assume $\angle XAY>\angle XBY.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); 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 = -13.65138319721991, xmax = 11.010604454866138, ymin = -8.745266251788848, ymax = 7.5882468817238795; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw(circle((-4.55,0.34), 2.851581410625324), linewidth(0.8) + blue); draw(circle((1.97,0.34), 5.126097756902516), linewidth(0.8) + blue); draw((-12.724182097089779,0.34)--(0.1817495182991372,5.144064781829826), linewidth(0.8)); draw((-12.724182097089779,0.34)--(0.1817495182991376,-4.464064781829825), linewidth(0.8)); draw((-12.724182097089779,0.34)--(1.97,0.34), linewidth(0.8) + dtsfsf); draw((1.97,0.34)--(0.1817495182991372,5.144064781829826), linewidth(0.8) + dtsfsf); draw((-4.55,0.34)--(-5.5447804495717,3.0124386613304805), linewidth(0.8) + dtsfsf); draw((-4.55,0.34)--(-2.6815154656362807,2.4941313994897123), linewidth(0.8) + dtsfsf); draw((-2.6815154656362807,2.4941313994897123)--(1.97,0.34), linewidth(0.8) + dtsfsf); draw((-12.724182097089779,0.34)--(-2.6815154656362807,2.4941313994897123), linewidth(0.8) + dtsfsf); /* dots and labels */ dot((-4.55,0.34),linewidth(2pt) + dotstyle); label("$A$", (-4.433459943653323,0.4457040098082481), NE * labelscalefactor); dot((-2.6815154656362807,2.4941313994897123),linewidth(2pt) + dotstyle); label("$X$", (-2.573703497758309,2.601943367367684), NE * labelscalefactor); dot((1.97,0.34),linewidth(2pt) + dotstyle); label("$B$", (2.0891641129639718,0.4457040098082481), NE * labelscalefactor); dot((-2.6815154656362807,-1.8141313994897124),linewidth(2pt) + dotstyle); label("$Y$", (-2.573703497758309,-1.710535347751188), NE * labelscalefactor); dot((-5.5447804495717,-2.332438661330481),linewidth(4pt) + dotstyle); dot((-5.5447804495717,3.0124386613304805),linewidth(4pt) + dotstyle); dot((-12.724182097089779,0.34),linewidth(2pt) + dotstyle); label("$T$", (-12.627169502379179,0.4457040098082481), NE * labelscalefactor); dot((-5.5447804495717,3.0124386613304805),linewidth(2pt) + dotstyle); label("$P$", (-5.430720646524562,3.11405021478805), NE * labelscalefactor); dot((0.1817495182991372,5.144064781829826),linewidth(2pt) + dotstyle); label("$Q$", (0.28331365100794387,5.243336580377993), NE * labelscalefactor); dot((0.1817495182991376,-4.464064781829825),linewidth(2pt) + dotstyle); label("$R$", (0.28331365100794387,-4.351928560761497), NE * labelscalefactor); dot((-5.5447804495717,-2.332438661330481),linewidth(2pt) + dotstyle); label("$S$", (-5.430720646524562,-2.222642195171554), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Because of similar triangles, we have \[\frac{AX}{XB}=\frac{AP}{BQ}=\frac{TA}{TB}\]therefore, $XT$ is the external angle bisector of $\angle AXB$ in $\triangle AXB,$ so, \[\angle XTA=\angle XAB-\angle TXA\\ =\angle XAB-\tfrac12(\angle XAB+\angle XBA)\\ =\tfrac12(\angle XAB-\angle XBA) \]Hence the lemma follows. $\square$ Claim 1. $IX$ is the angle bisector of $\angle BID.$ Proof. [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 = -15.029587282329945, xmax = 7.821125636107711, ymin = -3.9589162285562303, ymax = 11.29166970599573; /* image dimensions */ pen qqffff = rgb(0,1,1); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw(circle((-3.63,0.82), 2.561249694973139), linewidth(0.4) + green); draw(circle((-2.9301484511856355,1.6716958302436256), 1.8351371121061375), linewidth(0.4) + qqffff); draw(circle((-5.043050471867659,-0.23065722546216416), 1.007977746206019), linewidth(0.4) + qqffff); draw(circle((-4.9236315400142,1.7612603815200627), 1.9954940827501395), linewidth(0.4) + dtsfsf); draw(circle((-2.432709104490185,-1.005352803974168), 2.722872617954823), linewidth(0.4) + dtsfsf); draw((-6.537200028152899,2.935310366987723)--(-1.1613820360339426,3.824225153086133), linewidth(0.4)); draw((-1.1613820360339426,3.824225153086133)--(-0.939307930838531,-3.28214621316702), linewidth(0.4)); draw((-0.939307930838531,-3.28214621316702)--(-5.960033123256581,-0.9124689989877295), linewidth(0.4)); draw((-5.960033123256581,-0.9124689989877295)--(-6.537200028152899,2.935310366987723), linewidth(0.4)); draw((-11.757241194551257,9.3512014833134)--(-3.63,0.82), linewidth(0.4)); draw((-3.63,0.82)--(-1.1613820360339426,3.824225153086133), linewidth(0.4)); draw((-3.63,0.82)--(-5.960033123256581,-0.9124689989877295), linewidth(0.4)); draw((-4.773226070595338,-2.3967605892700847)--(-11.757241194551257,9.3512014833134), linewidth(0.4)); draw((-11.757241194551257,9.3512014833134)--(-0.8041446517700683,1.1768050997863675), linewidth(0.4)); draw((-6.537200028152899,2.935310366987723)--(-0.939307930838531,-3.28214621316702), linewidth(0.4)); draw(circle((-7.813705265614099,4.97120219354323), 5.893714397661739), linewidth(0.4) + blue+dashed); draw((-11.757241194551257,9.3512014833134)--(-5.043050471867659,-0.23065722546216416), linewidth(0.4)); draw((-2.9301484511856355,1.6716958302436256)--(-11.757241194551257,9.3512014833134), linewidth(0.4)); draw((-6.537200028152899,2.935310366987723)--(-5.043050471867659,-0.23065722546216416), linewidth(0.4)); draw((-6.537200028152899,2.935310366987723)--(-2.9301484511856355,1.6716958302436256), linewidth(0.4)); /* dots and labels */ dot((-3.63,0.82),linewidth(2pt) + dotstyle); label("$I$", (-3.5287328729400675,0.9232845491716251), NE * labelscalefactor); dot((-6.537200028152899,2.935310366987723),linewidth(2pt) + dotstyle); label("A", (-6.44798694622065,3.037227153961006), NE * labelscalefactor); dot((-1.1613820360339426,3.824225153086133),linewidth(2pt) + dotstyle); label("B", (-1.062466500685783,3.918036572623248), NE * labelscalefactor); dot((-5.960033123256581,-0.9124689989877295),linewidth(2pt) + dotstyle); label("D", (-5.869169328242603,-0.813168304762509), NE * labelscalefactor); dot((-0.939307930838531,-3.28214621316702),linewidth(2pt) + dotstyle); label("C", (-0.8359726501726344,-3.1787707434553876), NE * labelscalefactor); dot((-2.9301484511856355,1.6716958302436256),linewidth(2pt) + dotstyle); label("$I_d$", (-2.824085338010272,1.778927984443517), NE * labelscalefactor); dot((-5.043050471867659,-0.23065722546216416),linewidth(2pt) + dotstyle); label("$I_b$", (-4.938027942799659,-0.13368675322306528), NE * labelscalefactor); dot((-4.773226070595338,-2.3967605892700847),linewidth(4pt) + dotstyle); dot((-0.8041446517700683,1.1768050997863675),linewidth(4pt) + dotstyle); dot((-11.757241194551257,9.3512014833134),linewidth(2pt) + dotstyle); label("$X$", (-11.657345508023067,9.454552918500196), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] It is obvious that $B,I_d,I$ and $D,I_b,I$ are collinear. Therefore, \[\angle I_dII_b=\angle BID=360^{\circ}-\tfrac12\angle B-\tfrac12\angle C-\angle A=180^{\circ}-\tfrac{1}{2}(\angle A-\angle C).\]Note that $\angle I_d A I_b=\tfrac12 \angle A$ and $\angle I_dCI_b=\tfrac12\angle C,$ therefore by lemma 1, we get that $\angle I_d X I_b=\tfrac12(\angle A-\angle C).$ So, it follows that $XI_dII_b$ is cyclic, but it is obvious that $XI_d=XI_b,$ so $IX$ is the angle bisector of $\angle BID.$ Hence the claim. $\square$ Coming back to the original problem, it reduces to showing that the angle bisectors of $\angle BID$ and $\angle AIC$ are perpendicular. Note that $\angle AIB+\angle DIC=180^{\circ},$ hence it follows that $AI$ and $CI$ are isogonal with respect to $\angle BID,$ thus, it is obvious by simple angle chasing that $IX\perp IY.$ And we are done. $\blacksquare$

ISL 2017 G7 wrote: A convex quadrilateral $ABCD$ has an inscribed circle with center $I$. Let $I_a, I_b, I_c$ and $I_d$ be the incenters of the triangles $DAB, ABC, BCD$ and $CDA$, respectively. Suppose that the common external tangents of the circles $AI_bI_d$ and $CI_bI_d$ meet at $X$, and the common external tangents of the circles $BI_aI_c$ and $DI_aI_c$ meet at $Y$. Prove that $\angle{XIY}=90^{\circ}$. Solution: $CD+AB=AD+BC$ $\implies$ $CD+AC-AD=AC+BC-AB \implies I_BI_D \perp AC$ & Similarly, $I_AI_C \perp BD$. Obviously, $\Delta YI_AI_C$ & $\Delta XI_BI_D$ are Isosceles. Let $\stackrel{\longrightarrow}{YI_C}, \stackrel{\longrightarrow}{YI_A}$ $\cap$ $\odot (BI_AI_C)$ $=$ $E,F$. Let $O_1$ be center of $\odot (BI_AI_C)$ $$\angle I_CYI_A=180^{\circ}-2\angle I_AFE=\angle I_AO_1F-2\angle O_1FE=90^{\circ}-\angle I_AO_1Y-\angle O_1FE=\angle I_ADI_C-\angle I_AO_1Y$$$$\angle I_ADI_C-\angle I_AO_1Y=\angle I_ADI_C-\angle I_ABI_C=\tfrac{1}{2} (\angle ADC-\angle ABC)=\tfrac{1}{2}(360^{\circ}-2\angle ABC-\angle BCD-\angle DAB)=180^{\circ}-(\angle ABC+\angle BCI+\angle BAI)=180^{\circ}-\angle I_AII_C$$Hence, $YI_CI_AI$ is cyclic and Similarly, $XI_DII_B$ is cyclic $\implies$ $YI$ bisects $\angle AIC$ & $XI$ bisects $\angle BID$. Using, the fact $\angle CID$ $+$ $\angle AIB$ $=$ $180^{\circ}$ $\implies$ $\angle XIY$ $=$ $90^{\circ}$ $\qquad \blacksquare$

Not too hard but nice! fastlikearabbit wrote: A convex quadrilateral $ABCD$ has an inscribed circle with center $I$. Let $I_a, I_b, I_c$ and $I_d$ be the incenters of the triangles $DAB, ABC, BCD$ and $CDA$, respectively. Suppose that the common external tangents of the circles $AI_bI_d$ and $CI_bI_d$ meet at $X$, and the common external tangents of the circles $BI_aI_c$ and $DI_aI_c$ meet at $Y$. Prove that $\angle{XIY}=90^{\circ}$. Here's the key lemma: Lemma 1: We have $I_AI_C \perp BD.$ Proof: We just need to show $I_CB^2-I_CD^2=I_AB^2-I_AD^2.$ This is easy using cosine rule and Pitot's theorem. $\square$ Let $S,T$ be the centers of $BI_AI_C$ and $DI_AI_C$ respectively. Here are two corollaries of the above lemma: Corollary 1: The points $S,T$ lie on $IB,ID$ respectively. Proof: By the lemma, we get that $DT,DB$ are isogonal in $\angle I_ADI_C.$ But also $$\angle I_CDI_A=\frac{1}{2}\angle D=\angle IDC \implies \angle I_ADI=\angle I_CDC=\angle BDI_C$$So $DI, DB$ are also isogonal in $\angle I_ADI_C.$ Hence, $T \in DI.$ Similarly $S \in BI.$ $\square$ Corollary 2: We have $ST \parallel BD.$ Proof: Just see that both are perpendicular to $I_AI_C.$ $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); 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 = -4.999905125993694, xmax = 6.960844435565477, ymin = -3.562507651284475, ymax = 4.202883936685942; /* image dimensions */ pen qqzzff = rgb(0,0.6,1); pen ccwwff = rgb(0.8,0.4,1); pen ffwwzz = rgb(1,0.4,0.6); pen ccqqqq = rgb(0.8,0,0); draw((-2.4288478136593787,1.0901400430807582)--(-0.25955984491353046,1.54429430694683)--(4.125856827611618,-2.7638568707290907)--(-3.9416120490338984,-2.7595651023533248)--cycle, linewidth(0.5) + ccwwff); draw((-0.48603505340567665,0.7714704399390128)--(-1.9611616324499497,0.5364320276549644)--(-1.6407999043190942,-1.1925016900688314)--(-0.14871777663353036,-1.014156004968163)--cycle, linewidth(0.5) + green); /* draw figures */ draw(circle((-0.917136819796367,-0.6996209717618296), 2.0615528128088307), linewidth(0.5) + qqzzff); draw((-2.4288478136593787,1.0901400430807582)--(-0.917136819796367,-0.6996209717618296), linewidth(0.4)); draw((-0.917136819796367,-0.6996209717618296)--(-0.25955984491353046,1.54429430694683), linewidth(0.4)); draw((-0.917136819796367,-0.6996209717618296)--(4.125856827611618,-2.7638568707290907), linewidth(0.4)); draw((-3.9416120490338984,-2.7595651023533248)--(-0.917136819796367,-0.6996209717618296), linewidth(0.4)); draw((-2.4288478136593787,1.0901400430807582)--(-0.25955984491353046,1.54429430694683), linewidth(0.5) + ccwwff); draw((-0.25955984491353046,1.54429430694683)--(4.125856827611618,-2.7638568707290907), linewidth(0.5) + ccwwff); draw((4.125856827611618,-2.7638568707290907)--(-3.9416120490338984,-2.7595651023533248), linewidth(0.5) + ccwwff); draw((-3.9416120490338984,-2.7595651023533248)--(-2.4288478136593787,1.0901400430807582), linewidth(0.5) + ccwwff); draw(circle((-0.6399687175479255,0.24618729580528861), 1.3526983118026712), linewidth(0.5) + ffwwzz); draw(circle((-2.191950403003942,-1.567885566208114), 2.116935519201168), linewidth(0.5) + ffwwzz); draw((-0.48603505340567665,0.7714704399390128)--(-1.9611616324499497,0.5364320276549644), linewidth(0.5) + green); draw((-1.9611616324499497,0.5364320276549644)--(-1.6407999043190942,-1.1925016900688314), linewidth(0.5) + green); draw((-1.6407999043190942,-1.1925016900688314)--(-0.14871777663353036,-1.014156004968163), linewidth(0.5) + green); draw((-0.14871777663353036,-1.014156004968163)--(-0.48603505340567665,0.7714704399390128), linewidth(0.5) + green); draw((-2.191950403003942,-1.567885566208114)--(2.10703572759203,3.457092593501731), linewidth(0.4) + linetype("2 2")); draw((-3.9416120490338984,-2.7595651023533248)--(-0.25955984491353046,1.54429430694683), linewidth(0.5) + ccqqqq); 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label("$I$", (-0.8045471524049737,-0.9475148764188879), NE * labelscalefactor); dot((-2.835865702781308,0.05435480609801824),linewidth(4pt) + dotstyle); dot((-1.3395775135209669,1.318185723708646),linewidth(4pt) + dotstyle); dot((0.5275882124686249,0.7710148027067911),linewidth(4pt) + dotstyle); dot((-0.9182335337781975,-2.7611734928532625),linewidth(4pt) + dotstyle); dot((-2.4288478136593787,1.0901400430807582),linewidth(4pt) + dotstyle); label("$A$", (-2.6350420948108924,1.1672184110811086), NE * labelscalefactor); dot((-0.25955984491353046,1.54429430694683),linewidth(4pt) + dotstyle); label("$B$", (-0.2588095298243271,1.6788474322504625), NE * labelscalefactor); dot((4.125856827611618,-2.7638568707290907),linewidth(4pt) + dotstyle); label("$C$", (4.266264924073534,-3.0167700287038306), NE * labelscalefactor); dot((-3.9416120490338984,-2.7595651023533248),linewidth(4pt) + dotstyle); label("$D$", (-4.226776827337778,-3.0281395625075938), NE * labelscalefactor); dot((-0.48603505340567665,0.7714704399390128),linewidth(4pt) + dotstyle); label("$I_B$", (-0.4066134692732522,0.8033933293606791), NE * labelscalefactor); dot((-0.14871777663353036,-1.014156004968163),linewidth(4pt) + dotstyle); label("$I_C$", (0.0026897476622327163,-1.2999704243355539), NE * labelscalefactor); dot((-1.6407999043190942,-1.1925016900688314),linewidth(4pt) + dotstyle); label("$I_D$", (-1.6686317214909974,-1.4477743637844784), NE * labelscalefactor); dot((-1.9611616324499497,0.5364320276549644),linewidth(4pt) + dotstyle); label("$I_A$", (-2.2939560806979884,0.6328503223042278), NE * labelscalefactor); dot((0.615309870914632,-0.2578646738379675),linewidth(4pt) + dotstyle); dot((-1.3322473202308909,1.408315971564181),linewidth(4pt) + dotstyle); dot((2.10703572759203,3.457092593501731),linewidth(4pt) + dotstyle); label("$Y$", (2.1515316365735284,3.5434509760676636), NE * labelscalefactor); dot((-0.6399687175479255,0.24618729580528861),linewidth(4pt) + dotstyle); label("$S$", (-0.8386557538162641,0.40545964622895936), NE * labelscalefactor); dot((-2.191950403003942,-1.567885566208114),linewidth(4pt) + dotstyle); label("$T$", (-2.339434215913042,-1.4364048299807148), NE * labelscalefactor); dot((-1.2854592253228834,0.34514599057340617),linewidth(4pt) + dotstyle); dot((-3.275347307114448,0.25081373291180575),linewidth(4pt) + dotstyle); dot((-0.22747406173625173,-2.3567129852644157),linewidth(4pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Clearly $Y$ lies on $I_AI_C.$ If $r_b,r_d$ are the radii of $BI_AI_C, DI_AI_C$ respectively, then $$\frac{YS}{YT}=\frac{r_b}{r_d}=\frac{SB}{TD}=\frac{SI}{TI}$$This, by sine rule, implies that $YI$ bisects $\angle BID$ externally. We get a similar result for $XI.$ Now since $\angle AID+\angle BIC=\pi,$ hence we get $\angle XIY=\pi/2,$ so done. $\blacksquare$

Nice problem, posting for storage! fastlikearabbit wrote: A convex quadrilateral $ABCD$ has an inscribed circle with center $I$. Let $I_a, I_b, I_c$ and $I_d$ be the incenters of the triangles $DAB, ABC, BCD$ and $CDA$, respectively. Suppose that the common external tangents of the circles $AI_bI_d$ and $CI_bI_d$ meet at $X$, and the common external tangents of the circles $BI_aI_c$ and $DI_aI_c$ meet at $Y$. Prove that $\angle{XIY}=90^{\circ}$. Denote as $O_C$ the circumcenter of $CI_BI_D$ and $O_A$ the circumcenter of $AI_BI_D$. Claim 01. $I_BI_D \perp AC$. Proof. We'll prove that the incircles of $\triangle ABC$ and $\triangle ACD$ tangent together at a point on $AC$. To prove this, notice that $BC + AC - AB = CD + CA - AD$ by Pitot Theorem, and therefore the statement is true. Claim 02. $C, O_C, I$ are collinear. Proof. Notice that \[ \angle I C I_D = \angle ICD - \angle I_D C D = \frac{1}{2} \angle BCD - \frac{1}{2} \angle ACD = \frac{1}{2} \angle ABC = \angle I_B C A = 90^{\circ} - \angle CI_BI_D = \angle O_C C I_D \]Similarly, $A,O_A, I$ are collinear as well. Claim 03. $O_C O_A \parallel AC$ Proof. This immediately follows after notice that $O_C O_A \perp I_B I_D$ and $I_B I_D \perp AC$ by Claim 01. Claim 04. $XI$ is the external angle bisector of $\angle CIA$. Denote $R_C$ and $R_A$ as the radius of the circumcircle of $(CI_BI_D)$ and $(AI_BI_D)$. Now, notice that \[ \frac{XO_C}{XO_A} = \frac{R_C}{R_A} = \frac{O_C C}{O_A A} = \frac{O_C I}{O_AI} \]Therefore, by Sine Rule, \[ \frac{O_C X}{O_A X} = \frac{\frac{IO_C}{\sin \angle O_C XI}}{\frac{IO_A}{\sin \angle O_A X I}} = \frac{\frac{O_C X}{\sin \angle XIO_C}}{\frac{O_A X}{\sin XIO_A} }\]which concludes that $XI$ is the external angle bisector of $\angle CIA$. Similarly, $YI$ is the external angle bisector of $\angle BID$. Claim 05. $\angle CIX - \angle DIY = \frac{1}{2} ( \angle CID - \angle AIB)$ Proof. Notice that $\angle ACI = \angle ACI_D - \angle ICI_D = \angle ACI_D - \angle I_B C A = \frac{1}{2} ( \angle ACD - \angle ACB)$. Therefore, \[ \angle CIX - \angle DIY = \frac{1}{2} ( \angle ACI + \angle CAI - \angle BDI - \angle DBI) = \frac{1}{4} ( \angle ABC + \angle BAD - \angle BCD - \angle ADC) = \frac{1}{2} ( \angle CID - \angle AIB) \] Hence, \[ \angle XIY = \angle XID + \angle DIY = \angle CID - \angle CIX + \angle DIY = \frac{1}{2} (\angle AIB + \angle CID) = 90^{\circ} \]

Nice problem. Lemma: Circles $\omega_A$ and $\omega_C$, with the radius of $\omega_A$ less than the radius of $\omega_C$, intersect at two points $I_B$ and $I_D$. Point $A$ is chosen on the circumference of $\omega_A$ but not in the interior of $\omega_C$. Point $C$ is chosen on the circumference of $\omega_C$ but not in the interior of $\omega_A$. If the common external tangents of $\omega_A$ and $\omega_C$ intersect at $X$, then $\angle I_BXI_D=\angle I_BAI_D-\angle I_BCI_D$. Proof. Denote by $\bullet'$ the homothety at $X$ sending $\omega_C$ to $\omega_A$. Then $X=\overline{I_BI_B'}\cap\overline{I_DI_D'}$, so with arcs taken with respect to $\omega_A$, \[\angle I_BXI_D=\frac{\widehat{I_BI_D}-\widehat{I_B'I_D'}}2=\angle I_BAI_D-\angle I_B'C'I_D'=\angle I_BAI_D-\angle I_BCI_D,\]as claimed. [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pen pri=heavyblue; pen sec=heavycyan; pen tri=purple; pen qua=heavygreen; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; real t=53.75; pair I,WW,X,Y,Z,A,B,C,D,IB,ID,T,U1,U2; I=(0,0); WW=dir(150+t); X=dir(90+t); Y=dir(10+t); Z=dir(250+t); A=2*WW*X/(WW+X); B=2*X*Y/(X+Y); C=2*Y*Z/(Y+Z); D=2*Z*WW/(Z+WW); IB=incenter(A,B,C); ID=incenter(A,D,C); T=2*foot(circumcenter(I,IB,ID),I,incenter(I,IB,ID))-I; U1=intersectionpoint(circle( (circumcenter(C,IB,ID)+T)/2,length(circumcenter(C,IB,ID)-T)/2),circumcircle(C,IB,ID)); U2=reflect(T,circumcenter(C,IB,ID))*U1; pair O1,O2,IBp,IDp,Cp; O1=circumcenter(A,IB,ID); O2=circumcenter(C,IB,ID); IBp=T+(IB-T)*(O1-T)/(O2-T); IDp=T+(ID-T)*(O1-T)/(O2-T); Cp=T+(C-T)*(O1-T)/(O2-T); draw(U1--T--U2,qua); filldraw(circumcircle(A,IB,ID),sfil,sec); filldraw(circumcircle(C,IB,ID),sfil,sec); draw(IB--T--ID,pri+Dotted); draw(T--C,tri+Dotted); dot("$A$",A,NW); dot("$C$",C,E); dot("$C'$",Cp,SE); dot("$I_B$",IB,dir(110)); dot("$I_B'$",IBp,SE); dot("$I_D$",ID,dir(250)); dot("$I_D'$",IDp,NE); dot("$X$",T,W); [/asy][/asy] $\blacksquare$ Claim: $I_BII_DX$ is cyclic. Proof. Without loss of generality the radius of $(AI_BI_D)$ is less than the radius of $(CI_BI_D)$. By the lemma, $\angle I_BXI_D=\angle I_BAI_D-\angle I_BCI_D$. However \[\angle I_BII_D=360^\circ-A-\frac B2-\frac D2=180^\circ-\frac A2+\frac C2=180^\circ-\angle I_BAI_D+\angle I_BCI_D,\]so $I_BII_DX$ is cyclic. $\blacksquare$ [asy][asy] size(8cm); defaultpen(fontsize(10pt)); pen pri=heavyblue; pen sec=heavycyan; pen tri=purple; pen qua=heavygreen; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; real t=53.75; pair I,WW,X,Y,Z,A,B,C,D,IB,ID,T,U1,U2; I=(0,0); WW=dir(150+t); X=dir(90+t); Y=dir(10+t); Z=dir(250+t); A=2*WW*X/(WW+X); B=2*X*Y/(X+Y); C=2*Y*Z/(Y+Z); D=2*Z*WW/(Z+WW); IB=incenter(A,B,C); ID=incenter(A,D,C); T=2*foot(circumcenter(I,IB,ID),I,incenter(I,IB,ID))-I; U1=intersectionpoint(circle( (circumcenter(C,IB,ID)+T)/2,length(circumcenter(C,IB,ID)-T)/2),circumcircle(C,IB,ID)); U2=reflect(T,circumcenter(C,IB,ID))*U1; filldraw(circumcircle(IB,I,ID),tfil,tri); draw(U1--T--U2,qua); filldraw(circumcircle(A,IB,ID),sfil,sec); filldraw(circumcircle(C,IB,ID),sfil,sec); filldraw(A--B--C--D--cycle,fil,pri); draw(B--I--D,pri); draw(A--C,pri); //filldraw(circle(I,1),fil,pri); filldraw(incircle(A,B,C),fil,pri); filldraw(incircle(A,D,C),fil,pri); dot("$I$",I,E); dot("$A$",A,W); dot("$B$",B,N); dot("$C$",C,E); dot("$D$",D,S); dot("$I_B$",IB,dir(10)); dot("$I_D$",ID,dir(-10)); dot("$X$",T,W); [/asy][/asy] By symmetry, $XI_B=XI_D$, so $\overline{IX}$ bisects $\angle BID$. Analogously $\overline{IY}$ bisects $\angle AIC$. It is easy to check that $\angle AIB+\angle CID=180^\circ$, so $\overline{AI}$ and $\overline{CI}$ are isogonal with respect to $\angle BID$. Hence $\overline{IX}\perp\overline{IY}$, as desired.

fastlikearabbit wrote: A convex quadrilateral $ABCD$ has an inscribed circle with center $I$. Let $I_a, I_b, I_c$ and $I_d$ be the incenters of the triangles $DAB, ABC, BCD$ and $CDA$, respectively. Suppose that the common external tangents of the circles $AI_bI_d$ and $CI_bI_d$ meet at $X$, and the common external tangents of the circles $BI_aI_c$ and $DI_aI_c$ meet at $Y$. Prove that $\angle{XIY}=90^{\circ}$.

Nice problem. Claim: $I_BII_DX$ is cyclic. Proof: To sort configuration issues, suppose that the point $X$ lies closer to $(I_BCI_D)$ than $(I_BAI_D)$. Let $\overline{XI_B}$ and $\overline{XI_D}$ intersect $(I_BCI_D)$ again at points $N$ and $M$, respectively. Now $\angle I_BXI_D = 1/2 (\widehat{I_BI_D} - \widehat{MN}) = \angle I_BCI_D - 1/2( \widehat{MN}) = \angle I_BCI_D - \angle I_BAI_D = \angle C/2 - \angle A/2$, while $\angle I_BII_D = 360^\circ - \angle C - \angle B/2 - \angle D/2 = 180^\circ - \angle C/2 + \angle A/2$, implying the desired result. Similarly $I_AII_CY$ is cyclic, so by incenter-excenter lemma as $X$ is the arc midpoint of $\widehat{I_BXI_D}$ we find $\overline{IX}$ bisects $\angle BID$, and similarly $\overline{IY}$ bisects $\angle AIC$. Now $\angle AIC+ \angle BID = 180^\circ$ by easy angle chasing, so $\overline{IX}$ is the exterior angle bisector of $\angle AIC$. Thus $\angle XIY = 90^\circ$ as desired.

[asy][asy] size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.143752258207736, xmax = 2.0400521822554074, ymin = -1.9619602628430035, ymax = 1.6009262041238428; /* image dimensions */pen zzttff = rgb(0.6,0.2,1); pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); draw((0.2688991041530577,-0.20488190249924387)--(0.343196309386966,-0.1798436523355436)--(0.31815805922326573,-0.10554644710163531)--(0.24386085398935742,-0.13058469726533556)--cycle, linewidth(2) + qqwuqq); /* draw figures */draw((-1.724052928736652,-1.1476111770905153)--(0.10721516655121223,-0.030815959615788993), linewidth(0.8) + zzttff); draw((0.10721516655121223,-0.030815959615788993)--(0.5387819607844094,-1.0057182971017373), linewidth(0.8) + zzttqq); draw((0.10721516655121223,-0.030815959615788993)--(1.7547138624769931,0.1753890157021572), linewidth(0.8) + zzttff); draw((0.10721516655121223,-0.030815959615788993)--(-0.24717655734806881,1.3264942555025145), linewidth(0.8) + zzttqq); draw(circle((-0.01853208013662149,0.4507924159254678), 0.9050591189518615), linewidth(0.8) + qqwuqq); draw(circle((0.2603461018327108,-0.37673624065269123), 0.6878553299095975), linewidth(0.8) + qqwuqq); draw(circle((-0.7426495270039757,-0.5491042581479939), 1.1495056193662294), linewidth(0.8) + blue); draw(circle((0.8717951742236775,0.06488074836982934), 0.889807556281553), linewidth(0.8) + blue); draw((-0.24717655734806881,1.3264942555025145)--(-1.724052928736652,-1.1476111770905153), linewidth(0.8)); draw((-1.724052928736652,-1.1476111770905153)--(0.5387819607844094,-1.0057182971017373), linewidth(0.8)); draw((0.5387819607844094,-1.0057182971017373)--(1.7547138624769931,0.1753890157021572), linewidth(0.8)); draw((1.7547138624769931,0.1753890157021572)--(-0.24717655734806881,1.3264942555025145), linewidth(0.8)); draw((-0.24717655734806881,1.3264942555025145)--(-0.42721849587105754,-0.3567392691445728), linewidth(0.8) + fuqqzz); draw((-0.42721849587105754,-0.3567392691445728)--(0.5387819607844094,-1.0057182971017373), linewidth(0.8) + fuqqzz); draw((0.5387819607844094,-1.0057182971017373)--(0.7955617418711429,0.055339185104539945), linewidth(0.8) + fuqqzz); draw((0.7955617418711429,0.055339185104539945)--(-0.24717655734806881,1.3264942555025145), linewidth(0.8) + fuqqzz); draw((-0.42721849587105754,-0.3567392691445728)--(0.7955617418711429,0.055339185104539945), linewidth(0.8) + zzttqq); draw((-0.24717655734806881,1.3264942555025145)--(0.5387819607844094,-1.0057182971017373), linewidth(0.8) + zzttqq); /* dots and labels */dot((0.10721516655121223,-0.030815959615788993),dotstyle); label("$I$", (0.12185915699006185,0.007975926880366944), NE * labelscalefactor); dot((-0.24717655734806881,1.3264942555025145),linewidth(4pt) + dotstyle); label("$A$", (-0.2329511135792507,1.3569941431074404), NE * labelscalefactor); dot((-1.724052928736652,-1.1476111770905153),linewidth(4pt) + dotstyle); label("$B$", (-1.8554689133701694,-1.1118939896040254), NE * labelscalefactor); dot((0.5387819607844094,-1.0057182971017373),linewidth(4pt) + dotstyle); label("$C$", (0.5542841742464115,-0.9751441978221029), NE * labelscalefactor); dot((1.7547138624769931,0.1753890157021572),linewidth(4pt) + dotstyle); label("$D$", (1.8072079421942964,0.08928661388583438), NE * labelscalefactor); dot((-0.42721849587105754,-0.3567392691445728),linewidth(4pt) + dotstyle); label("$I_{b}$", (-0.5840654438301329,-0.3468343436889455), NE * labelscalefactor); dot((0.3983693400869136,-0.6885284809256835),linewidth(4pt) + dotstyle); label("$I_{c}$", (0.41383844214605864,-0.6572933304370939), NE * labelscalefactor); dot((0.7955617418711429,0.055339185104539945),linewidth(4pt) + dotstyle); label("$I_{d}$", (0.7982162352628138,-0.05855099885137914), NE * labelscalefactor); dot((0.017367587756149494,0.31329770974530463),linewidth(4pt) + dotstyle); label("$I_{a}$", (0.09598757476104948,0.34061055553909736), NE * labelscalefactor); dot((-0.01853208013662149,0.4507924159254678),linewidth(4pt) + dotstyle); label("$O_a$", (-0.15533636689221358,0.4995359892316019), NE * labelscalefactor); dot((-0.7426495270039757,-0.5491042581479939),linewidth(4pt) + dotstyle); label("$O_b$", (-0.8501731467571173,-0.48358413547086804), NE * labelscalefactor); dot((0.8717951742236775,0.06488074836982934),linewidth(4pt) + dotstyle); label("$O_d$", (0.886918802905142,0.09298255420426473), NE * labelscalefactor); dot((0.2603461018327108,-0.37673624065269123),linewidth(4pt) + dotstyle); label("$O_c$", (0.27339271004570576,-0.3468343436889455), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] CLAIM 1. $I_bI_d\perp AC$ Proof. Suppose the incircle of $\triangle ABC$ and $ADC$ touches $AC$ at $X_B$ and $X_D$ respectively, then $$CX_B=\frac{AC+CB-AB}{2}=\frac{AC+CD-AD}{2}=CX_D$$hence $X_B=X_D$, therefore, $I_bI_d\perp AC$, similarly $I_aI_c\perp BD$ $\blacksquare$ CLAIM 2. The circumcenter of $\triangle CI_bI_d$ and $\triangle AI_bI_d$ lies on $AI$ and $CI$ respectively. Proof. Let $I_bI_d\cap AC=Z$ Notice that $$\angle AI_bZ=90^{\circ}-\angle I_bAC=\angle II_bC$$similarly, $\angle AI_dZ=\angle CI_dI$. Notice that $I$ is the isogonal conjugate of itself in $ABCD$, therefore, $$\angle AII_b+\angle CII_d=\angle AIB+\angle CID=180^{\circ}$$Therefore, $I$ has an isogonal conjugate in $AI_bCI_d$, from above $(I_bI,I_bZ)$ and $(I_dI,I_dZ)$ are isogonal lines. Therefore $(AI,AZ)$ and $(CI,CZ)$ are isogonal lines as well. As a result, the circumcenter of $\triangle CI_bI_d$ and $\triangle AI_bI_d$ lies on $AI$ and $CI$ as desired. $\blacksquare$. [asy][asy] size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -26.137041507663255, xmax = 5.310509315849689, ymin = -12.35753595063471, ymax = 14.141973659388436; /* image dimensions */pen zzttff = rgb(0.6,0.2,1); pen qqwuqq = rgb(0,0.39215686274509803,0); pen ffvvqq = rgb(1,0.3333333333333333,0); pen zzttqq = rgb(0.6,0.2,0); /* draw figures */draw(circle((-14.135509563806407,0), 4.881560784356673), linewidth(0.8) + zzttff); draw(circle((-9.136957241044206,0), 3.1553593420097803), linewidth(0.8) + zzttff); draw((0,0)--(-14.135509563806407,0), linewidth(0.8) + blue); draw((-18.882061092484275,-1.1401246760660608)--(-11.049708756628332,0.7412113034972723), linewidth(0.8) + qqwuqq); draw((-11.049708756628332,0.7412113034972723)--(-6.1947797650523775,-1.1401246760660608), linewidth(0.8) + qqwuqq); draw((-11.049708756628332,0.7412113034972723)--(0,0), linewidth(0.8) + ffvvqq); draw((0,0)--(-12.44971001170649,4.581235156757253), linewidth(0.8) + ffvvqq); draw((-12.44971001170649,-4.581235156757253)--(0,0), linewidth(0.8) + ffvvqq); draw((-18.882061092484275,-1.1401246760660608)--(-6.1947797650523775,-1.1401246760660608), linewidth(0.8) + blue); draw((-10.248497292213486,-2.953095205348955)--(-10.248497292213486,2.953095205348955), linewidth(0.8) + zzttqq); /* dots and labels */dot((0,0),dotstyle); label("$Y$", (0.11506604168540333,0.28745826161699845), NE * labelscalefactor); dot((-9.136957241044206,0),dotstyle); label("$O_{d}$", (-9.038810203270724,0.28745826161699845), NE * labelscalefactor); dot((-14.135509563806407,0),dotstyle); label("$O_{b}$", (-14.014340534613245,0.28745826161699845), NE * labelscalefactor); dot((-10.248497292213486,2.953095205348955),linewidth(4pt) + dotstyle); label("$I_{a}$", (-10.138374917379569,3.173815636152715), NE * labelscalefactor); dot((-10.248497292213486,-2.953095205348955),linewidth(4pt) + dotstyle); label("$I_{c}$", (-11.073004924372087,-3.2861270592367453), NE * labelscalefactor); dot((-18.882061092484275,-1.1401246760660608),dotstyle); label("$B$", (-18.769957923133997,-0.867084688197288), NE * labelscalefactor); dot((-6.1947797650523775,-1.1401246760660608),linewidth(4pt) + dotstyle); label("$D$", (-6.097474593029567,-0.9220629239027303), NE * labelscalefactor); dot((-11.049708756628332,0.7412113034972723),linewidth(4pt) + dotstyle); label("$I$", (-10.935559335108481,0.9746862079350261), NE * labelscalefactor); dot((-12.44971001170649,-4.581235156757253),linewidth(4pt) + dotstyle); label("$E$", (-12.584906406271747,-5.320321780338107), NE * labelscalefactor); dot((-8.047284572720484,-2.961233873424497),linewidth(4pt) + dotstyle); label("$F$", (-8.021712842720044,-3.670974709174841), NE * labelscalefactor); dot((-12.44971001170649,4.581235156757253),linewidth(4pt) + dotstyle); label("$G$", (-12.337504345597258,4.795673589463259), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] CLAIM 3. $IY$ is the external angle bisector of $\angle BID$ Proof. Notice that $Y$ is the external homothetic center of $(I_aI_cD)$ and $(I_aI_cB)$, therefore, $$\frac{YO_d}{YO_b}=\frac{O_dF}{O_bE}=\frac{O_dD}{O_bB}=\frac{IO_d}{IO_b}$$This proves the CLAIM by angle bisector theorem. $\blacksquare$ We now finish the problem by angle chasing. $$\angle XIY=\angle YID+\angle XIC-\angle CID=180^{\circ}-\frac{1}{2}(\angle BID+\angle AIC)-\angle CID=180^{\circ}-\frac{1}{2}(\angle BIC+\angle AID)=90^{\circ}$$

100th HSO post! fastlikearabbit wrote: A convex quadrilateral $ABCD$ has an inscribed circle with center $I$. Let $I_a, I_b, I_c$ and $I_d$ be the incenters of the triangles $DAB, ABC, BCD$ and $CDA$, respectively. Suppose that the common external tangents of the circles $AI_bI_d$ and $CI_bI_d$ meet at $X$, and the common external tangents of the circles $BI_aI_c$ and $DI_aI_c$ meet at $Y$. Prove that $\angle{XIY}=90^{\circ}$. Solved with A-Thought-Of-God. First, we note that the intouch points of $\odot(I_b)$ and $\odot(I_d)$ on $\overline{AC}$ coincide. Proof. Let $\odot(I_b)$ and $\odot(I_d)$ touch $\overline{AC}$ at $E$ and $E'$ respectively. By Pitot's $(*)$, we know that $$AE = \frac{AB-BC+CA}{2} \overset{(*)}{=} \frac{DA-CD+CA}{2} = AE'.$$And similarly, we get $CE=CE'$, whence $E \equiv E'.$ So, we get $\overline{I_bI_d} \perp \overline{AC}$. Now, we let $F$ and $G$ denote the centers of $\odot(AI_bI_d)$ and $\odot(CI_bI_d)$. Since, by radical axis, $\overline{FG} \perp \overline{I_bI_d}$, thus $FG$ is parallel to $AC$. Next, as $\angle DAI=\angle I_bAI_d=\tfrac{\angle A}{2}$, on taking out the common part, we have $$\angle IAI_b = \angle DAI_d = \angle I_dAE = 90^{\circ} - \angle AI_dI_b,$$so, $F$ lies on $\overline{IA}$, and similarly, we get that $G$ lies on $\overline{IC}.$ Hence, in $\triangle GIF$, we have $\overline{FG} \parallel \overline{AC}$, with $F \in \overline{IA}$ and $G \in \overline{IC}.\ (\clubsuit)$ Now, it's well known that there exists a homothety at $X$, that takes $\odot(AI_bI_d)$ to $(CI_bI_d),$ so, we have: $$\frac{XF}{XG} = \frac{AF}{CG} \overset{(\clubsuit)}{=} \frac{IF}{IG},$$where the second equality is ratio of radii, by definition. So, $IX$ is the external bisector of $\angle FIG$, which is $\angle AIC.$ Similarly, we have $IY$ as the external bisector of $\angle BID.$ Lastly, we observe that, \begin{align*} \angle XIY &= \tfrac{(180^{\circ} - \angle AIC)}{2} + \angle AID + \tfrac{(180^{\circ} - \angle BID)}{2} \\&= \tfrac{180^{\circ} - (\angle AID + \angle DIC)}{2} + \angle AID + \tfrac{180^{\circ} - (\angle BIA + \angle AID)}{2} \\&=180^{\circ} - \cancel{\tfrac{\angle AID}{2}} - \tfrac{\angle CID}{2} + \cancel{\angle AID} - \tfrac{\angle AIB}{2} - \cancel{\tfrac{\angle AID}{2}} \\&= 180^{\circ} - \left(\tfrac{\angle AIB + \angle CID}{2}\right) \\&= 180^{\circ} - \tfrac{180^{\circ}}{2} = 90^{\circ}. \end{align*}since, $\angle AIB + \angle CID = \left(180^{\circ} - \tfrac{\angle A}{2}-\tfrac{\angle B}{2}\right) + \left(180^{\circ} - \tfrac{\angle C}{2} - \tfrac{\angle D}{2}\right) = 180^{\circ}.$ And we're done. $\ \blacksquare$

Solved with (for emotional support) Alex Zhao, Amol Rama, Eric Shen, Luke Robitaille, Max Lu, Raymond Feng, William Yue, Yuchan Yang Swap $I_A, I_C$ and $I_B, I_D$ or something because I can't draw diagrams. First, by Pitot, we have $BC + AD = AB + CD \Rightarrow BC + BD - CD = AB + BD - AD$. Therefore, the incircle centered at $I_C$ and $I_A$ have the same foot to $BD$, which means $I_AI_C \perp BD$. Similarly, $I_BI_D\perp AC$. Now, observe that \[\angle IDI_C = \frac{1}{2}(\angle ADC - \angle ADB) = \angle BDI_A = 90 - \angle I_CI_AD\]Therefore, the circumcenter of $(DI_AI_C)$ is on $DI$. Similarly, the circumcenter of $(BI_AI_C)$ is on $BI$. Let these two circumcenters be $F,E$ respectively. Then, $Y\in EF$. Now, since $EF\perp I_BI_D$, we have $EF || BD$, which means \[\frac{IE}{IF} = \frac{EB}{FD} = \frac{EY}{FY}\]so $YI$ is the angle bisector of $\angle BID$. Similarly, $XI$ is the angle bisector of $\angle AIC$. Now, we have \[\angle XIY = \angle YIB + \angle BIC + \angle CIX = 90 + 90 - \frac{1}{2}(\angle BID + \angle AIC) + \angle BIC\]\[= 180 - \frac{1}{2}(360 - C - \frac12B - \frac12D + 360 - B - \frac12A - \frac12C) + 180 - \frac12B - \frac12C = \frac{1}{4}(A + B + C + D) = 90\]

Let $O_A,O_B,O_C,$ and $O_D$ be the circumcenters of triangles $DAB,ABC,BCD$ and $CDA.$ Let $(I_a)$ and $(I_c)$ touch $\overline{BD}$ at $T_a$ and $T_c.$ By Pitot, $$DT_a=\frac{-AB+AD+BD}{2}=\frac{CD-BC+BD}{2}=DT_c$$so $T_a=T_c.$ Hence, $\overline{I_aI_c}\perp\overline{BD}$ and similarly $\overline{I_bI_d}\perp\overline{AC}.$ Since $\angle I_dCI_b=\tfrac{1}{2}\angle B$ and $$\angle I_dCA=90-\angle I_bI_cC=\angle O_CCI_b,$$we know $$\angle O_CCB=\angle I_bCB+\angle I_dCA=\tfrac{1}{2}\angle B$$so $O_C$ lies on $\overline{IC}.$ Similarly, $O_A,O_B,$ and $O_D$ lie on $\overline{IA},\overline{IB},$ and $\overline{ID}.$ Notice $\overline{O_AO_C}\parallel\overline{AC}$ as $\overline{AC}$ and $\overline{I_bI_d}$ are the radical axes of $\{(I_b),(I_d)\}$ and $\{(O_A),(O_C)\}.$ Thus, $$\frac{XO_A}{XO_C}=\frac{O_AO}{O_CO}=\frac{IO_A}{IO_C}$$and $\overline{XI}$ is the external angle bisector of $\angle AIC.$ Similarly, $\overline{YI}$ is the external angle bisector of $\angle BID.$ Also, $$\angle AID=\tfrac{1}{2}(\angle A+\angle D)=180-\tfrac{1}{2}(\angle B+\angle C)=180-\angle BIC=\angle AIB$$so $\overline{XI}$ is the internal angle bisector of $\angle BID.$ $\square$

When you FINALLY figure out what the end condition means... Let $O_a, O_b, O_c, O_d$ be the circumcenters and let $R_a, R_b, R_c, R_d$ be the radii of $DAB, ABC, BCD, CDA$ respectively. Claim: $X, I_b, I_d, I$ lie on an Appollonius circle of $O_bO_d$. Proof: Clearly, $I_b, I_d$ lie on the circle with ratio $\frac{R_b}{R_d}$. Since $X$ lies on the perpendicular bisector of $I_bI_d$, $\frac{XO_b}{XO_d} = \frac{R_b}{R_d}$. Now notice $\frac{d(A, I)}{d(C, I)} = \frac{\sin(\frac{C}{2})}{\sin(\frac{A}{2}} = \frac{R_b}{R_d}$ by LoS, and so $\frac{O_bI}{O_dI} = \frac{d(A, I)-R_b}{d(C, I)-R_d} = \frac{R_b}{R_d}$ and we are done. Since $XO_b = XO_d$, it follows that $XI$ is the angle bisector of $I_bII_d$, which is $AIC$. Similarly, $YI$ is the angle bisector of $BID$, so let $a, b, c, d$ be the angle measures of $IA, IB, IC, ID$, then $\angle (XI, YI) = \frac{\frac{a+c}{2}+\frac{b+d}{2}+\pi}{2}-\frac{\frac{a+c}{2}+\frac{b+d}{2}}{2} = \frac{\pi}{2}$ so we are done.

Claim. $X$ lies on external bisector of angle $AIC.$ Proof. Denote by $O_A,O_C$ circumcenters of $AI_BI_D,CI_BI_D$ respectively. Let $K$ be projection of $I_B$ onto $AC.$ $$|AK|=\frac{|AB|+|AC|-|BC|}{2}=\frac{|AD|+|AC|-|CD|}{2}\implies I_D\in I_BK.$$$$\angle I_BAC=\frac{\angle BAC}{2}=\frac{\angle BAD-\angle CAD}{2}=\angle IAD-\angle I_DAD=\angle IAI_D\stackrel{AC\perp I_BI_D}{\implies} O_A\in AI,O_C\in CI.$$$$O_AO_C\perp I_BI_D\implies \overline{XO_AO_C}\parallel AC\implies \frac{|XO_A|}{|XO_C|}=\frac{|AI_A|}{|CI_C|}=\frac{|II_A|}{|II_C|}\text{ } \Box$$ Analogously $Y$ lies on external bisector of $BID.$ Hence $$\measuredangle XIY\stackrel{\text{mod } \pi}{\equiv} \measuredangle XIA+\measuredangle AIB+\measuredangle BIY=\frac{\measuredangle CIA+\measuredangle AIB+\measuredangle DIC+\measuredangle BID}{2}=\frac{\pi}{2}.$$

This was going to be the last geo for me in a while and I am happy that I could solve a G7. ISL 2017 G7 wrote: A convex quadrilateral $ABCD$ has an inscribed circle with center $I$. Let $I_a, I_b, I_c$ and $I_d$ be the incenters of the triangles $DAB, ABC, BCD$ and $CDA$, respectively. Suppose that the common external tangents of the circles $AI_bI_d$ and $CI_bI_d$ meet at $X$, and the common external tangents of the circles $BI_aI_c$ and $DI_aI_c$ meet at $Y$. Prove that $\angle{XIY}=90^{\circ}$. https://photos.app.goo.gl/CY68dhhvzBkHAMAX9 Say the incircle of $\triangle ACD$ and $\triangle BAC$ touch $AC$ at $J$ and $J'$ respectively,we claim that $J \equiv J'$ For the proof,we can just note that $$CJ=\frac{AC+CD-AD}{2}\stackrel{\text{By Pitot,} CD-AD=CB-AB}=\frac{AC+CB-AB}{2}=CJ'\implies J\equiv J'$$So we get that $I_bI_d \perp AC$,similarly $I_aI_c \perp BD$ Claim : The centre of $\odot(\triangle AI_bI_d)$ lies on $AI$ and similarly other symmetric cases hold. Proof :Just by angle chasing note that $$\angle CAI_d=\frac{\angle CAD}{2}=\frac{\angle A-\angle BAC}{2}=\angle IAB-\angle BAI_b=\angle IAI_b$$Or $AI,AC$ are isogonal w.r.t $\angle I_bAI_d$ and the proof follows easily $\blacksquare$ Again if the centres of $\odot(\triangle I_bAI_d),\odot(\triangle I_bCI_d)$ are $E,F$ respectively, we have $EF\perp I_bI_d$ or $EF \parallel AC$. Thus we have, $$\frac{IE}{IF}=\frac{EA}{FC}=\frac{I_bE}{I_bF}=\frac{I_cE}{I_cF}=\frac{XE}{XF}$$Or $X,I_b,I,I_d$ lie on the appolonian circle of $EF$ So keeping in mind that $XI_b=XI_d$,(an angle chase wrt above figure,bcoz directed angles is a headache) $$\angle XII_d=\angle XI_bI_d=90-\frac{\angle I_bXI_d}{2}=\frac{\angle I_bII_d}{2}=90-\frac{\angle A-\angle C}{4}$$Similarly $\angle YII_c=90-\frac{\angle B-\angle D}{4}$ Finally, $$\angle XIY=\angle XII_d+\angle YII_c-\angle I_cII_d=180-\frac{\angle A+\angle B-\angle C-\angle D}{4}-180+\frac{\angle C+\angle D}{2}=\frac{\angle A+\angle B+\angle C+\angle D}{4}=90\text{ }\blacksquare$$

Let the perpendiculars from $I_A$ and $I_C$ to $BD$ be $H_A$ and $H_C$, respectively. Note that \begin{align*} BH_A+DH_C &= \frac{AB+BD-AD}{2} + \frac{BD+CD-BC}{2} \\ &= BD + \frac{AB-AD}{2} + \frac{CD-CB}{2} \\ &= BD \end{align*}so $H_A=H_C$. Thus, $I_AI_C\perp BD$ and thus $I_BI_D\perp AC$. Also note that \begin{align*} \measuredangle I_AAI_D &= \angle I_AAD - \angle I_DAD \\ &= \frac{1}{2}\angle BAD - \frac{1}{2} CAD \\ &= \frac{1}{2} \angle BAC\\ &= \angle I_BAC \end{align*}which implies that lines $AC$ and $AI_A$ are isogonal conjugates of $\triangle AI_BI_D$. Since $AC$ is the height, the circumcenter lies on $AI_A$ and analogously for the others. If $O_A$ is the circumcenter, and $r_A$ is the radius of $\triangle AI_BI_D$ and similarly for the others , then $O_AO_C\perp I_BI_D\implies O_AO_C\parallel AC$. Of course, \[\frac{XO_A}{XO_C} = \frac{r_A}{r_C} = \frac{IO_A}{IO_C}\]so $XI$ is external angle bisector of $\angle AIC$. The rest is easy.

yo necesito draw better diagrams. solved with hints... so i was apparently trolling and this basically means a few things... the otis hints are really trivial ($I_bI_d\perp AC$ and $A,O_a,I_a$ collinear) and are guessable by GOOD DIAGRAM then question is TRIVIAL!! but i figure 9 circles is TOO MUCH of course this is all an excuse anyhow grammar mode activate Note that $I_bI_d\perp AC$ and also $O_a,A,I_a$ collinear by simple angle chase. Now it's actually really easy to finish the problem, just let $\triangle IAC$ be reference triangle and we can very easily prove that $IX$ is the external angle bisector of $\angle IAC$ and this finishes. ?????

Easy for G7

Let $O_A$ and $O_C$ denote the centers of $(AI_BI_D)$ and $(CI_BI_D)$ Let $F = YI_B \cap (CI_BI_D)$, $G = YI_C \cap (CI_BI_D)$ Key Claim : $YI_BI_DI$ is cyclic

Claim 1: $\measuredangle BIY = \measuredangle YID$

Claim 2:External angle bisector of $\measuredangle AIC$ is internal angle bisector $\measuredangle BID$

Let $H = AC\cap I_bI_d$ Claim 1: $AH \perp I_bI_d$ Let $B'$ be point where incircle of $\triangle ABC$ touch $AC$, and $D'$ be point where incircle of $\triangle ADC$ touch $AC$. we have $AB' = \frac{AB+AC-BC}{2}$ and $AD' = \frac{AD+AC-CD}{2}$ , but from $AB+CD = BC+AD$ we have $AB'=AD' \implies B'=D'=H$ Now let $G$ and $J$ be center of $\triangle AI_bI_d$ and $CI_bI_d$. Claim 2: $GJ \parallel AC$ Note from $\measuredangle HAI_d + \measuredangle DAI_b= \measuredangle I_bAG + \measuredangle DAI_b = \measuredangle DAI$ hence we have $A,G,I$ collinear. Similarly $C,J,I$ are collinear. As $I_bI_d$ is radical axis of $(AI_bI_d)$ and $(CI_bI_d) \implies GJ \perp I_bI_d$ hence $GJ \parallel AC$. Observe that $AG$ and $JC$ are radius of $(AI_bI_D)$ and $(CI_bI_d)$ and $X$ is exsimilicenter of these circles. $$\frac{XG}{XJ} = \frac{AG}{CJ} = \frac{IG}{IJ}$$$XI$ is external angle bisector of $\measuredangle CIA$. Similarly $YI$ is external angle bisector of $\measuredangle BID$ Now just by angle chaseing we have $\angle XIY = 90$.

2017 g7 Great problem, although easy for G7. Let $T$ be the intersection of $AC$ with $I_bI_d$. We use the following lemma. Lemma: $(I_b)$ and $(I_d)$ are tangent to each other at $AC$. Which is provable by pitot and easy length chase. Now for the main idea, we want to show that $YI$ is the external angle bisector of $\angle BID$. Let $O_1$ and $O_2$ be the centres of $(AI_bI_d)$ and $(CI_bI_d)$. We start off with the following claim. Claim: $AC \| O_1O_2$ Proof: We first show that $A-O_1-I$ are collinear by angle chase, now since $I_bI_d$ is the radicle axis of these 2 circles, this implies that $O_1O_2 \perp I_bI_d $, hence we proved our claim. $\blacksquare$ Finally we apply Thales theorem, this will give us that $Y$ lies on the Appolinian circle of $IO_1P_2$ hence we are done.