Dear Mathlinkers, to begin... 1. F' the second point of intersection of FI with the incircle 2. K the second point of intersection of AF' with the incircle 3. Tk the tangent to the incircle at K Sincerely Jean-Louis

Let $P$ be the concurrence point of $C_1I, CM, A_1B_1$ where $M$ is the midpoint of $AB$. Then $CP\perp KI, KP\perp CI$, so $IP\perp CW$, and since $IP=IC_1\perp AB$, this means $CK||AB$ as desired. (note: the fact that the above lines concur is proven for example here: http://yufeizhao.com/olympiad/geolemmas.pdf )

Dear Mathlinkers, can we imagine a proof with polar reciprocity... Sincerely Jean-Louis

Let the Median intersect the incircle at $X$,and $Y$ It is easy to observe that the quadrilateral $B'XA'Y$ is harmonic $\Rightarrow$ tangents at $X$ and $Y$ should intersect on $A'B'$ But they intersect on $IK$ as well $\Rightarrow K$ is the point where the tangents to the incircle at $X$ and $Y$ should intersect.Let $CX$ intersect $A'B'$ at $R \Rightarrow (B',A':R,K)$ is harmonic.Now let the line through $C$ parallel to $BA$ intersect $A'B'$ at $L$.It is well known that $IC'$ passes through $R \Rightarrow IR \perp CL$,Let $IR$ intersect $CL$ at $J \Rightarrow C,B',A',I ,J$ all lie on a single circle $\Rightarrow \angle ICB'=\angle IJB'$ and $\angle IJA'=\angle A'CI \Rightarrow JI$ bisects $\angle B'J'A$ .But since $\angle IJL=90$ we have $(B',A':R,L)$ is harmonic $\Rightarrow L \equiv K$ as desired

Let $\omega$ be the circumcircle of $\triangle A'B'C$ and let $K'$ be the intersection of line $A'B'$ with the line through $C$ parallel to $AB$. Furthermore, let $Z$ be the foot of the perpendicular from $I$ to $CM$ and observe that $Z \in \omega$. It suffices to prove that $\angle K'ZL$ is right, because this will imply $K'=K$. Let $P_\infty$ be the point at infinity on line $AB$. Then the quadruple $(A,B;M,P_\infty)$ is clearly harmonic. Taking perspectivity from $C$ onto line $A'B'$ we observe that $(B',A';L,K')$ is harmonic. Now consider point $Z$. Observe that $ZL$ is an angle bisector of $\angle BZA'$, since $B'C = A'C$ implies the arcs $B'C$ and $A'C$ are equal. Then we conclude that $LZ \perp K'Z$ as desired. [asy][asy] /* DRAGON 0.0.9.6 Homemade Script by v_Enhance. */ import olympiad; import cse5; size(11cm); real lsf=0.8000; real lisf=2011.0; defaultpen(fontsize(10pt)); /* Initialize Objects */ pair C = (-3.0, 2.5); pair A = (-3.5, -1.0); pair B = (2.0, -1.0); pair I = incenter(C,A,B); pair C_prime = foot(I,A,B); pair A_prime = foot(I,B,C); pair B_prime = foot(I,C,A); pair M = midpoint(A--B); pair K_prime = IntersectionPoint(Line(C,relpoint(Line(A,B,lisf),0.5-10/lisf)-relpoint(Line(A,B,lisf),0.5+10/lisf)+C,lisf),Line(A_prime,B_prime,lisf)); path w = circumcircle(C,A_prime,B_prime); path KC_prime = K_prime--B_prime; path AM = C--M; pair Z = IntersectionPoint(w,AM,1); pair L = IntersectionPoint(KC_prime,AM); /* Draw objects */ draw(C--A, rgb(0.6,0.2,0.0)); draw(A--B, rgb(0.6,0.2,0.0)); draw(B--C, rgb(0.6,0.2,0.0)); draw(incircle(C,A,B), rgb(0.0,0.0,0.6) + linetype("4 4")); draw(w, rgb(0.0,0.8,0.8) + linewidth(1.0)); draw(C--K_prime); draw(KC_prime); draw(AM); draw(B_prime--Z, rgb(0.0,0.6,0.0) + linetype("4 4")); draw(A_prime--Z, rgb(0.0,0.6,0.0) + linetype("4 4")); draw(K_prime--Z, rgb(0.0,0.6,0.2) + linetype("4 4")); /* Place dots on each point */ dot(C); dot(A); dot(B); dot(I); dot(C_prime); dot(A_prime); dot(B_prime); dot(M); dot(K_prime); dot(Z); dot(L); /* Label points */ label("$C$", C, lsf * dir(120)); label("$A$", A, lsf * dir(225)); label("$B$", B, lsf * dir(-45)); label("$I$", I, lsf * dir(-45)); label("$C'$", C_prime, lsf * dir(270)); label("$A'$", A_prime, lsf * dir(45)); label("$B'$", B_prime, lsf * dir(225)); label("$M$", M, lsf * dir(270)); label("$K'$", K_prime, lsf * dir(45)); label("$Z$", Z, lsf * dir(0)); label("$L$", L, lsf * dir(70)); [/asy][/asy]

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=503311 After that is easy to show that $ \{Q\}=A'K\cap CM $ is the orthocenter of $\triangle ABC$.

This is sort of a combination of other solutions Let $K'$ be the intersection of the line parallel to $AB$ and $A'B'$. Then $C(K', M; A, B)$ is harmonic, and intersecting it with $A'B' \implies (K', L; A', B')$ is harmonic. Let $CM \cap w = X, Y$ ($w$ is the incircle of $\triangle ABC$). Then $K'X, K'Y$ are tangents to $w$ because of harmonic quadrilateral $A'XB'Y$. Thus, $CM$ is the polar of $K'$, so $IK' \perp CM$, or $K = K'$.

Let the $C-$median meet the incircle at $X$ and $Y$. Then clearly $B'A'XY$ is harmonic. Therefore, $K$ must be the intersections of the tangents at $X$ and $Y$. It follows that $(B',A';L,K')$ is harmonic. Now since $(A,B;M,P_\infty)$ is harmonic, perspectivity from $C$ gives us that $CK\cap AB=P_\infty$ as desired.

Everybody is so OP with projective geometry, here's an angle chasing solution.

Let $C-$median meet $A'B'$ at $X$, and $IK$ at $Y$. Now as $YC$ bisects $\angle A'YB'$ and $\angle XYK=90^{\circ}$, $$-1=(A',B';Y,K) \stackrel{C}{=} (A,B;CY \cap AB, CK \cap AB)$$This gives that $CK \cap AB=P_{\infty}$, as needed. $\square$

Redefine $K$ as the point where the line through $C$ parallel to $AB$ meets $A'B'$. Then, from here, we get that $CM$ is the polar of $K$, where $M$ is the midpoint of $AB$. This gives $IK \perp CM$, proving the given problem.

Let $H = CM \cap DK$. It is well-known that $F,I,H$ are collinear, and since $HI \perp AB$, it suffices to show that $HI \perp CK$. Key Claim: $H$ is the orthocenter of $\triangle IKC$. Proof: We are given that $CH \perp KI$, and also $KH \perp CI$, proving the desired claim.

Math-lover123 wrote: The incircle of triangle $ABC$ touches $BC$, $CA$, $AB$ at points $A_1$, $B_1$, $C_1$, respectively. The perpendicular from the incenter $I$ to the median from vertex $C$ meets the line $A_1B_1$ in point $K$. Prove that $CK$ is parallel to $AB$. Note that $CI \perp A'B'$ and $CM \perp IK \implies$ the concurrency point of $\overline{A'-B'-K}$ and $\overline{CM}$ is the orthocenter of $\triangle CIK$. Let us denote it by $H'$. Recall that by The Incircle Concurrency Lemma we have $CM, IC'$ and $A'B'$ concurrent. As $H'= \overline{A'B'} \cap \overline{CM} \implies IC'$ passes through $H$ as well, and hence $\overline{C'-I-H} \perp CK$; also, $IC' \perp AB \implies CK \parallel AB$ $\quad \blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.5700816154896442, xmax = 13.918412567381893, ymin = -5.5571388455225, ymax = 4.215176998877735; /* image dimensions */ pen ttffqq = rgb(0.2,1,0); pen ffqqff = rgb(1,0,1); pen xfqqff = rgb(0.4980392156862745,0,1); draw((0.5441213316161794,-3.364632085560908)--(10.98,-3.52)--(1.82,3.22)--cycle, linewidth(1.2) + ttffqq); draw((3.116387237859491,-0.09753252381222777)--(3.4258162355067316,0.023382584236986304)--(3.3049011274575175,0.3328115818842274)--(2.9954721298102767,0.2118964738350133)--cycle, linewidth(1.2) + red); draw((3.9231599989277535,-0.33446498399287555)--(3.637245834982063,-0.5036390328146919)--(3.8064198838038794,-0.7895531967603824)--(4.09233404774957,-0.620379147938566)--cycle, linewidth(1.2) + red); /* draw figures */ draw((0.5441213316161794,-3.364632085560908)--(10.98,-3.52), linewidth(1.2) + ttffqq); draw((10.98,-3.52)--(1.82,3.22), linewidth(1.2) + ttffqq); draw((1.82,3.22)--(0.5441213316161794,-3.364632085560908), linewidth(1.2) + ttffqq); draw((1.82,3.22)--(5.7620606658080895,-3.4423160427804538), linewidth(1.2) + ffqqff); draw(circle((3.465608194146705,-0.9912098500350361), 2.416649118419325), linewidth(1.2)); draw((1.82,3.22)--(10.367729732368556,3.0927425850848), linewidth(1.2) + red); draw((1.0930872120859763,-0.5314956439912129)--(10.367729732368556,3.0927425850848), linewidth(1.2) + blue); draw((1.82,3.22)--(3.465608194146705,-0.9912098500350361), linewidth(1.2) + ffqqff); draw((3.465608194146705,-0.9912098500350361)--(10.367729732368556,3.0927425850848), linewidth(1.2) + blue); draw((3.4863758082071103,0.40372617230037616)--(3.4296334438482328,-3.4075911897448643), linewidth(1.2) + xfqqff); /* dots and labels */ dot((0.5441213316161794,-3.364632085560908),linewidth(3pt) + dotstyle); label("$A$", (0.6067643819007943,-3.2706675101339844), NE * labelscalefactor); dot((10.98,-3.52),linewidth(3pt) + dotstyle); label("$B$", (11.036832254289514,-3.4272751358455267), NE * labelscalefactor); dot((1.82,3.22),linewidth(3pt) + dotstyle); label("$C$", (1.8752861501642872,3.306852769750789), NE * labelscalefactor); dot((5.7620606658080895,-3.4423160427804538),linewidth(3pt) + dotstyle); label("$M$", (5.821798318095154,-3.3489713229897555), NE * labelscalefactor); dot((3.465608194146705,-0.9912098500350361),linewidth(3pt) + dotstyle); label("$I$", (3.5353269827066356,-0.8902315993185425), NE * labelscalefactor); dot((3.4296334438482328,-3.4075911897448643),linewidth(3pt) + dotstyle); label("$C'$", (3.488344694993173,-3.317649797847447), NE * labelscalefactor); dot((4.897857047534576,0.9552885916612395),linewidth(3pt) + dotstyle); label("$A'$", (4.9604563766816705,1.051702959504581), NE * labelscalefactor); dot((1.0930872120859763,-0.5314956439912129),linewidth(3pt) + dotstyle); label("$B'$", (1.1548910718911924,-0.43606948475507007), NE * labelscalefactor); dot((10.367729732368556,3.0927425850848),linewidth(3pt) + dotstyle); label("$K$", (10.426062514014498,3.1815666691815556), NE * labelscalefactor); dot((3.4863758082071103,0.40372617230037616),linewidth(3pt) + dotstyle); label("$H'$", (3.55098774527779,0.5035762695141833), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

Idea same as IMO SL2005/G6 But easier version

Let $M$ be the midpoint of $AB$, and let $P=A'B'\cap CM$. Notice that $C$ lies on the polar of $K$, so $CM$ is the polar of $K$. Thus, $CK$ is the polar of $P$. Since $PI\perp AB$, we have $CK\parallel AB$, as desired.

Let $M$ be the midpoint of $AB$ and $D$ be the intersection of the c-median and $B_1A_1$. By the incenter concurrency lemma, $D, I, C_1$ are collinear and $\overline{DIC_1} \perp AB$ since $C_1$ is the point of tangency. Let $P$ be the intersection of $A_1B_1$ and $CI$ and point $Q$ be the foot of the perpendicular from $I$ to $CM.$ It is easy to see that $IC \perp A_1B_1$ (just angle chase to get that $\triangle B_1PI \cong \triangle A_1PI$). This means that $D$ is the orthocenter of $\triangle CIK$ and hence $ID \perp CK$. Since $CK$ and $AB$ are both perpendicular to $DIC_1,$ $CK \parallel AB$.

Let $M$ be the midpoint of $AB$ and let $CM\cap \overline{A_1B_1}=N$ Also define $IK\cap CM=D$ Since, $\angle IA_1C+\angle IB_1C=180 \Longrightarrow I,A_1,B_1,C$ are concyclic. But $ID\perp CM \Longrightarrow \angle IDC=90 \Longrightarrow D,I,A_1,B_1,C$ are concyclic $\Longrightarrow \angle A_1DN=\angle B_1DN$ and also $\angle KDN=\angle KDC=180-\angle IDC=180-90=90$ $\therefore$ By a well known lemma $(A_1,B_1;N,K)=-1$ Now projecting $\overline {A_1B_1}$ to $\overline {AB}$ by taking perspectivity at $C$ we find, $-1=(A_1,B_1;N,K)\stackrel{C}{=} (B,A,M,CK\cap AB)$ Since $M$ is the midpoint of $AB \Longrightarrow CK\cap AB$ msut be the point at infinity along $AB$ $\therefore CK\parallel AB$

Let $M$ be the midpoint of $AB$. Let $CM\cap A'B' = D$ and $CM\cap IK = E$. We have $CA'IEB'$ and $C'IEM$ is cyclic. Let $(C'IEM)$ intersect the incircle again at $F$. By Radical Axis Theorem, we have $C' , F , K$ are collinear. We have $\angle IFM = 90 \implies MF$ is tangent to the incircle. So $K$ lies on the polar of $M$ so $M$ also lies on the polar of $K$. And also $K$ lies on the polar of $A$ so $A$ also lies on the polar of $K$. So $AM$ is the polar of $K$ gives $-1 = (A',B';D,K) \stackrel{A} = (B,A;M,CK\cap AB) \implies CK\parallel AB$.$\square$

Let $l$ be the line through $C$ parallel to $AB$, $M$ be the midpoint of $BC$, $IK \cap CM = E$, $C_1$ be the antipode of $C'$ (with respect to the incircle), and $l \cap C'C_1 = D$. Since $C'C_1 \perp AB$ and $l \parallel AB$, we conclude $C'C_1 \perp l$. By basic properties of tangent, parallel, and perpendicular lines, we know $$\angle CA'I = \angle CB'I = 90^{\circ} = \angle CDI = \angle CEI$$so $A'EIB'CD$ is cyclic. Let $F \ne C$ lie on the incircle of $ABC$ such that $MF$ is tangent to the incircle. Now, it's easy to see $C'IEFM$ is cyclic, as $$IK \perp CM \implies \angle MEI = 90^{\circ} = \angle MC'I = \angle MFI.$$Now, define $X = C_1F \cap AB$. Claim: $MC' = MX$. (An equivalent statement is $CC'$ and $CX$ are isotomic with respect to $ABC$.) Proof. Because $\angle C'FX = 180^{\circ} - \angle C_1FC' = 90^{\circ}$, it's easy to see $M$ is the center of $(C'FX)$, implying the result. $\square$ Now, the Diameter of the Incircle Lemma implies $X$ is the extouch point of the $C$-excircle on $AB$. But this yields $C, C_1, X$ are collinear by the same lemma, so we conclude $C, C_1, F, X$ are all collinear. Claim: $CDFC'$ is cyclic. Proof. Since $$\angle CDC' = \angle CDI = 90^{\circ} = \angle C_1FC' = \angle CFC'$$the result easily follows. $\square$ Claim: $l = CD$, $A'B'$, and $C'F$ are concurrent. Proof. Notice these $3$ lines are the pairwise Radical Axes of $(A'EIB'CD)$, $(CDFC')$, and $(A'B'C'F)$, so they concur at the Radical Center of these $3$ circles. $\square$ Claim: $A'B'$, $IE$, and $C'F$ are concurrent. Proof. Observe these $3$ lines are the pairwise Radical Axes of $(A'EIB'CD)$, $(A'B'C'F)$, and $(C'IEFM)$, so they concur at the Radical Center of these $3$ circles. $\square$ Thus, $l = CD$, $A'B'$, $C'F$, and $IE$ are all concurrent. This means $K = A'B' \cap IE$ lies on $L$, and the result follows easily. $\blacksquare$ Remark: The inspiration for my solution came from GOTTEM Geometry Mock $2019$/$1$ https://artofproblemsolving.com/community/c6h1979299p13748522. The result from this particular question led me to conjecture $T_c \in C_1F$ (extouch point), and the rest of my solution fell apart from this point on. https://www.geogebra.org/m/v6bfstja (GGB).

Finally able to use projective on this problem ! Let $M$ be the midpoint of $AB$. Obviously, it suffices to show $(A, B; M, AB \cap CK) = -1$. By the Incircle Concurrency Lemma, $C'I$, $CM$, and $A'B'$ concur at some point $P$. Claim: The polar of $K$ with respect to the incircle is $CM$. Proof. Since $K \in A'B'$, which is the polar of $C$, we know $C$ lies on the polar of $K$ by La Hire's. Because $IK \perp CM$, the result easily follows. $\square$ Thus, $(B', A'; P, K) = -1$ as $P \in CM$. Finishing, we find $$- 1 = (B', A'; P, K) \overset{C}{=} (A, B; M, CK \cap AB)$$as desired. $\blacksquare$ https://www.geogebra.org/m/v6bfstja (All the extra lines/circles in this diagram are from my previous solution.)

Relatively easy and seemingly original angle chase. Hopefully no config issues Replace $A_1,B_1,C_1$ with $A',B',C'$. Now let $M$ be the midpoint of $\overline{AB}$, $D$ be the foot of the $C$-altitude, $P=\overline{CI} \cap \overline{A'B'},X=\overline{CM}\cap\overline{A'B'}$, and $Q$ be the the foot of the perpendicular from $I$ to $\overline{CM}$. Clearly $P$ is the midpoint of $\overline{A'B'}$ and $IPXQ$, $CA'QIB'$ are cyclic. By well-known incircle configurations, $C',I,X$ are collinear. As $\angle CPK=\angle CQK=90^\circ$, $CPQK$ is also cyclic with diameter $\overline{CK}$. Thus, $\overline{CK} \parallel \overline{AB}$ if and only if $\overline{CD}$ is tangent to $(CPQ)$. Now, WLOG $AC<AB$. Since $$\measuredangle DCP=\measuredangle ACI+\measuredangle DCA=\measuredangle ACI+\measuredangle DAC+90^\circ,$$and $$\measuredangle CQP=\measuredangle XQP=\measuredangle XIP=\measuredangle XIC=\measuredangle C'IC=\measuredangle ACI+90^\circ+\measuredangle C'AC=\measuredangle ACI+90^\circ+\measuredangle DAC,$$hence $\measuredangle DCP=\measuredangle CQP$ and $\overline{CD}$ is tangent to $(CPQ)$ as desired. $\blacksquare$ Edit: oops it's still overcomplicated $X$ is just the orthocenter

A cute problem! The classic get 2 the third free! Let $A_1K\cap CM=X$. Clearly, $\angle CPA_1=90^{\circ},$ and $\angle CQI=90^{\circ}$ is given. Then, $IX=C_1X$ is the third altitude in $\triangle CIK,$ and thus we are done.

Let $M$ be the midpoint of $\overline{AB}.$ Notice that $\overline{CI}\perp\overline{A_1B_1}$ so $H=\overline{CM}\cap\overline{A_1B_1}$ is the orthocenter of $\triangle CIK.$ But $H$ lies on $\overline{C_1I}$ by lemma 4.17, so $\overline{C_1I}\perp\overline{CK},$ which implies $\overline{AB}\parallel\overline{CK}.$ $\square$

Quite easy but nevertheless I am going to post the solution.I took the perpendicular on $A$-median guys so please bear with me By the well known Incircle configuration,$B'C'$,$A'I$ and the $A$-median are concurrent on,say $L$.Notice that $\triangle AC'L\cong \triangle AB'L$ ,hence $KC'\perp AI$ This gives us that $L$ is the orthocentre of $\triangle AKI.$ and hence $A'I\perp AK$ This gives us that $AK||BC$ and hence our solution is completed

Mogmog8 wrote: Let $M$ be the midpoint of $\overline{AB}.$ Notice that $\overline{CI}\perp\overline{A_1B_1}$ so $H=\overline{CM}\cap\overline{A_1B_1}$ is the orthocenter of $\triangle CIK.$ But $H$ lies on $\overline{C_1I}$ by lemma 4.17, so $\overline{C_1I}\perp\overline{CK},$ which implies $\overline{AB}\parallel\overline{CK}.$ $\square$ Are you allowed to state EGMO lemmas in olympiads? Like would it count as a valid sol?

Let $M$ the midpoint of $AB$, Let $N=A_1B_1 \cap MC$ Then: $(B_1,A_1;N,K)=(A,B;M,KC \cap AB)=-1$ $\implies KC \parallel AB$

$CM$ is the polar of $K \implies K$ lies on the polar of $M$.Extend $C'$ to meet the incircle at $D$.Then $CD$ meets the side $AB$ at the excircle touchpoint.It is well known that if $DE$ meets the circle at ,say, $E$ , then $C'E$ is the polar of $M$.Also , let $CI$ intersect $A'B'$ at $X$.Then $\angle CXK =\angle CEK \implies (CXEK).$ Now ,$$CB'^2=CX.CI =CD.CE \implies \angle IXE=\angle IDE =\angle EC'M $$But $$\angle IXE =\angle XCE + \angle XEC = \angle XKE+ \angle XKC = \angle CKE =\angle CKC' \implies CK \parallel C'M= AB$$

First posting as #27 and again posting on #31 after 5 months Notice that the pole of the $C$-median lies on $IK$. By La-Hire,note that this point also lies on the polar of $C$ or $A'B'$.Thus the pole of the $C$ median is $K$.Thus $(K,C$-median$\cap A'B';B',A')$=-1.Taking perspectivity from C with respect to $AB$ gives $CK||AB$

IAmTheHazard wrote: Relatively easy and seemingly original angle chase. Hopefully no config issues Replace $A_1,B_1,C_1$ with $A',B',C'$. Now let $M$ be the midpoint of $\overline{AB}$, $D$ be the foot of the $C$-altitude, $P=\overline{CI} \cap \overline{A'B'},X=\overline{CM}\cap\overline{A'B'}$, and $Q$ be the the foot of the perpendicular from $I$ to $\overline{CM}$. Clearly $P$ is the midpoint of $\overline{A'B'}$ and $IPXQ$, $CA'QIB'$ are cyclic. By well-known incircle configurations, $C',I,X$ are collinear. As $\angle CPK=\angle CQK=90^\circ$, $CPQK$ is also cyclic with diameter $\overline{CK}$. Thus, $\overline{CK} \parallel \overline{AB}$ if and only if $\overline{CD}$ is tangent to $(CPQ)$. Now, WLOG $AC<AB$. Since $$\measuredangle DCP=\measuredangle ACI+\measuredangle DCA=\measuredangle ACI+\measuredangle DAC+90^\circ,$$and $$\measuredangle CQP=\measuredangle XQP=\measuredangle XIP=\measuredangle XIC=\measuredangle C'IC=\measuredangle ACI+90^\circ+\measuredangle C'AC=\measuredangle ACI+90^\circ+\measuredangle DAC,$$hence $\measuredangle DCP=\measuredangle CQP$ and $\overline{CD}$ is tangent to $(CPQ)$ as desired. $\blacksquare$ Edit: oops it's still overcomplicated $X$ is just the orthocenter Projective my beloved Relabel points so the problem is $A$-indexed. Let $M$ be the midpoint of $\overline{BC}$, $P=\overline{B'C'} \cap \overline{AM}$, and $Q=\overline{IK} \cap \overline{AM}$. Note that since $\angle AB'I=\angle AC'I=\angle AQI=90^\circ$, $AB'C'QI$ is cyclic. Further, $AC'IB'$ is clearly harmonic, hence $$-1=(C',B';A,I)\stackrel{Q}{=}(C',B';P,K)\stackrel{A}{=}(B,C;M,\overline{AK} \cap \overline{BC}),$$from which we conclude that $\overline{AK} \cap \overline{BC}=P_\infty$, i.e. $\overline{AK} \parallel \overline{BC}$. $\blacksquare$

Let $M$ be the midpoint of $AB, X=CM\cap A^\prime B^\prime$, then it´s well know that $X,I,C^\prime$ are collinear; since $IC\perp A^\prime B^\prime$ and $IK\perp QC$, then $X$ is the orthocenter of $\triangle IKC$, clearly $B A^\prime X C^\prime$ and $IQ A^\prime C$ are cyclic, therefore $$\angle C^\prime B A^\prime= \angle A^\prime IX\hspace{1cm} \angle A^\prime IQ=\angle A^\prime CQ \hspace{1cm} \angle XIQ=\angle KCX$$ Combining these equalities we get that $\angle ABC=\angle KCB$ or $CK\parallel AB$, as desired .

Switch to $A$-indexing and let $K’$ be a point on line $B’C’$ such that $AK\parallel BC$. We will show that $IK\perp AM$ which establishes $K=K’$, solving the problem. Employ barycentric coordinates on $\triangle ABC$ with $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Of course, $M=(0,1/2,1/2),I=\frac{1}{a+b+c}(a:b:c),P_\infty=(0:-1:1)$ where $P_\infty$ is the point at infinity along line $BC$. Since $K’$ is on cevian $AP_\infty$, it can be parameterized by $(t:-1:1)$. Since $K’$ is also on $B’C’$, we must have $$\begin{vmatrix}t&-1&1\\ s-c&0&s-a\\ s-b&s-a&0\end{vmatrix}=0 \iff t=\frac{b-c}{s-a}$$so $K’=\frac{1}{b-c}(b-c:-(s-a):s-a).$ Note that $$\overrightarrow{AM}=(-2:1:1)$$and $$\overrightarrow{IK’}=((a+b+c)(b-c):(a+b+c)(s-a):(a+b+c)(s-a))-(a(b-c):b(b-c):c(b-c))=((b+c)(b-c):-\frac12 (-a^2+3b^2+c^2):\frac12 (-a^2+b^2+3c^2))$$where we subtracted normalized $A,M$ and $I,K’$. Now, EFFT gives $$a^2(-b^2+c^2)+b^2(b^2-c^2+a^2-b^2-3c^2)+c^2(-a^2+b^2+c^2+b^2-c^2)=-a^2b^2+a^2c^2+a^2b^2-4b^2c^2-a^2c^2+4b^2c^2=0$$so $IK’\perp AM$ which completes the problem.

Can anyone complete this radical axis solution? First, relabel some points, switch to $A$-index, (ie basically replace $C$ with $A$) and let $\Delta DEF$ be the contact triangle. Let M be the midpoint of $BC$. Now, let perpendicular from $I$ meet the $A$ median, $AM$ at $P$. Ok so here is where the radical axes come in. Consider the circles with diameter $IM$, say $\omega_1$ and circle with diameter $AI$ call it $\omega_2$. So, as $\angle IPM = \angle IDM = 90^{\circ}, \ \omega_1= (IPMD)$, and similarly, $\angle API= \angle AEI= \angle AFI= 90^{\circ}, \implies \omega_2= (AEPIF)$. So Line through $I$ perpendicular to median is radical axis of $\omega_1$ and $\omega_2$. Now, let line parallel to $BC$ at $A$ meet $\omega_2$ at $L$. We have $L,I,D$ collinear. Clearly the radical axis of $\omega_2$ and any circle passing through $A$ and $L$ would be $AL$, ie. the line parallel to $BC$ at $A$. So now, we just want to find a circle passing through $A$ and $L$ such that the radical axis of that circle and $\omega_1$ is $EF$. Let's try to construct such a circle. We know radical axis is perpendicular to line joining centers, and center of $\omega_1$ is simple midpoint of $IM$, say $X$. As $AI \perp EF$, the center of the third circle must simply lie on line through $X$ perpendicular to $EF$ which is simple the $M$-midline of $\Delta AIM$. But as $A$ and $L$ also lie on that circle, the center must also lie on perpendicular bisector of $AL$, and so the center of the third circle is simple intersection of $M$ midline of $AIM$ and perpendicular bisector of $AL$ (which if we say the foot of perpendicular from $A$ to $BC$ is $N$, then it is basically the perpendicular bisector of $ND$. And drawing this on geogebra, it is indeed true that $EF$ is radical axis of $\omega_1$ and the third circle but I have not been able to prove it for sooooo long, so I would really appreciate if someone more experienced than me could prove it, because it seems very doable, like it seems that it should succumb to trig but it has not yet for me... So basically to rephrase my problem: Let $\Delta ABC$ be a triangle with contact triangle $DEF$ and incenter $I$. Let M be midpoint of $BC$ and let $N$ be the foot of $A$ altitude. Let the perpendicular bisector of $ND$ and $M$ midline of $\Delta AIM$ meet at $O$. let $\Omega$ be the circle with center $O$ passing through $A$ and let $\omega_1$ be the circle with diameter $IM$. Prove that the radical axis of $\Omega$ and $\omega_1$ is $EF$. Here is the geogebra diagram: https://www.geogebra.org/m/enfefthj

Since $A_1C=B_1C$, and $CI$ is a bisector of $\angle A_1CB_1$, we have $CI\perp A_1B_1$. We are given that $IK\perp CM$. Let $P$ be the intersection of $A_1B_1$ and $CM$. Thus, $P$ is the orthocenter of $\triangle KIC$. Furthermore, it it well know that $C_1,I,P$ are collinear. Thus, $C_1I\perp CK$, and thus $AB\parallel CK$ as desired.

Pretty sure projective geometry kills this problem. Let $K'$ be the point on $A'B'$ such that $CK' \parallel AB$. Let $M$ be the midpoint of $AB$ and $X$ the intersection of $CM$ and $A'B'$. Then we have that \[ (A,B; M, \infty) = (B', A'; XK') = -1 \]which implies that $X$ must lie on the polar of $K'$. However, we also have that $C$ is the pole of $A'B'$, so it follows that $C$ must lie on the polar of $K'$, from which we see that $CM$ is the polar of $K'$, implying that $K'$ must lie on the line perpendicular to $CX$ passing through $I$, so $K' = K$, as desired. $\blacksquare$

Let $M$ be the midpoint of $AB$. Let $D$ be the point at which $A'B'$, $C'I$, and $CM$ are known to concur. Then, since $CM \perp IK$ and $A'B' \perp CI$ it follows that $D$ is the orthocenter of $\triangle CIK$, so $C'D \perp CK$ and thus $AB \parallel CK$.

Letting M be the midpoint of AB, notice A'B' perp. CI, CM perp. IK; it's well known as an incircle concurrency that C'I also passes through the intersection point of A'B' and CM, whence C'I perp. CK since it passes through the orthocenter of CIK. It follows that AB perp. CI and CK perp. CI so they're parallel.

Let $X$ denote the perpendicular from $I$ to $C$-median $CM$, where $M$ is the midpoint of $BC$. Also denote by $Y$ the intersection of the $C$-median with $A_1B_1$. Make the trivial observation that $C, B_1, I, X, A_1$ are concyclic points. Let $K'$ be the point so that $CK' \parallel AB$. It suffices to show $K=K'$. We have $(A, B; M, P_{\infty})$ is harmonic. Taking perspectivity at $C$ onto $K'B'$, \[-1 = C(A, B; M, P_{\infty}) \stackrel{K'B'}{\longrightarrow} (B_1, A_1; Y, K').\] However we also similarly have that $(B_1, A_1; Y, K)$ is harmonic, as $CX$ bisects angle $\angle A_1XB_1$, and in particular, the point of intersection of the perpendicular at $X$ to $A_1B_1$ forms a harmonic bundle with the remaining points on $A_1B_1$. Thus $-1 = (B_1, A_1; Y, K') = (B_1, A_1; Y, K)$, which is enough to prove that $K=K'$, as desired. $\blacksquare$

Lemma. Let $ABC$ be a triangle with incenter $I$ and contact triangle $DEF$. If $M$ is the midpoint of $BC$, then $EF$, $AM$ and ray $DI$ concur. Proof. Let $X = DI \cap FE$, and $B'$ and $C'$ be where the line parallel to $BC$ through $X$ meet with $AB$ and $AC$, respectively. Since $XI \perp BC$ and $B'C' \parallel BC$, we get $XI \perp B'C'$, so $\overline{FXE}$ is the Simson line of $\Delta AB'C'$, and it follows that $I$ lies on the $(AB'C')$. Thus, $$\angle IB'C' = \angle IAC = \angle IAB = \angle IC'B'$$so $\Delta IB'C'$ is isoceles, and $X$ is the midpoint of $B'C'$. The result follows after taking a homothety centered at $A$ that sends $B'C'$ to $BC$, which sends $X$ to the midpoint of $BC$. $\blacksquare$ Returning to the problem and applying the lemma, let $E$ be the concurrency point of the $C$-median, $C'I$, and $A'B'$. We claim that $E$ is the orthocenter of $\Delta IKC$. This directly follows from $A'B' \perp CI$ and $CE \perp IK$. Therefore, $IE \perp KC$, which gives $C'E \perp KC$. Since $C'E \perp AB$, the desired condition follows. $\blacksquare$

Let $K'$ be the point where $CK' \parallel AB$ and $K' \in A'B'$ and let $CM \cap K'A' = X$. We will try to show that $K'I \perp CM$. First note that $CK' \cap AB = \infty$ so $-1 = (A, B; M, AB_\infty) \overset{C}= (B', A'; X, K')$. So it follows that $X'$ lies on the polar of $K$. However $C$ is the pole of line $A'B'$ which passes through $K'$ so by La Hire's we get that $C$ lies on the polar of $K' \implies$ $CX$ is the polar of $K'$. Note that $CX$ must be perpendicular to $K'I$ so $K' = K$ as desired.

yay my first proj geo post! Let $A’B’ $ the parallel to $AB$ through $C$ in $K’$.Let $M$ be the midpoint of $AB$. Let $CM$ meet $A’B’$ in $D$ and $\odot(A’B’C’)$ in $E,N$ where $E$ is closer to $C$ than $N$. Let $K’N$ meet $CB$ in $F$ and $\odot(A’B’C’)$ in $P$. We shall show that $CM$ is the polar of $K’$. Indeed, note that $(B’,A’;D,K’) \stackrel{C}{=} (A,B;M,CK’ \cap AB)=-1$ and $(B,P;A’,N) \stackrel{B’}{=} (CB’ \cap K’N,F;K’,N) \stackrel{C}{=}(B’,A’;D,K’)=-1$, so the polar of $K’$ passes through $DN$ and hence, $CM$. As a result, $K \equiv K’$ and $IK \perp CM$. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -11.08, xmax = 7.64, ymin = -5, ymax = 8.2; /* image dimensions */ pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); /* draw figures */ draw((-3.66,-0.72)--(4.5,-0.98), linewidth(0.8)); draw((4.5,-0.98)--(-2.3,4.54), linewidth(0.8)); draw((-2.3,4.54)--(-3.66,-0.72), linewidth(0.8)); draw(circle((-1.1802459584353864,1.1377394472443558), 1.935768837771497), linewidth(0.8)); draw((-2.3,4.54)--(0.42,-0.85), linewidth(0.8)); draw((0.1340879647641478,-0.2834316654701312)--(5.094825925720433,4.304380546484397), linewidth(0.8)); draw((-3.0543843594553306,1.6223075509301212)--(1.7738107537818089,1.2330242116359436), linewidth(0.8)); draw((-2.3,4.54)--(5.094825925720433,4.304380546484397), linewidth(0.8) + uququq); draw((5.094825925720433,4.304380546484397)--(-3.0543843594553306,1.6223075509301212), linewidth(0.8)); draw((-4.703996813085826,-4.757811203552534)--(-3.66,-0.72), linewidth(0.8)); draw((-4.703996813085826,-4.757811203552534)--(0.1340879647641478,-0.2834316654701312), linewidth(0.8)); /* dots and labels */ dot((-3.66,-0.72),dotstyle); label("$A$", (-3.58,-0.52), NE * labelscalefactor); dot((4.5,-0.98),dotstyle); label("$B$", (4.58,-0.78), NE * labelscalefactor); dot((-2.3,4.54),dotstyle); label("$C$", (-2.22,4.74), NE * labelscalefactor); dot((-1.1802459584353855,1.137739447244356),linewidth(4pt) + dotstyle); label("$I$", (-1.1,1.3), NE * labelscalefactor); dot((-1.241893582022601,-0.7970475084159467),linewidth(4pt) + dotstyle); label("$C_{1}$", (-1.16,-0.64), NE * labelscalefactor); dot((0.03976982247757295,2.640657438224088),linewidth(4pt) + dotstyle); label("$A_{1}$", (0.12,2.8), NE * labelscalefactor); dot((-3.0543843594553306,1.6223075509301212),linewidth(4pt) + dotstyle); label("$B_{1}$", (-2.98,1.78), NE * labelscalefactor); dot((5.094825925720433,4.304380546484397),linewidth(4pt) + dotstyle); label("$K_{1}$", (5.18,4.46), NE * labelscalefactor); dot((0.42,-0.85),linewidth(4pt) + dotstyle); label("$M$", (0.5,-0.68), NE * labelscalefactor); dot((-1.144780876641393,2.2507973989327605),linewidth(4pt) + dotstyle); label("$D$", (-1.06,2.42), NE * labelscalefactor); dot((-1.5426777063067008,3.0392767783062933),linewidth(4pt) + dotstyle); label("$E$", (-1.46,3.2), NE * labelscalefactor); dot((0.1340879647641478,-0.2834316654701312),linewidth(4pt) + dotstyle); label("$N$", (0.22,-0.12), NE * labelscalefactor); dot((1.7738107537818089,1.2330242116359436),linewidth(4pt) + dotstyle); label("$F$", (1.86,1.4), NE * labelscalefactor); dot((0.7473172493445819,1.3157873994937745),linewidth(4pt) + dotstyle); label("$P$", (0.82,1.48), NE * labelscalefactor); dot((-4.703996813085826,-4.757811203552534),linewidth(4pt) + dotstyle); label("$G$", (-4.62,-4.6), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]