this a little long solution but nice : let the line perpendicular to $BK$ intersects $CD$ at point $S$ and let $BK \cap AL=M$ we will prove that $XM \cap AB=T$ let $XM \cap AB=T'$ we can easily see that $X$ is orthocenter of $ASB$ and $\angle SKB=\angle SLA=90$ $ \frac{AT'}{BT'}=\frac{sin(\angle AXT')}{sin(\angle BXT'}.\frac{AX}{XB}=\frac{sin(\angle KXM)}{sin(\angle LXM}.\frac{cos(SAB)}{cos(SBA)} =\frac{sin(\angle XKM)}{sin(\angle XLM}.\frac{\frac{AD}{SA}}{\frac{BD}{SB}}=\frac{sin(\angle KSB)}{sin(\angle LSA}.\frac{AD}{BD}.\frac{SB}{SA} =\frac{\frac{KB}{SB}}{\frac{LA}{SA}}.\frac{AD}{BD}.\frac{SB}{SA}=\frac{BC}{AC}.\frac{AD}{BD}=\frac{BC}{AC}.\frac{AD.AB}{BD.AB} =\frac{BC}{AC}.\frac{AC^2}{BC^2}=\frac{AC}{BC} \Longrightarrow T' \equiv T $ so by harmonic division we have $(B,A,T,N)=-1 $ where $(KL \cap BC=N)$ so we have $CT,CN$ are the internal and external bisectors of $C$ we want to prove that $NL.NK=ND.NT$ $ \angle NCT=\angle CDT=90 \Longrightarrow NC^2=ND.NT $ so we need to prove that $NC$ is tangent to the circumcircle of $CLK$ and this is not hard

for the last part we can do like this : we want to prove $NC$ is tangent to circumcircle of triangle $CLK$ so we need to prove that $\angle CKL=\angle NCL$ but we had $\angle NCB=45$ so we need to prove that $ \angle CKL-\angle BCL=45 $ $ \angle CKL-\angle BCL=\angle CAL+\angle KCA+\angle KLA-\angle LCB$ let $J=CB \cap SL$ the we have $JCLA$ is cyclic ,$AC=AL, JC=JL \Longrightarrow \angle BCL=\frac{\angle CAL}{2}$ so we need to prove that $\angle LSK+\angle LAC+\angle KBC=90$ $\Longleftrightarrow \angle LSB+\angle CBS=\angle LAC$ and this obvious ,so we are done

sco0orpiOn wrote: so by harmonic division we have $(B,A,T,N)=-1 $ where $(KL \cap BC=N)$ so we have $CT,CN$ are the internal and external bisectors of $C$ Doesn't it follow that $CA$ and $CB$ are bisectors of $\angle NCT$?

Lyub4o wrote: Doesn't it follow that $CA$ and $CB$ are bisectors of $\angle NCT$? yes ,and i used it ($\angle NCB=45$)

We know $X$ is on the radical axis $CD$ of the orthogonal circles $\omega_A = (A,AC)$ and $\omega_B = (B,BC)$, so letting $\{K,K'\} = AX\cap \omega_B$ and $\{L,L'\} = BX\cap \omega_A$, we see that $K,K',L,L'$ lie on a circle $\omega$ centered at $O$. Since $OB\perp XA$ and $OA\perp XB$, $O$ must be the orthocenter of $\triangle{XAB}$; in particular, $O\in CXD$. Furthermore, $AL^2 = AL'^2 = AC^2 = AK\cdot AK'$, so $BLXL'$ is the polar of $A$ (with respect to $\omega$) and $AKXK'$ is the polar of $B$. Hence $\triangle{XAB}$ is self-polar. Since $D$ lies on the polar of $X$ and $O,X,D$ are collinear, $D$ is the inverse of $X = KK'\cap LL'$. It follows from the standard Yufei configuration that $D$ is the spiral center taking $LK$ to $K'L'$ (and $LK'$ to $KL'$), and $T = LK'\cap KL'$ (note that $DK'L'T$ is also cyclic, and that $LK'\cap KL'$ lies on the polar of $X$). To complete the diagram, let $T' = LK\cap K'L'$ so $\triangle{XTT'}$ is self-polar by Brokard's theorem (by duality, we should have $CT'$ the external bisector of $\angle{BCA}$). But $-1 = K(K,K';L,L') = K(B,A;T',T)$, so it suffices to show $\angle{TCT'} = 90^\circ$. However, $O$ has fixed power $OX\cdot OD$ with respect to any polar triangle $\triangle{XYZ}$ through $X$, and thus lies on the radical axis of the circles with diameters $AB,TT'$. Since line $ABTT'$ is perpendicular to $OCD$, we conclude that $\angle{TCT'} = 90^\circ$, as desired. (Since $L\in (BX)$, $K\in (AX)$, we have $T'\notin (BA)\implies T\in (BA)$.) Comment. After the first two paragraphs, it's straightforward to complex bash with unit circle $\omega$ ($(k+k')(l+l') = 2(kk'+ll')$ or $(k+k'-l-l')^2 = (k-k')^2 + (l-l')^2$ from $KLK'L'$ harmonic), with the observation that $c,2d-c$ are inverses with respect to $\omega$: we get $1-\frac{c}{d} = \pm\frac{(k-k')(l-l')}{2(kk'-ll')}$.

An easy exercise if you've seen IMO 2012 #5 before.. From that problem it follows that $K, L$ are pairs of antihomologous points w.r.t. the two orthogonal circles centered at $A, B$ with radii $AC, BC$; hence the line $KL$ goes through the exsimilicenter $Se$ of the two circles, and our goal is to prove that $T$ is the insimilicenter $Si$.. For this we consider the inversion centered at $Se$ that swaps the two circles, and it remains to show that the points $Si$ and $D$ are inverses.. Indeed, $CSi$ is tangent to the circle of inversion since it is perpendicular to $CSe$.. hence the two points are indeed inverses and the problem is finished.

v_Enhance wrote: Let $ABC$ be a scalene triangle with $\angle BCA = 90\dg$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK = BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL = AC$. The circumcircle of triangle $DKL$ intersects segment $AB$ at a second point $T$ (other than $D$). Prove that $\angle ACT = \angle BCT$. As shown in the figure below. $ E, F $ are the intersection points of $AX, BX$ with the circumcircle of $Rt\triangle ABC$. $G$ is the intersection of $AF, BE$. It's obvious $X$ is the orthocenter of $\triangle ABG$ and $G$ is on $CD$. we have $\angle DKB = \angle DAE = \angle DGE$ so $B,G,K,D$ are concyclic. $\angle GKB=\angle GDB=90$, $GK^2= GE\times GB$. Also $\angle DLA = \angle DBF = \angle DGA$ so $A,D,L,G$ are concyclic. $\angle ALG=\angle ADG=90$, $GL^2= GF\times GA=GE\times GB =GK^2$. So $GK=GL$, Thus $SK=SL$, $KN=ML$ As it's quite obvious that $TM//BX$, $TN//AX$, we have $\frac{AM}{ML} =\frac{AT}{TB}=\frac{KN}{NB}$, so $ML\times KN=AM\times NB=(AL-ML)(BK-KN)$ $AL\times BK= AL\times KN + BK\times ML=(AC+BC)\times ML$ $ML=KN=\frac{AC\times BC}{AC+BC}$ $AM=AL-ML=\frac{AC^2}{AC+BC}$ Consequently $\frac{AT}{TB}=\frac{AM}{ML}=\frac{AC}{BC}$ so $CT$ is the bisector of $\angle ACB$, namely $\angle ACT = \angle BCT$.

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My solution same with @Andrew 's solution

As it's quite obvious that $TM//BX$, $TN//AX$, Can you show me why we have that?

We proved that A,D,L,G are concyclic so $\angle ALD = \angle AGD = \angle ABF$. And $\angle ATM = \angle ALD$ since MTDL is an cyclic quadrilateral. Therefore, $\angle ATM = \angle ABF$. So $TM//BX$

v_Enhance wrote: Let $ABC$ be a scalene triangle with $\angle BCA = 90^{\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK = BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL = AC$. The circumcircle of triangle $DKL$ intersects segment $AB$ at a second point $T$ (other than $D$). Prove that $\angle ACT = \angle BCT$. WLOG $CA<CB$. Let $\omega_A, \omega_B$ be circles centered at $A, B$ respectively, passing through point $C$. Let $\overline{BL}$ meet $\omega_A$ again at $L'$ and $\overline{AK}$ meet $\omega_B$ again at $K'$. Observe that $X$ lies on the radical axis of $\omega_A, \omega_B$ hence $XL \cdot XL'=XK \cdot XK'$ implying $KLK'L'$ is cyclic with circumcircle $\omega$. Now let $T=\overline{KL'} \cap \overline{K'L}$ and $S=\overline{KL} \cap \overline{K'L'}$. Let $Y$ be the orthocenter of $\triangle XAB$. Then $\overline{AY}$ is the perpendicular bisector of $\overline{LL'}$ and $\overline{BY}$ is the perpendicular bisector of $\overline{KK'}$. Thus, $Y$ is the center of $\omega$. Observe that $A=\overline{LL} \cap \overline{LL'}$ while $B=\overline{KK} \cap \overline{K'K'}$ proving $A,B,T,S$ are collinear and $(AB, ST)=-1$. Now $\angle ACB=90^{\circ}$ so $\overline{CA}, \overline{CB}$ are bisectors of angle $TCS$. Now by Brokard's Theorem, $Y$ is the orthocenter of $\triangle TXS$. Thus, $DT \cdot DS=DX \cdot DY=DA \cdot DB=DC^2$ proving $\angle TCS=90^{\circ}$. Hence $\angle ACT=\angle BCT$. Finally, we show $T$ lies on $\odot(DKL)$; or equivalently $D$ is the spiral center of $\overline{KL'} \mapsto \overline{LK'}$. Let $N=\overline{AY} \cap \overline{BX}$ then $OL^2=ON \cdot OA=OX \cdot OD$ proving $X, D$ are inverses in $\omega$. Hence $D$ is the desired spiral center. Thus, our proof is complete! $\blacksquare$ Remark: This is an example of "completing the picture". By appending $Y$ we are able to see things from $\triangle AYB$ and $X$ suddenly becomes more natural. This gives us a foothold to start with and subsequent work is more motivated.

Eh sorry if I'm reasoning circularly, but isn't it exceedingly trivial after $\sqrt{ab}$ inversion at $C$ ? Let the ray $CD$ hit the circumcircle of $\omega_{ABC}$ again at $H$. Perform a $\sqrt{CA \cdot CB}$ inversion at $C$ followed by the reflection at $C$-angel bisector. Now the problem reads: USA 2013 TST P3, inverted wrote: Let $\Delta CAB$ be a right angled traingle with $\angle CAB = 90$. Let $H$ be the circumcentre of $\omega_{ABC}$, $D$ be the $C$-antipode and $E$ be the midpoint of arc $AB$ not containing $C$. Define $\ell_A, \ell_B$ to the lines through $H$ passing through the midpoint of $\bar{AC}, \bar{AB}$ respectively. For a point $X \in \vec{CH}$ outside $\Delta ABC$, define $K_X \in \{ \ell_B \cap \omega_{ACH} \} , L_X \in \{ \ell_A \cap \omega_{BCH} \}$ such that $K_X, A$ are in the same side of $\vec{CH}$ (similar for $L_X$). Then $K_X, L_X, D, E$ are concyclic. Claim $HK_X = HL_X$ Proof Define $K_Z$ to be the other intersection of $\ell_B$ with $\omega_{ACH}$. $\ell_B$ is clearly perpendicular to $BC$, so the circumcentre of $\omega_{ACH}$ lies on $\ell_A$ (and so does $H$). So $HK_X = HK_Z$ by symmetry. By PoP $HA \cdot HX = HK_X \cdot HK_Z = HK_X^2 \Rightarrow HK_X = \sqrt{HA \cdot HX}$. Similary $HL_X = \sqrt{HA \cdot HX}$. The conclusion follows. $\blacksquare$ Claim $K_X, L_X, D, E$ is an isosceles trapezium. Proof (Angles are NOT directed) Assume $AC < AB$, and let $G$ be the antipode of $I$. Then $$ \angle KHE = \angle GH \ell_B = \angle GHB = \angle \ell_B HB = 90 - \frac{\angle CHB}{2} = 90 - \angle CAB = \angle CBA = \angle CH\ell_A = \angle LHD$$. Coupled with $HI = HD$ and $HK = HL$, we get $\Delta KHE \sim \Delta KHD$, from which the conclusion follows. $\blacksquare$ Since any isosceles trapezium is a cyclic quad, we're done (?)

[asy][asy] unitsize(0.9inches); pair B=dir(180); pair A=dir(0); pair C=dir(117); pair CC=2*foot(C,A,B)-C; pair D=(C+CC)/2; pair X=0.77*D+(1-0.77)*C; pair K=intersectionpoints(A--((X-A)*10+A),circle(B,abs(B-C)))[0]; pair KK=intersectionpoints(A--((X-A)*10+A),circle(B,abs(B-C)))[1]; pair L=intersectionpoints(B--((X-B)*1000+B),circle(A,abs(A-C)))[0]; pair LL=intersectionpoints(B--((X-B)*1000+B),circle(A,abs(A-C)))[1]; pair T=extension(KK,L,A,B); pair S=extension(K,L,A,B); pair BB=2*foot(0,B,X)-B; pair AA=2*foot(0,A,X)-A; pair J=extension(B,AA,A,BB); draw(A--B); draw(circumcircle(A,B,C)); draw(circle(B,abs(C-B))); draw(circle(A,abs(C-A))); draw(B--LL); draw(A--KK); draw(S--B); draw(J--CC); draw(J--B); draw(J--A); draw(LL--T); draw(KK--T); draw(LL--S); draw(K--S); dot("$A$",A,dir(A)); dot("$B$",B,dir(180+45)); dot("$C$",C,2*dir(C)); dot("$C'$",CC,dir(CC)); dot("$D$",D,dir(270-45)); dot("$X$",X,1.5*dir(60)); dot("$K$",K,1.5*dir(70)); dot("$K'$",KK,dir(130)); dot("$L$",L,dir(L)); dot("$L'$",LL,dir(LL)); dot("$S$",S,dir(S)); dot("$T$",T,dir(-90)); dot("$A'$",AA,1.6*dir(120)); dot("$B'$",BB,dir(65)); dot("$J$",J,dir(90)); [/asy][/asy] Let $\omega_A$ be the circle with center $A$ passing through $C$ and let $\omega_B$ be the circle with center $B$ passing through $C$. Let $C'$ be the reflection of $C$ over $AB$. Let $\{K,K'\}=AX\cap\omega_B$ and $\{L,L'\}=BX\cap\omega_A$ with $K$ and $L$ inside $(AB)$. Also, let $A'$ and $B'$ be the second intersections of lines $AX$ and $BX$ with $(AB)$ respectively. Note that $\angle BA'X=\angle BDX=\pi/2$ so $(A'XDB)$ is cyclic, and similarly $(B'XDA)$ is also cyclic. By radical axis along with $(AB)$, we see that $A'B$, $B'A$, and $CC'$ concur at some point $J$ with \[JC\cdot JC'=JX\cdot JD=JA'\cdot JB=JB'\cdot JA.\]\ Let $T=K'L\cap KL'$ and $S=KL\cap K'L'$. We'll show that $(DLK)\cap AB=T$ and that $\angle BCT=\angle ACT$. Note that \[XK\cdot XK'=XC\cdot XC'=XL\cdot XL'\]so $\omega\equiv(KLK'L')$ is cyclic. We see that $BA'$ is the perpendicular bisector of $KK'$ and $AB'$ is the perpendicular bisector of $LL'$, so $J$ is the center of $\omega$. Since $JC\cdot JC'=JX\cdot JD$ and $D$ is the midpoint of $CC'$, we have that $(CC';JX)=-1$. Therefore, the polar of $J$ with respect to $\omega_B$ passes through $X$, but is also perpendicular to $BJ$. Thus, the polar of $J$ with respect to $\omega_B$ is $KK'$, so \[JK^2=JC\cdot JC'.\]Therefore, inversion in $\omega$ swaps the pairs $(C,C')$, $(X,D)$, $(B,A')$, and $(A,B')$. By Brokard's theorem on $KLK'L'$, we see that $ST$ is the polar of $X$ with respect to $\omega$. Since $JX\cdot JD=JK^2$ and $DA\perp JX$, we see that $AB$ is the polar of $X$ with respect to $\omega$, so $S$ and $T$ lie on $AB$. Furthermore, \[(ST;AB)\stackrel{L}{=}(KK';XA)=-1\]since $CC'$ is the polar of $A$ with respect to $\omega_B$ (fundamentally this is because $\omega_A$ and $\omega_B$ are orthogonal). Thus, $S$ and $T$ are on $AB$ and are in fact harmonic conjugates with respect to $AB$. We see that by the same application of Brokard's theorem that $S$ is on the polar of $T$ with respect to $\omega$ and vice versa, so by a well known lemma, we see that $(ST)\perp\omega$. This means that if $\{C_1,C_1'\}\equiv(ST)\cap CC'$, then $(C_1,C_1')$ are swapped under inversion about $\omega$, since $(ST)$ should be fixed under the inversion. Therefore, \[JC_1\cdot JC_1'=JK^2=JC\cdot JC',\]so in fact $\{C_1,C_1'\}=\{C,C'\}$. Thus, $\angle SCT=\pi/2$. Therefore $CS$ and $CT$ are harmonic conjugates with respect to lines $CB$ and $CA$, and furthermore $CS\perp CT$. This can only happen if $CT$ and $CS$ are the internal and external angle bisectors of $CB$ and $CA$, so we have $\angle BCT=\angle ACT$, as desired. All that's left to show is that $DTKL$ is cyclic. We see that $X$ and $D$ are inverses with respect to $\omega$, and since $X=KK'\cap LL'$, this implies that $D$ is the Miquel point of $KLK'L'$. Thus, $D\in(TLK)$, as desired. This completes the proof.

Nice blend of rich geometry configuration and projective geometry. Take $H$ as the orthocenter of $\triangle AXB$. $\textbf{Claim 01.}$ $(H, X; C, C') = -1$. $\textit{Proof.}$ Notice that $HX \cdot HD = HA' \cdot HB = HC \cdot HC'$ $\textbf{Claim 02.} LKL'K'$ is cyclic. $\textit{Proof.}$ We have $XK' \cdot XL = XC \cdot XC' = XL \cdot XL'$ $\textbf{Claim 03.}$ $H$ is the center of $KLK'L'$. $\textit{Proof.}$ Notice that the center of $KLK'L'$ must lie on both the perpendicular bisector of $KK'$ and $LL'$ which are $BA'$ and $AB'$. This then follows by the fact that $AB' \cap A'B = H$. $\textbf{Claim 04.}$ $HK$ is tangent to $(K'CKC')$. $\textit{Proof.}$ Notice that $(H, X; C, C') = -1 $ and the fact that $K'C'CK$ is harmonic (This follows by the fact that $AC$ tangent to $(K'CKC')$.) Hence \[-1 = K(K', K; C, C') = (X, KK \cap CB; C, C') \]It follows that $HK$ is tangent to $BC$. Redefine $T = K'L \cap KL'$. and $L'K' \cap LK = Y$. By Brocard Theorem on $K'L'KL$, we have that $H$ is the orthocenter of $\triangle TXY$. Furthermore this gives us $HX \perp TY$, or $TY \parallel AB$. $\textbf{Claim 05.}$ $A,B,T,Y$ are collinear. In fact, $(A,B;T,Y) = -1$. $\textit{Proof.}$ Pascal on $LLL'K'KK$ gives us $A,B, L'K' \cap LK = Y$ are collinear. Since we have $TY \parallel AB$, then $A,B,T,Y$ are collinear. Now, notice that \[ -1 = X(Y, TX \cap K'L' ; K', L') = (Y, T; A, B) \] $\textbf{Claim 06.}$ $\angle YCD = 90^{\circ}$. $\textit{Proof.}$ To prove this, notice that \[ DY \cdot DT = DX \cdot DH = DA \cdot DB = DC^2 \]Since $\angle YCT = 90^{\circ}$ and $(A,B;T,Y) = -1$. Then, we must have $CT$ bisects $\angle ACB$, which is what we wanted. It suffices to prove that $D,T,K,L$ is cyclic. $\textbf{Claim 07.}$ $D$ is the Miquel Point of $K'L'KL$. $\textit{Proof.}$ Notice that $(HL')^2 = HC \cdot HC' = HX \cdot HD$. Since $X = KK' \cap LL'$ and $H$ is the center of $(KK'LL')$, it follows that $D$ is the Miquel Point of $K'L'KL$. Therefore, this gives us $DTKL$ being cyclic.

Let $\overline{AL}$ and $\overline{BK}$ intersect at a point $M$. Let $\omega_A$ and $\omega_B$ be the circles centered at $A$ and $B$ with radius $AC$ and $BC$, respectively. Suppose that lines $AK$ and $BL$ intersect $\omega_B$ and $\omega_A$ at points $P$ and $Q$, respectively. Since $KX \cdot XP = CX \cdot FX = LX \cdot QX$, we find that $KLPQ$ is cyclic. Note that $\omega_A$ and $\omega_B$ are orthogonal, so $AK \cdot AP = AC^2 = AL^2$, so $\overline{AML}$ is tangent to $(KLPQ)$. Similarly, $\overline{BMK}$ is tangent to $(KLPQ)$. Note that $D$ is the miquel point of cyclic quadrilateral $KLPQ$, so $T$ lies on line $PL$ and $QK$. Furthermore, lines $KL$ and $PQ$ intersect a point $S$ on line $AB$ by Brokard's theorem. Finally, $-1 = (PK;XA) \stackrel{Q}{=} (ST;AB)$. Now consider the inversion around $T$ with radius $TC=TF$. Note that this inversion swaps $\omega_A$ and $\omega_B$, so it swaps $L$ and $P$. Now $TD \cdot TS = TL \cdot TP = TC^2$, so $\angle TCS = 90^\circ$. This implies $\overline{CT}$ bisects $\angle ACB$, as desired.

Let $C'$ be the reflection of $C$ over $AB$. Then, let $E$ and $M$ be where $AX$ hits the circle centered at $B$ with radius $BC$, and let $F$ and $N$ be where $BX$ hits the circle centered at $A$ with radius $AC$. Let $AE>AM$ and $BE>BM$. Note that $EX \cdot XM = C'X \cdot XC = FX \cdot XN$ and so $FENM$ is cyclic. Now, we see that the circle through $C$ with center $B$ has two tangents $AC$ and $AC'$, meaning that $(AX; ME) = -1$ and $(BX; NF) = -1$ too. So, $EF$, $MN$ and $AB$ concur at a point because such a point is needed to map $(AX; ME)$ to $(BX; NF)$. Now, with one of those, let's say $(BX; NF) = -1$, taking perspectivity with respect to $E$ gets that since $EM$ and $FN$ hit $AB$ at $T$, $(PT;AB) = -1$. Because symmetry holds meaning $C'$ is on the circumcircle of $PCT$, then $\angle PCT = \angle PC'T = 90^{\circ}$. Finally, $(PT;AB) = -1$ and $\angle PCT = 90^{\circ}$ gets that $\angle ACT = \angle BCT$ and we are done.

v_Enhance wrote: Let $ABC$ be a scalene triangle with $\angle BCA = 90^{\circ}$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let $K$ be the point on the segment $AX$ such that $BK = BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL = AC$. The circumcircle of triangle $DKL$ intersects segment $AB$ at a second point $T$ (other than $D$). Prove that $\angle ACT = \angle BCT$. Solution. Claim 1. If $\overline{AKX}\cap(\omega_B) = K' \ne K$ and $\overline{BLX}\cap(\omega_A) = L' \ne L$, where $\omega_A$ and $\omega_B$ denotes the circles with radii $AC = AL$ and $BC = BK$ respectively. Then, we have $K,L,K',L'$ concyclic. Proof: Since, $\{X\} \in \overline{CD}$ which's perpendicular to $\overline{AB}$, so, $X$ lies on the radical axis of $\omega_A$ and $\omega_B$. Hence, $$|\text{pow}_{(\omega_A)}X| = |\text{pow}_{(\omega_B)}X| \implies XL \cdot XL' = XK \cdot XK'.\ \square$$Let $\Omega$ denote the circle here. We re-define $T$ as: $\overline{K'L} \cap \overline{L'K} = \{T\} - (1)$, we wish to show $\odot(DKL) \cap \overline{AB} =T \ne D.$ Next, we let $\overline{KL} \cap \overline{L'K'} = \{R\} - (2)$ and $J$ to be the orthocenter of $\triangle XAB.\ (\spadesuit)$ Claim 2. $J$ is the center of $\Omega.$ Proof. Since, $\overline{LL'}$ and $\overline{KK'}$ are the radical axes of $\left\{\omega_A, \Omega\right\}$ and $\left\{\omega_B, \Omega\right\}$ respectively, hence, $\overline{AJ}$ and $\overline{BJ}$ are perpendicular bisectors of $\overline{LL'}$ and $\overline{KK'}$ respectively, and we're through. $\square$ From $(1)$ and $(2)$, by Brokard's, we get $\overline{RT}$ as the polar of $X.$ Now, we observe that $\overline{K'K'} \cap \overline{KK}=\{B\}$ and $\overline{L'L'} \cap \overline{LL}=\{A\}$, hence, $\overline{AB}$ is the polar of $X$. Thus, $R, B, T, A$ are collinear and $(A,B;T,R) = -1\ (\diamondsuit)$. Since, $\angle BCA = 90^{\circ}$, ergo, $\overline{CA}, \overline{CB}$ are bisectors of $\angle TCS$. Again by Brokard's, $J$ is the orthocenter of $\triangle TRX.$ Hence, by PoP WRT foot of altitude $D$, with $J$ as the orthocenter of $\triangle TRX\ (\heartsuit)$ and $\triangle XAB\ (\spadesuit)$ gives: $$DR \cdot DT \overset{(\heartsuit)}{=} DX \cdot DJ \overset{(\spadesuit)}{=} DA \cdot DB.\qquad(\clubsuit).$$At last, since $\angle BCA = 90^{\circ}$, whence $DA \cdot DB = DC^2$. So, from $(\clubsuit)$, we have $DT \cdot DR=DC^2$ and thus, $\angle TCR = 90^{\circ}$; this along with $(\diamondsuit)$, gives $\angle ACT = \angle BCT$, and we're good. Lemma.[Well known] Let $ABCDEF$ be a complete quadrilateral, where $\overline{AB}\cap\overline{CD} = \{E\}, \overline{AD}\cap\overline{BC} = \{F\}$ and $\overline{AC}\cap\overline{BD}=\{K\}.$ Then, $M$, the Miquel point of the quadrilateral is the inverse of $K$ WRT $\odot(ABCD)$. (Proof. Brokard suffices.) Hence, we just need to show $T, K, L, D$ are concyclic. But before that, we note that, due to Orthocentric system surrounding $(\spadesuit)$, we have $A, D, X$ and $\overline{AJ} \cap \overline{BX}$ concyclic, since $\overline{AJ} \perp \overline{LL'} \equiv \overline{BX}$; say $\overline{AJ} \cap \overline{BX} = \{Q\}$. So, $$JX \cdot JD = JQ \cdot JA = JL^2,$$and thus, $X, D$ are inverses of each other WRT $\Omega$. Since, $\overline{LL'} \cap \overline{KK'}=\{X\}$ thus, by our lemma, we get $D$ to be the Miquel point of $KL'K'L$; hence, it belongs to $\odot(TLD)$, and we're done. $\blacksquare$

anantmudgal09 wrote: Let $N=\overline{AY} \cap \overline{BX}$ then $OL^2=ON \cdot OA=OX \cdot OD$ proving $X, D$ are inverses in $\omega$. Hence $D$ is the desired spiral center. What's $O$ here?? (By usual notations, it denotes the circumcenter of $\triangle ABC$, but then how does that Power of Point hold?) It will be the center of $KL'KL'$ (in other words, the orthocenter of $XAB$), I guess ; unless it's PoP WRT something else. Anyways, nice sol.

Indeed, the solution relies on IMO 2012/5. By inversion at $C$, we get the following problem: $ACBD$ is a rectangle, $X$ lies on the extension of line $CD$ past $D$, $(XAC)$ intersects the perpendicular bisector of $BC$ again at $K$ and $(XBC)$ intersects the perpendicular bisector of $AC$ again at $L$, show $(DKL)$ passes through the arc midpoint of arc $AB$ not passing through $C$. Let the perpendicular bisectors of $AC$ and of $BC$ intersect again at $O$, which is the center of $(ABC)$. Remark that the statement of IMO 2012/5 after this inversion would imply that a circle passing through $K$ and $L$ is tangent to both perpendicular bisectors, so we can eschew the information about $X$ and instead write $OK=OL$. If we define $T$ as the arc midpoint of arc $AB$ not passing through $C$, it suffices to show $DTKL$ is an isosceles trapezoid. As $OT=OD,OK=OL$, it is sufficient to show $DT\parallel KL$. Indeed, note that by a rotation by $90^\circ$ about $O$, we have \[\measuredangle KOT=\measuredangle LOA=\measuredangle COL=\measuredangle DOL,\]hence we are done.

Let $K^*$ and $L^*$ be the inverses of $K$ and $L$ with respect to the circles they do not lie on. Observe that $K^*$ lies on the circle at $B$ and $L^*$ lies on the circle at $A$ because the two circles are orthogonal. In fact, radical axis implies $KL^*K^*L$ is cyclic. Next, we cliam that the center of this circle is $\overline{LL} \cap \overline{KK}$. First we will show that this point also lies on $\overline{CD}$. Let $O$ be the polar of $\overline{BX}$, which obviously lies on $\overline{CD}$. Now $O$ lies on the polar of $L$ as well, implying $\overline{OL}$ is tangent to the circle at $A$. This implies that $X$ is the orthocenter of $\triangle OAB$, because $\overline{OX}$ is the polar of $B$ with respect to the circle with center $A$. Hence, $\overline{AX} \perp \overline{BO}$, and because $B$ lies on the polar of $A$, $\overline{BO}$ is the polar of $A$, and thus $\overline{AX}$ is the polar of $O$. Hence $\overline{OK}$ is tangent to the circle at $B$ as well, proving the concurrence. Now $\overline{AX}$ is the polar of $O$ with respect to the circle with center $B$, implying that $OK = OK^*$. Similarly, $OL = OL^*$ and thus $O$ is the center of $(K^*LKL^*)$ as required. Next, denote $S = \overline{LK} \cap \overline{L^*K^*}$. Since $\measuredangle ODB= \measuredangle OKB = 90^{\circ}$, $BDKOK^*$ is cyclic, and so is $ADLLOL^*$. It follows that $D$ is the Miquel Point of complete quadrilateral $KLK^*L^*$. Henceforth both $\overline{KL^*} \cap \overline{LK^*}$ and $\overline{KL} \cap \overline{K^*L^*}$ lie on $\overline{AB}$. Observe that the first point msut be $T$ because it lies on $(DLK)$; call the second point $S$. The main claim is now the following. Claim. $(ST;BA)=-1$. Project through $X$. Then we need $(S, XT \cap \overline{L^*K^*}; L^*K^*) = -1$ which is obvious. $\blacksquare$ Hence it suffices to show $\angle SCT = 90^{\circ}$ by the bisectors lemma. Notice that $O$ lies on the radical axis of $(ABC)$ and $(ST)$, and thus $\overline{OD}$ is the radical axis of these two circles, which passes through $C$. Thus $(ST)$ passes through $C$ as well, and we are done.

Nice projective geometry. First, let $R=\underbrace{\Gamma(A,AC)}_{\omega_a}\cap BX\ne L$ and $S=\underbrace{\Gamma(B,BC)}_{\omega_b}\cap AX\ne K$. Since $\angle ACB=90^{\circ},$ we must have $\omega_a$ and $(ABC)$ orthogonal circles, and the same for $\omega_b$ and $(ABC)$. This implies that $$(V,SL\cap AB,A,B)\overset{S}{=}(R,L,X,B)\overset{C}{=}(R,L,C,E)=-1,$$where $V=RS\cap AB$. If we do the same for $(V,RK\cap AB,A,B)$ we can easily conclude that $SL$, $RK$ and $AB$ concur. Furthermore, we have $$(LK\cap AB,T',A,B)\overset{K}{=}(L,R,X,B)=-1,$$from which we get $LK$, $AB$ and $RS$ concur at $V$. Also, if we let $CD\cap (ABC)=E\ne C$, we have $XR\cdot XL=XC\cdot XE=XD\cdot XK$, concluding that $RSKL$ is a cyclic quadrilateral. Since $X$ lies on $AD$, we have $XD\perp VT'$ and therefore $D$ is $RSLK$'s Miquel Point, and thus $$KLDT'\text{ is cyclic}\implies T=T'.$$Finally, see that $XAB$ is an auto polar triangle w.r.t (RSLK) and, by Brokard's theorem, $XVT$ is also an auto polar triangle. If we let $O$ to be the center of $(RSLK)$, we conclude that $X$ is the orthocenter of $OVT$ and $OAB$. This implies that $O$ lies on the radical axis of the circunferences with diameter $AB$ and $TV$, which implies that $CEVT$ is cyclic. Since $E$ is the reflection of $C$ w.r.t to $AB$, we must have $\angle VCT=90^{\circ}$, and since $(V,T,A,B)=-1$ it follows that $CT$ bisects $\angle BCA.$ $\blacksquare$

Attachments:

WLOG suppose that $CD = 1$, and invert about the circle with diameter $CD$, denoting the inverse of a point $P$ with $P'$. Then, let $C$ be the origin and $D = (\cos\theta, \sin\theta)$ in the Cartesian plane. We then deduce that \begin{align*} A' &= (\cos\theta,0)\\ B' &= (0,\sin\theta)\\ X' &= (r\cos\theta, r\sin\theta),\quad r > 1. \end{align*} Claim: The equation of the inverse of the circle centered at $A$ with radius $AC$ is \[x = \frac{1}{2}\cos\theta,\]and similarly, the inverse of the circle centered at $B$ with radius $BC$ has equation \[y = \frac{1}{2}\sin\theta.\]Proof. Let $C^\ast$ be the reflection of $C$ over $A$. Then the inverse of $C^\ast$ is the midpoint of $CA'$, and the image of the circle goes through this point and is perpendicular to line $AC$, which implies the result. The other half follows analogously. $\blacksquare$ Now, we will compute the circumcenter and circumcircle of $(B'CX')$. The perpendicular bisector of $CX'$ has equation \[-\cot\theta(x - \frac{1}{2} r\cos\theta) = y - \frac{1}{2} r\sin\theta,\]and the perpendicular bisector of $B'C$ is $y = \frac{1}{2}\sin\theta$, so the $x$-coordinate of their intersection is given by \[-\cot\theta(x - \frac{1}{2} r\cos\theta) = \frac{1}{2}(1 - r)\sin\theta,\]which, upon solving, yields the point \[\left(\frac{(r - 1)\sin^2\theta}{2\cos\theta}{2\cos\theta} + \frac{r\cos\theta}{2}, \frac{\sin\theta}{2}\right).\]We can clean up the $x$-coordinate a little bit by rewriting it as \[\frac{r\sin^2\theta - \sin^2\theta + r\cos^2\theta}{2\cos\theta} = \frac{r - 1 + \cos^2\theta}{2\cos\theta} = \frac{r - 1}{2\cos\theta} + \frac{\cos\theta}{2}.\]Then, the equation of the corresponding circle is \[\left(x - \left(\frac{r - 1}{2\cos\theta} + \frac{\cos\theta}{2} \right)\right)^2 + \left(y - \frac{\sin\theta}{2}\right)^2 = \left(\frac{r - 1}{2\cos\theta} + \frac{\cos\theta}{2}\right)^2 + \frac{\sin^2\theta}{4}.\]To compute $L'$, we want to intersect this circle with $x = \frac{1}{2}\cos\theta$, which gives us the following: \begin{align*} \left(\frac{\cos\theta}{2} - \left(\frac{r - 1}{2\cos\theta} + \frac{\cos\theta}{2} \right)\right)^2 + \left(y - \frac{\sin\theta}{2}\right)^2 &= \left(\frac{r - 1}{2\cos\theta} + \frac{\cos\theta}{2}\right)^2 + \frac{\sin^2\theta}{4}\\ \left(\frac{1 - r}{2\cos\theta}\right)^2 + \left(y - \frac{\sin\theta}{2}\right)^2 &= \left(\frac{r - 1}{2\cos\theta}\right)^2 + \frac{r - 1}{2} + \frac{\cos^2\theta}{4} + \frac{\sin^2\theta}{4}\\ \left(y - \frac{\sin\theta}{2}\right)^2 &= \frac{r - 1}{2} + \frac14 = \frac{2r - 1}{4}\\ y &= \frac{\pm\sqrt{2r-1}}{2} + \frac{\sin\theta}{2}. \end{align*}There is now a positive root and a negative root; $L'$ corresponds to the positive root because it's above line $CD$, not below. Thus, $L'$ has coordinates \[\left(\frac{\cos\theta}{2}, \frac{\sqrt{2r-1}}{2} + \frac{\sin\theta}{2}\right).\]Similarly, we can find that $K'$ has coordinates \[\left(\frac{\sqrt{2r-1}}{2} + \frac{\cos\theta}{2}, \frac{\sin\theta}{2}\right).\]Now, let $\widehat{T}$ in the original picture be the intersection of the bisector of $\angle BCA$ with $AB$ (so we want to show that $D$, $K$, $L$, $\widehat T$ are concyclic). Then $\widehat T'$ is the midpoint of arc $\widehat{A'DB'}$ on the circle with diameter $CD$; since $A'B'$ is a diameter of this circle too, this corresponds to a counterclockwise $90^{\circ}$ rotation of $A'$ about the center. This means \[\widehat T' = \left(\frac{\sin\theta + \cos\theta}{2}, \frac{\sin\theta + \cos\theta}{2}\right).\]It now suffices to prove that $\widehat T'$, $D$, $K'$, and $L'$ are concyclic. We will show that $D\widehat T'$ and $K'L'$ share a perpendicular bisector, implying that they are the bases of an isosceles trapezoid. To do this, observe that the midpoints of the aforementioned segments are \begin{align*} M &= \left(\frac{\sin\theta + \cos\theta}{4} + \frac{\cos\theta}{2}, \frac{\sin\theta + \cos\theta}{4} + \frac{\sin\theta}{2}\right),\\ N &= \left(\frac{\sqrt{2r-1}}{4} + \frac{\cos\theta}{2}, \frac{\sqrt{2r-1}}{4} + \frac{\sin\theta}{2}\right) \end{align*}respectively; then vector $\overrightarrow{MN}$, which is \[\overrightarrow{NM} = \left(\frac{\sqrt{2r - 1}}{4} - \frac{\sin\theta + \cos\theta}{4}, \frac{\sqrt{2r - 1}}{4} - \frac{\sin\theta + \cos\theta}{4}\right),\]has a slope of 1, while segments $D\widehat{T}'$ and $K'L'$ both have slopes of $-1$, so we are done!

Wait inversion at $C$ trivializes this. What? Perform a $\sqrt{AC\cdot BC}$ inversion centered at $C$ followed by a reflection over the $C$ internal angle bisector. This map swaps $A$ and $B$, it swaps $AB$ and $ABC$. It swaps the point $D$ and the $C$ antipode $C'$, it swaps $X$ with a point $X'$ on $CO$, it swaps the circle centered at $B$ with radius $BC$ that is tangent to $AC$ at $C$ to a line $\ell_a$ parallel to $BC$ passing through the center $O$ of $(ABC)$, and it swaps the circle centered at $A$ with radius $AC$ to a similarly defined line $\ell_b$. Let $K_1$ and $K_2$ be the intersections of $(BXC)$ with $K_1$ being the point outside of triangle $ABC$ that $K$ inverts to. Define $L_1$ and $L_2$ similarly. The picture now looks like a rectangle. Since the center of $BXC$ must lie on the bisector of $BC$ which is $\ell_a$, we must have that $OK_1=OK_2$. Now PoP gives $CO\cdot XO = K_1O \cdot K_2O=K_1O^2$. We can similarly get that $CO\cdot XO = L_1O^2$. Thus $K_1O=L_1O$, and $K_1$ and $L_1$ are on $\ell_b$ and $\ell_a$, which are parallel to the sides $AC$ and $AB$. We now apply Cartesian coordinates with $\ell_a$ and $\ell_b$ as the $x$ and $y$ axes. Now set $A=(x,y)$. Since we have a rectangle, we compute $C=(-x,y)$, $B=(-x,-y)$, $C'=(x,-y)$. Since $M$ is a $90$ degree rotation of $A$ about the origin, it has coordinates $(y,-x)$. Since $OK_1=OL_1$, we can set $L_1=(a,0)$ and $K_1=(0,-a)$. Now we can easily check that $K_1L_1$ and $C'M$ have the same slopes so are parallel, and we can also check that $K_1M=L_1C'$ and $K_1C'=L_1M$ with the distance formula. These imply that $K_1L_1C'M$ is a cyclic isosceles trapezoid. Inverting back, we get that $LKDT'$ is cyclic, where we define $T$ as the intersection of the $C$ bisector and $AB$. Thus, we are done.

Let $\omega_A$ be the circle with center $A$ and radius $AC$, and $\omega_B$ be the circle with center $B$ and radius $BC$. Denote by $K^{\ast}$ the second intersection of $KX$ with $\omega_B$, and $L^{\ast}$ the second intersection of $LX$ with $\omega_A$. Then because $\omega_A$ and $\omega_B$ are both symmetric about line $AB$, we find that $CD$ is their radical axis, so $KK^{\ast}LL^{\ast}$ is cyclic. Moreover, since $\omega_A$ and $\omega_B$ are orthogonal, we find that $\overline{BK}$ and $\overline{BK^{\ast}}$ are tangent to $\omega = (KK^{\ast}LL^{\ast})$, so $KK^{\ast}$ is the polar of $B$. Similarly, we see that $LL^{\ast}$ is the polar of $A$, and since $X = \overline{LL^{\ast}} \cap \overline{KK^{\ast}}$, by La Hire's we find that $AB$ is the polar of $X$. Thus, $D$ is the inverse of $X$ wrt $\omega$, so it must be the Miquel point of $KLK^{\ast}L^{\ast}$, and $T = \overline{KL^{\ast}} \cap \overline{LK^{\ast}}$. We also find that $S = \overline{KL} \cap \overline{K^{\ast}L^{\ast}}$ lies on $AB$. Now, by Brokard, we find that $S$ lies on the polar of $T$ wrt $\omega$, so if $O$ is the center of $\omega$, then $\text{Pow}_{(ST)}(O) = R_{\omega}^2$. We also find that $A$ lies on the polar of $B$, so $\text{Pow}_{(AB)}(O) = R_{\omega}^2$, and hence line $OCD$ is the radical axis of $(ST)$ and $(AB)$. Therefore $C$ lies on $(ST)$, so $\angle SCT = 90^{\circ}$. However, since $$-1 = (K, K^{\ast}; L, L^{\ast}) \overset{K}{=} (B, A; S, T),$$by the right angles and bisectors lemma we indeed have $\angle ACT = \angle BCT$.

Let $K^*$ and $L^*$ be the inverses of $K$ and $L$ wrt $\omega_A$ and $\omega_B$, the circles through $C$ centered at $A$ and $B$, respectively. $\omega_A$ and $\omega_B$ are orthogonal, so $K^*$ lies on $\omega_B$ and $L^*$ lies on $\omega_A$. $X$ lies on the radax of $\omega_A$, and $\omega_B$, so $KLK^*L^*$ is cyclic. Let its circumcircle be $\gamma$. $KLK^*L^*$ is also harmonic, with tangents from $K$ and $K^*$ meeting at $B$ and tangents from $L$ and $L^*$ meeting at $A$, as \[AK \cdot AK^* = AC^2 = AL^2.\] Thus, Brocard on $KLK^*L^*$ tells us the polar of $X$ is $AB$. Let $O$ be the center of $\gamma$ and $P = K^*L^* \cap KL$. Now we note $O$ is the orthocenter of $\triangle XPT$, so $O$ lies on $CD$. Master Miquel says $D$ is the Miquel point of $KLK^*L^*$, so $T = KL^* \cap K^*L$. Self-polar triangles tell us $\gamma$ is orthogonal to both $(PT)$ and $(AB)$, so the radical axis of $(PT)$ and $(AB)$ is the altitude from $O$ to $AB$. Therefore $(ST)$ passes through $C$. We finish using Apollonius, as $(PT;AB)=-1$ and $\angle PCT=90$. $\blacksquare$

math154 wrote: We know $X$ is on the radical axis $CD$ of the orthogonal circles $\omega_A = (A,AC)$ and $\omega_B = (B,BC)$, so letting $\{K,K'\} = AX\cap \omega_B$ and $\{L,L'\} = BX\cap \omega_A$, we see that $K,K',L,L'$ lie on a circle $\omega$ centered at $O$. Since $OB\perp XA$ and $OA\perp XB$, $O$ must be the orthocenter of $\triangle{XAB}$; in particular, $O\in CXD$. Furthermore, $AL^2 = AL'^2 = AC^2 = AK\cdot AK'$, so $BLXL'$ is the polar of $A$ (with respect to $\omega$) and $AKXK'$ is the polar of $B$. Hence $\triangle{XAB}$ is self-polar. Since $D$ lies on the polar of $X$ and $O,X,D$ are collinear, $D$ is the inverse of $X = KK'\cap LL'$. It follows from the standard Yufei configuration that $D$ is the spiral center taking $LK$ to $K'L'$ (and $LK'$ to $KL'$), and $T = LK'\cap KL'$ (note that $DK'L'T$ is also cyclic, and that $LK'\cap KL'$ lies on the polar of $X$). To complete the diagram, let $T' = LK\cap K'L'$ so $\triangle{XTT'}$ is self-polar by Brokard's theorem (by duality, we should have $CT'$ the external bisector of $\angle{BCA}$). But $-1 = K(K,K';L,L') = K(B,A;T',T)$, so it suffices to show $\angle{TCT'} = 90^\circ$. However, $O$ has fixed power $OX\cdot OD$ with respect to any polar triangle $\triangle{XYZ}$ through $X$, and thus lies on the radical axis of the circles with diameters $AB,TT'$. Since line $ABTT'$ is perpendicular to $OCD$, we conclude that $\angle{TCT'} = 90^\circ$, as desired. (Since $L\in (BX)$, $K\in (AX)$, we have $T'\notin (BA)\implies T\in (BA)$.) Comment. After the first two paragraphs, it's straightforward to complex bash with unit circle $\omega$ ($(k+k')(l+l') = 2(kk'+ll')$ or $(k+k'-l-l')^2 = (k-k')^2 + (l-l')^2$ from $KLK'L'$ harmonic), with the observation that $c,2d-c$ are inverses with respect to $\omega$: we get $1-\frac{c}{d} = \pm\frac{(k-k')(l-l')}{2(kk'-ll')}$. Another finish is to look at inversions at $A$, $B$, $T'$ and $T$ fixing $\omega$. The fourth is the composition of the first three, so it fixes $C$. This gives $\angle TCT'=90^{\circ}$.