A-angle bisector cuts $DE$ at $T$ and the circumcircle $(O)$ of $\triangle ABC$ again at $F.$ Since $KL \parallel BC,$ then arcs $BK,CL$ are equal $\Longrightarrow$ $AT$ also bisects $\angle KAL.$ $P$ lies on the ray $AK$ such that $AP=AL.$ Since $AT$ is self isogonal with respect to $\angle KAL,$ it follows that $AK \cdot AL=AT \cdot AF$ $\Longrightarrow$ $AK \cdot AP=AT \cdot AF$ $\Longrightarrow$ inversion with center $A$ and power $AK \cdot AP$ takes $PT$ into $(O)$ and $\ell \equiv DE$ into the circle $\odot(APF),$ congruent to $(O).$ Now, due to conformity, $(I_1) \equiv \gamma_1$ is taken into the circle $(U_1)$ tangent to $AD,PT$ and internally tangent to $\odot(APF).$ Consequently, $(I_2) \equiv \gamma_2 \cong (U_1)$ $\Longrightarrow$ $AI_1 \equiv AU_1,AI_2$ are isogonals with respect to $\angle BAC.$ If $(I_1),(I_2)$ touch $AB,AC$ at $M,N,$ then the right triangles $\triangle AMI_1$ and $\triangle ANI_2$ are similar. Let line $I_1I_2$ cut $AF$ at $S.$ By angle bisector theorem we have then $\frac{_{SI_1}}{^{SI_2}}=\frac{_{AI_1}}{^{AI_2}}=\frac{_{I_1M}}{^{I_2N}}$ $\Longrightarrow$ $S$ is the insimilicenter of $(I_1) \sim (I_2),$ i.e common internal tangents of $(I_1),(I_2)$ intersect on the A-internal bisector. Remark: Likewise, common external tangents of $(I_1),(I_2)$ intersect on the A-external bisector.

For people who are not tricky enough to find solutions like the one above.

dnkywin: Shouldn't $1-\cos(2x)=2\sin^2(x)$? Well, it doesn't seem like a real issue anyway. In the same spirit, but with absolutely zero ingenuity... (maybe I should've thought of this during the test darn). Let $\triangle{L'MN}$ be the medial triangle of $\triangle{ABC}$, with $R_1=(I_1)\cap(O)$ and $S_1=(I_1)\cap AB$ (define $R_2,S_2$ similarly). Using directed angles modulo $180^\circ$, define $\angle{L'OR_i}=\theta_i$. Then $\angle{OI_1S_1}=\angle{I_1ON}=-(B+\theta_1)$, so by a simple trigonometric calculation, we have \[r_1=R\frac{\cos C+\cos(B+\theta_1)}{\cos(B+\theta_1)-1}\implies R-r_1=R\frac{1+\cos C}{1-\cos(B+\theta_1)},\]so \[AS_1=R\sin C+(R-r_1)\sin(B+\theta_1)=R\frac{\sin C+\sin(B+\theta_1)+\sin(B+\theta_1-C)}{1-\cos(B+\theta_1)}\]and \[\frac{r_1}{AS_1}=-\frac{2\cos\frac{C}{2}\sin\frac{B+\theta_1}{2}}{\cos\frac{B+C+\theta_1}{2}},\]with (distance defined up to some constant sign) \begin{align*} d(O,KL) &=r_1+(R-r_1)\sin(\theta_1-90^\circ)\\ &=R\frac{\cos C+\cos(B+\theta_1)+\cos\theta_1+\cos C\cos\theta_1}{\cos(B+\theta_1)-1}\\ &=R+R\frac{(1+\cos C)(1+\cos\theta_1)}{\cos(B+\theta_1)-1}\\ &=R-2R\frac{\cos^2\frac{C}{2}\cos^2\frac{\theta_1}{2}}{\sin^2\frac{B+\theta_1}{2}}. \end{align*}By symmetry, \[\frac{\cos^2\frac{C}{2}\cos^2\frac{\theta_1}{2}}{\sin^2\frac{B+\theta_1}{2}}=\frac{\cos^2\frac{B}{2}\cos^2\frac{\theta_2}{2}}{\sin^2\frac{C+\theta_2}{2}}\implies \frac{\cos\frac{C}{2}\cos\frac{\theta_1}{2}}{\sin\frac{B+\theta_1}{2}}=\frac{\cos\frac{B}{2}\cos\frac{\theta_2}{2}}{\sin\frac{C+\theta_2}{2}}\](it's not difficult to see each individual thing is positive by the definitions of the angles). As in the previous two solutions, it suffices to show that $r_1/AS_1=r_2/AS_2$, or \[\frac{\cos\frac{C}{2}\sin\frac{B+\theta_1}{2}}{\cos\frac{B+C+\theta_1}{2}}=\frac{\cos\frac{B}{2}\sin\frac{C+\theta_2}{2}}{\cos\frac{B+C+\theta_2}{2}}.\]Let $x=B/2$, $y=C/2$, $z=\theta_1/2$, and $w=\theta_2/2$. We can easily verify that \begin{align*} &\cos y\cos z\sin(y+w)-\cos x\cos w\sin(x+z)\\ &=\cos y\sin(x+z)\cos(x+y+w)-\cos x\sin(y+w)\cos(x+y+z), \end{align*}so we're done. Edit: Accidentally used $L$ twice.

Here's another problem which uses the inversion idea used to solve this (i.e., inverting and invoking symmetry): http://www.artofproblemsolving.com/Forum/viewtopic.php?t=318917.

Let $f$ be the transformation consisting of an inversion about $A$ with radius $\sqrt{AK*AL}$ followed by a reflection about $l$, the common angle bisector of $\angle BAC$ and $\angle KAL$. Then $f(KL)=w, f(AB)=AC, f(AC)=AB$, and so $f(\gamma_1)=\gamma_2$ and visa versa. Let $N$ be the intermal center of similarity of $\gamma_1,\gamma_2$ and for notation's sake, define $X'=f(X)$ for any point $X$. Let $R,S,T,U$ be the points of tangency of the two common inner tangents of $\gamma_1,\gamma_2$. Then $(AR'T')$ is internally tangent to $\gamma_2$ and externally to $\gamma_1$ at $R,'T'$ respectively, so $R',T',N$ are collinear, as are $S',U',N$. This means that $N'$ is the second intersection of $(ART)$ and $(ASU)$. But clearly $N$ is on the radical axis of $(ART)$ and $(ASU)$, so $A,N,N'$ are collinear, which means that $N$ lies on $l$.

Let $X$ be the point of the common inner tangents to $\gamma_1$ and $\gamma_2$. $\gamma_1$ (center $O_1$ and radius $r_1$) touch $KL$ at $F$ and $\omega$ at $G$, $\gamma_2$ (center $O_2$ and radius $r_2$) touch $KL$ at $P$ and $\omega$ at $Q$, if $M \equiv PQ \cap FG$, then $M$ is the midpoint of the arc $KAL$. Now $KC \cap BL \equiv Y$, since $KY=LY$ we get the circle tangent to the segments $KY,LY$ and $\omega$ is tangent to $\omega$ at $M$. Using A concyclic problem we obtain $A,M,P,F$ are concyclic, Now $AM, KL, GQ$ are concurrent at $Z$, from Radical Axis Theorem $\odot (GFPQ), \odot (AMPF), \omega$, The parallel to $AB$ and $BC$ cut $AM$ at $B'$ and $C'$ respectively. Since $\angle O_1B'A=\angle O_2C'A \Rightarrow \frac{r_1}{r_2}=\frac{B'A}{C'A}$ and $\frac{B'O_1}{C'O_2}=\frac{ZO_1}{ZO_2}=\frac{r_1}{r_2} \Rightarrow \triangle O_1B'A \sim \triangle O_2C'A$ using $(Z,X,O_1,O_2)=-1 \Rightarrow AX$ is bisector angle of $\angle O_1AO_2 \Rightarrow AM$ is bisector angle of $\angle BAC$

Cleaner proof (possibly similar to other solutions here). mavropnevma wrote: A triangle $ABC$ is inscribed in a circle $\omega$. A variable line $\ell$ chosen parallel to $BC$ meets segments $AB$, $AC$ at points $D$, $E$ respectively, and meets $\omega$ at points $K$, $L$ (where $D$ lies between $K$ and $E$). Circle $\gamma_1$ is tangent to the segments $KD$ and $BD$ and also tangent to $\omega$, while circle $\gamma_2$ is tangent to the segments $LE$ and $CE$ and also tangent to $\omega$. Determine the locus, as $\ell$ varies, of the meeting point of the common inner tangents to $\gamma_1$ and $\gamma_2$. (Russia) Vasily Mokin and Fedor Ivlev Let $X_1, Y_1$ and $I_1$, be the touch-points of $\gamma_1$ with $\omega, KL$ and the center of $\gamma_1$, respectively. Likewise, define $X_2, Y_2$ and $I_2$. Let $M$ be the mid-point of arc $BAC$ and $T$ be the insimilicentre of $\gamma_1$ and $\gamma_2$. We claim that $T$ lies on the angle bisector $\overline{AN}$ in $\triangle ABC$. Notice that Inversion at $A$ of radius $\sqrt{AK \cdot AC}$ plus reflection along line $AM$ maps $\gamma_1$ to $\gamma_2$; so $\overline{AI_1}, \overline{AI_2}$ are isogonal in angle $BAC$. By Monge's Theorem on $\gamma_1, \gamma_2, \omega$, we see $\overline{X_1X_2}$ passes through $Z$. Now remark that $(AI_1, AI_2; AM, AN)=-1$ and $(I_1, I_2, Z, T)=-1$ so $T$ lies on $\overline{AN}$, as desired. $\blacksquare$

Let $\delta (F,m)$ denote the unsigned distance from point $F$ to line $m$. Let $X,Y$ be the tangency points of $\gamma_1, \gamma_2$ respectively with line $DE$. Let $S,T$ be the tangency points of $\gamma_1, \gamma_2$ with lines $AB,AC$ respectively. Let $P,Q$ be the centers of $\gamma_1, \gamma_2$ respectively, and let $p,q$ be the radii of $\gamma_1, \gamma_2$ respectively. Let $I$ be the incenter of $\triangle ABC$. Let $N$ be the intersection of the common internal tangents to $\gamma_1$ and $\gamma_2$. I claim that the locus of $N$ is the segment of the internal angle bisector of $\angle BAC$ that lies within $\triangle ABC$. First, we will prove that $N$ always lies on the angle bisector of $\angle BAC$. Since $N = \frac{q}{p+q}P + \frac{p}{p+q}Q$, it suffices to show that $\frac{\delta (P, AI)}{p} = \frac{\delta (Q,AI)}{q}$. This is equivalent to $\frac{p}{AS} = \frac{q}{AT}$. Let $I_1, I_2$ be the incenters of triangles $BKL, CKL$ respectively. It's well known that lines $XS, YT$ pass through $I_1, I_2$ respectively. Furthermore, since $KL \parallel BC$, we have $I_1I_2 \parallel BC$ and $BI_1 = CI_2$. Let $y = BI_1 = CI_2$. Let $\alpha, \beta, \gamma$ denote angles $\frac{A}{2}, \frac{B}{2}, \frac{C}{2}$ respectively. It's clear that $BS = y \cdot \frac{\sin (\alpha + \beta)}{\sin (\alpha + \gamma)}$ by Law of Sines in triangle $BSI_1$. Similarly $CT = y \cdot \frac{\sin (\alpha + \gamma)}{\sin (\alpha + \beta)}$. Note that $P$ lies on a fixed parabola with respect to triangle $ABC$. We deduce that \[p = BS \cdot AS \cdot \frac{\tan \gamma}{c}.\]Hence \[\frac{p}{AS} = BS \cdot \frac{\tan \gamma}{c} = y \cdot \frac{\sin (\alpha + \beta)}{\sin (\alpha + \gamma)} \cdot \frac{\tan \gamma}{c} = y \cdot \frac{\cos \gamma}{\cos \beta} \cdot \frac{\tan \gamma}{c}.\]It follows immediately that \[\frac{p}{AS} = \frac{q}{AT},\]as desired. Now we will prove that we can choose $\ell$ such that $N$ is any point on the angle bisector of $\angle BAC$ and inside the triangle $ABC$. Simply note that $N$ is a continuous function of the position of $\ell$. If we choose $\ell$ close to $BC$, then $N$ approaches $BC$ as well. If we choose $\ell$ to pass through $A$, we see that $N$ is at $A$. Thus $N$ traces the entire segment of the angle bisector inside the triangle. $\blacksquare$

We need AI1,AI2 isogonal. Let I1J,I1I perp DB,DK, F,G,H midarc BC,CA,AB. by Sawayanma incenter X lies on IJ. Now I1J/R.JA=IJ/IG.JA=JB/IG.JG=JB/GA^2 thus we need TC/JB=(AH/AG)^2. From JA/JB=XA/XB.sinAYL/sin(90+K/2-AYL)=cos(ACL/2)/sin(180-C/2-90-ACL/2) and TC/TA=YC/YA.sinTYC/sinAXK=sin(180-B/2-90-ABK/2)/cosABK/2 and AT/AJ=AT/AY.AY/AX.AX/AJ=AN/AG.AY/AX.AH/AM=AH/AG.AY/AM.AN/AX=sinAXK/AG.AH/sinAYL.AY/AX=AK/AX.AY/AL.AY/AX so TC/JB=AY/AX.AK/AL.YC/XB.cosACL/2:cosABK/2=sinABX:AX.AY:sinACY.YC/XB=(YC/XB)^2 so we left to YC/XB=AH/AG or YC/CA.CA/AB.AB/XB=sinC/2:sinB/2.AC/AB=AH/AG, done!

really nice problem. [asy][asy] size(10cm); defaultpen(fontsize(10pt)); pen pri=blue; pen sec=purple; pen tri=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair O,T1,U1,U2,T2,Q,S,M,M1,M2,O1,O2,X,Y,W,Z; O=(0,0); T1=dir(160); U1=dir(130); U2=dir(50); T2=dir(350); Q=intersectionpoint(T1--(T1+(U1-T1)*100),T2--(T2+(U2-T2)*100)); S=intersectionpoint(U2--(U2+(U1-U2)*100),T2--(T2+(T1-T2)*100)); M=foot(O,Q,S); M1=(U1+T1)/2; M2=(U2+T2)/2; W=circumcenter(Q,T1,T2); O1=intersectionpoint(T1--W,O--M1); O2=intersectionpoint(T2--W,O--M2); X=intersectionpoint(T1--U2,T2--U1); Y=intersectionpoint(O--X,O1--O2); Z=foot(X,O,S); draw(T1--U1--U2--T2--T1,pri); draw(U1--S,pri); draw(T1--S,pri); draw(U1--Q,pri); draw(U2--Q,pri); draw(Q--S,pri); draw(O--M1,sec); draw(O--M2,sec); draw(O1--T1,sec); draw(O1--U1,sec); draw(O2--T2,sec); draw(O2--U2,sec); draw(O--M,sec); draw(S--O2,sec); draw(U1--T2,pri); draw(U2--T1,pri); draw(O--S,sec); draw(Q--Z,sec); filldraw(circle(O,1),fil,pri); filldraw(circumcircle(Q,M,O),sfil,sec); filldraw(circle(O1,distance(O1,T1)),tfil,tri); filldraw(circle(O2,distance(O2,T2)),tfil,tri); label("$O$",O,dir(270)); label("$T_1$",T1,dir(210)); label("$U_1$",U1,dir(120)); label("$T_2$",T2,dir(330)); label("$U_2$",U2,dir(60)); label("$Q$",Q,dir(90)); label("$S$",S,dir(180)); label("$M$",M,dir(120)); label("$M_1$",M1,dir(140)); label("$M_2$",M2,dir(0)); label("$O_1$",O1,dir(270)); label("$O_2$",O2,dir(270)); label("$X$",X,dir(60)); label("$Y$",Y,dir(270)); [/asy][/asy] $O_1, O_2$ = centers of $\gamma_1, \gamma_2$; $U_1, U_2$ = tangency points on $KL$; $T_1, T_2$ = tangency points on $(ABC)$ $Q$ = midpoint of $\widehat{BAC}$ $M_1, M_2$ = midpoints of $T_1U_1, T_2U_2$ $S = U_1U_2 \cap O_1O_2$ $X = U_1T_2 \cap U_2T_1$ We claim that the locus is the bisector of $\angle{BAC}$. Claim 1: $Q = T_1U_1 \cap T_2U_2$, and $T_1U_1U_2T_2$ is cyclic. The former is true by homothety at $T_1, T_2$ taking $\gamma_1, \gamma_2$ to $(ABC)$, and the latter because $\triangle{QU_1K} \sim \triangle{QKT_1} \implies QU_1 \cdot QT_1 = QK^2 = QL^2 = QU_2 \cdot QT_2$. Claim 2: $MY, MS$ are internal and external bisectors of $\angle{O_1MO_2}$. First, by Monge on $\gamma_1, \gamma_2, (ABC)$, $S, T_1, T_2$ are collinear, thus $S$ is the exsimilicenter and if $Y$ is the insimilicenter then $S, O_1, Y, O_2$ is harmonic. By Brocard, $O$ is the orthocenter of $\triangle{XSQ}$, and if $M$ is the foot of the altitude from $O$ to $QS$ then $S, T_1, QX \cap T_1T_2, T_2$ is harmonic. Projecting from $Q$ onto $(OQ)$, $M, M_1, QX \cap (OQ), M_2$ is harmonic, and finally projecting from $O$ onto $O_1O_2$, we get that $O, Y, X, M$ are collinear, and since $\angle{SMY} = 90^{\circ}$, $MY$ bisects $\angle{O_1MO_2}$. Clearly $MY$ passes through the antipode of $Q$ wrt $(ABC) \implies MO_1, MO_2 \cap (ABC)$ lie on a line parallel to $BC$, and $\frac{MO_1}{MO_2} = \frac{O_1U_1}{O_2U_2} \implies$ the tangents from $M$ to $\gamma_1, \gamma_2$ intersect $(ABC)$ at points which are reflections over the perp bisector of $BC \implies M = A$, and the insimilicenter $Y$ does indeed lie on the $A$-bisector. $\square$

I claim that the insimillicenter of $\gamma_1$ and $\gamma_2$ lies on the $A$-angle bisector. Let $\gamma_1$ meet $\omega$ at $T_1$ and $\ell$ at $P_1$, and let $\gamma_2$ meet $\omega$ at $T_2$ and $\ell$ at $P_2$. Let $Z$ be the arc midpoint of $\widehat{BAC}$. Then $T_1-P_1-Z$ and $T_2-P_2-Z$. Invert around $A$ with radius $\sqrt{AD\cdot AC}=\sqrt{AE\cdot AB}$ and reflect around the angle bisector. Then $\ell$ and $\omega$ swap, so $\gamma_1$ goes to $\gamma_2$. This means $T_1$ goes to $P_2$ and $P_1$ goes to $T_2$, so $\triangle T_1AP_1\sim \triangle T_2AP_2$. By spiral similarity, $A,L,P_1,P_2$ are cyclic. Uninverting means that $\overline{P_1P_2}\cap\overline{T_1T_2}$ is the image of $L$ under the inversion+reflection, so $\overline{P_1P_2}\cap\overline{T_1T_2}$ is on the $A$-external bisector. By Monge, this point is the exsimillicenter of $\gamma_1$ and $\gamma_2$. Let $K$ be the intersection of the $A$-angle bisector and $\overline{BC}$. The centers of $\gamma_1,\gamma_2$, which can be called $O_1,O_2$, satisfy $\angle O_1AK=\angle KAO_2$ because $\gamma_1,\gamma_2$ swap under the inversion. $A(L,K,O_1,O_2)=-1$ so the insimillicenter of $\gamma_1,\gamma_2$ lies on $\overline{AK}$, as desired.

[asy][asy] unitsize(1cm); pair A, B, C, D, E, O1, O2, L; A=(2,3sqrt(5)); B=(0,0); C=(8,0); D=(A+B)/2; E=(A+C)/2; pair O=circumcenter(A,B,C); pair A1=intersectionpoints(circumcircle(A,B,C),O--5*O-2*(B+C))[0]; pair B1=intersectionpoints(circumcircle(A,B,C),O--5*O-2*(C+A))[0]; pair C1=intersectionpoints(circumcircle(A,B,C),O--5*O-2*(A+B))[0]; pair F=extension(A1,A1+(A1-O)*(0,1),C1,C1+(C1-O)*(0,1)); pair G=extension(A1,A1+(A1-O)*(0,1),B1,B1+(B1-O)*(0,1)); pair X1=intersectionpoints(circumcircle(A,B,C),D--3*D-2*F)[0]; pair Y1=intersectionpoints(circumcircle(A,B,C),E--3*E-2*G)[0]; pair X2=extension(D,E,A1,X1); pair Y2=extension(D,E,A1,Y1); pair X=extension(A,B,C1,X1); pair Y=extension(A,C,B1,Y1); draw(A--B--C--A); draw(intersectionpoints(circumcircle(A,B,C),2*D-E--2*E-D)[0]--intersectionpoints(circumcircle(A,B,C),2*D-E--2*E-D)[1]); draw(circumcircle(A,B,C)); draw(circumcircle(X,X1,X2)); draw(circumcircle(Y,Y1,Y2)); O1=circumcenter(X,X1,X2); O2=circumcenter(Y,Y1,Y2); draw(O1--A--O2--O1); L=extension(A,2*O-A1,B,C); draw(A--L); label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, SE); label("$E$", E, SW); label("$O_1$", O1, W); label("$O_2$", O2, dir(0)); label("$L$", L, SW); [/asy][/asy] Let $O_1$ and $O_2$ be the centers of $\gamma_1$ and $\gamma_2$, respectively, and let the radii of $\gamma_1$ and $\gamma_2$ be $r_1$ and $r_2$, respectively. The inversion centered at $A$ with radius $\sqrt{AD\cdot AC}=\sqrt{AE\cdot AB}$ followed by a reflection over the angle bisector of $\angle BAC$ maps $\gamma_1$ to $\gamma_2$, so $AO_1$ and $AO_2$ are isogonal. This implies $\frac{AO_1}{AO_2}=\frac{r_1}{r_2}$, and the angle bisector of $\angle BAC$ is the angle bisector of $\angle O_1AO_2$. Therefore, the angle bisector of $\angle BAC$ splits $O_1O_2$ into the ratio $AO_1:AO_2=r_1:r_2$, so the insimilicenter of $\gamma_1$ and $\gamma_2$ lies on the angle bisetor of $\angle BAC$. Let the angle bisector of $\angle BAC$ intersect $BC$ at $L$. When $\gamma_1$ and $\gamma_2$ approach $B$ and $C$ respectively, the insimilicenter approaches $L$. When one of $\gamma_1$ or $\gamma_2$ approaches $A$, the insimilicenter approaches $A$. By continuity, every point between those two points is achievable. Therefore, the locus is the segment $AL$ not including $A$ or $L$.