Inversion centered at $E$ $+$ easy similarities/trig.

Could you elaborate that please, I attempted that but inevitably failed. @below, there are $8$(!) possible configurations without accounting for the changes in radii.

TheDarkPrince wrote: Inversion centered at $E$ $+$ easy similarities/trig. Dubious? I think you get the Curvilinear Configuration and that is probably harder than the problem itself. TheDarkPrince wrote: Let $E$ be a common point of circles $\omega _1$ and $\omega _2$. Let $AB$ be a common tangent to these circles, and $CD$ be a line parallel to $AB$, such that $A$ and $C$ lie on $\omega _1$, $B$ and $D$ lie on $\omega _2$. The circles $ABE$ and $CDE$ meet for the second time at point $F$. Prove that $F$ bisects one of arcs $CD$ of circle $CDE$. Consider the configuration where angle $AEB$ is contained within angle $CED$. All others follow similarly (except for a minor plot twist). Now let $\odot(EAB)$ meet $EC, ED$ at points $P,Q$ respectively. Then we show $CP=DQ$ which will then prove $M$; the midpoint of arc $CED$ to be the rotation center mapping $CP$ to $DQ$ and the claim will follow. Now observe that $\angle EPB=\angle EAB=\angle ECA$ hence $BP \parallel CA$. Thus, $$\frac{CP}{AB}=\frac{\sin AEC}{\sin ACE}=\frac{AC}{AE}$$and similarly for the other length. Thus, $$\frac{CP}{DQ}=\frac{EB}{EA} \cdot \frac{AC}{BD}$$Now let $CD$ meet $EA, EB$ at $X,Y$ respectively. Since $\angle XCA=\angle CEX$ we conclude that $AC^2=AX \cdot AE$. Now we get $$\frac{AC^2}{BD^2}=\frac{AX \cdot AE}{BY \cdot BE}=\frac{EA^2}{EB^2}$$hence $CP=DQ$ as desired. Plot Twist: We actually need to use directed lengths because if $M$ is the midpoint of minor arc then we want exactly one of $P$ or $Q$ on their respective segment! @2below, it's correct (I think?) though as I said, it's harder than the problem itself

Invert at $E$ and use Pascal and Sawayama's lemma (or its extraversion).

anantmudgal09 wrote: TheDarkPrince wrote: Inversion centered at $E$ $+$ easy similarities/trig. Dubious? I think you get the Curvilinear Configuration and that is probably harder than the problem itself. Probably done a mistake [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 = -11.028, xmax = 15.812, ymin = -7.806, ymax = 7.132; /* image dimensions */ pair A = (-2.5,1.936491673103709), B = (1.25,2.904737509655563), C = (-3.0687169584707705,-1.69051591612649), D = (4.975016055061397,0.38636701739010193), F = (0.31789784934139714,1.808244801330289); /* draw figures */ draw(circle((-2.,0.), 2.), linewidth(1.)); draw(circle((2.,0.), 3.), linewidth(1.)); draw(A--B, linewidth(1.)); draw((-0.625,1.4523687548277813)--B, linewidth(1.)); draw((-0.625,1.4523687548277813)--A, linewidth(1.)); draw(C--D, linewidth(1.)); draw((-0.625,1.4523687548277813)--C, linewidth(1.)); draw((-0.625,1.4523687548277813)--D, linewidth(1.)); draw(circle((1.4842901654479481,-2.7091732140946783), 4.665569264426202), linewidth(1.)); draw(circle((-1.,3.8729833462074166), 2.449489742783178), linewidth(1.)); /* dots and labels */ dot((-0.625,1.4523687548277813),linewidth(4.pt) + dotstyle); label("$E$", (-0.85,0.7), NE * labelscalefactor); dot(A,linewidth(4.pt) + dotstyle); label("$A$", (-3.4,1.874), NE * labelscalefactor); dot(B,linewidth(4.pt) + dotstyle); label("$B$", (1.446,3.084), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (-3.7,-2.108), NE * labelscalefactor); dot(D,linewidth(4.pt) + dotstyle); label("$D$", (5.054,0.554), NE * labelscalefactor); dot(F,linewidth(4.pt) + dotstyle); label("$F$", (0.5,1.3), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Consider an inversion centered at $E$ with arbitrary radius $r$. Suppose $A,B,C,D,F$ goes to $A^*,B^*,C^*,D^*,F^*$ after inversion. Clearly $AB$ maps to circle $(A^*EB^*)$, $CD$ maps to circle $(C^*ED^*)$. Also $\omega_1$ maps to $A^*C^*$ and $\omega_2$ maps to $B^*D^*$. As $AB$ is tangent to $\omega_1$, $A^*C^*$ is tangent to $(A^*EB^*)$. Similarly $B^*D^*$ is tangent to $(A^*EB^*)$. As $AB||CD$, $AB$ and $CD$ do not intersect, so $(A^*EB^*)$ is tangent to $(C^*ED^*)$ at $E$. We need to show \[CF=DF \implies \frac{r^2}{EC^*\cdot EF^*}\cdot C^*F^*=\frac{r^2}{ED^*\cdot EF^*}\cdot D^*F^*\]or \[\frac{C^*F^*}{EC^*}=\frac{D^*F^*}{ED^*}.\] We can now rephrase the question to: Consider circles $\tau_1$ and $\tau_2$ centered at $O_1$ and $O_2$ tangent to each other at $E$. Points $A^*$ and $B^*$ lie on $\tau_1$. Tangents at $A^*$ and $B^*$ at $\tau_1$ intersect $\tau_2$ at $C^*$ and $D^*$ respectively. $A^*B^*$ and $C^*D^*$ intersect at $F^*$. Prove that \[\frac{C^*F^*}{EC^*}=\frac{D^*F^*}{ED^*}.\] We'll try to solve this problem. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(8.cm); 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 = -10.96, xmax = 13.42, ymin = -4.22, ymax = 9.3; /* image dimensions */ pair A = (-2.88,3.32), B = (4.,3.), O_1 = (0.6331914893617019,4.733617021276596), O_2 = (-0.38965630114566263,-1.2976104746317514), C = (-1.7757861831953674,0.5757527779834299), D = (1.925262573843098,-1.0292887863901314), F = (-8.731270215898373,3.5921521030650414); /* draw figures */ draw(circle(O_1, 3.7869285078763255), linewidth(1.)); draw(circle(O_2, 2.3304175433085077), linewidth(1.)); draw(F--A, linewidth(1.)); draw(F--B, linewidth(1.)); draw(F--C, linewidth(1.)); draw(F--D, linewidth(1.)); draw(C--(0.,1.), linewidth(1.)); draw((0.,1.)--D, linewidth(1.)); draw((0.,1.)--A, linewidth(1.)); draw((0.,1.)--B, linewidth(1.)); draw(F--(0.,1.), linewidth(1.)); draw(O_2--O_1, linewidth(1.)); draw(C--A, linewidth(1.)); draw(D--B, linewidth(1.)); /* dots and labels */ dot(A,dotstyle); label("$A*$", (-4,3.48), NE * labelscalefactor); dot(B,dotstyle); label("$B*$", (4.16,2.98), NE * labelscalefactor); dot((0.,1.),dotstyle); label("$E$", (-0.6,0.3), NE * labelscalefactor); dot(O_1,linewidth(4.pt) + dotstyle); label("$O_1$", (0.72,4.9), NE * labelscalefactor); dot(O_2,dotstyle); label("$O_2$", (-0.74,-2.2), NE * labelscalefactor); dot(C,linewidth(4.pt) + dotstyle); label("$C*$", (-3,0.1), NE * labelscalefactor); dot(D,linewidth(4.pt) + dotstyle); label("$D*$", (2.06,-1.12), NE * labelscalefactor); dot(F,linewidth(4.pt) + dotstyle); label("$F*$", (-9.8,3.56), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $\angle A^*B^*E=\alpha, \angle B^*A^*E=\beta, \angle D^*EO_2=\delta, \angle C^*EO_2=\theta$. Extend $EO_2$ to meet $\tau_2$ at $G$, so $EG$ is the diameter. So $\angle EC^*G=90^{\circ}\implies \angle C^*GE=90^{\circ}-\theta$. So $\angle ED^*C^*=90^{\circ}-\theta$ and similarly $\angle EC^*D^*=90^{\circ}-\delta$. Firstly, we look to simplify the condition to prove. From law of sines in $\triangle F^*EC^*$ and $\triangle F^*ED^*$, \[\frac{C^*F^*}{EC^*}=\frac{D^*F^*}{ED^*}\Leftrightarrow \frac{\sin \angle C^*EF^*}{\sin \angle EFC^*}=\frac{\sin \angle D^*EF^*}{\sin \angle EFD^*}\Leftrightarrow \sin \angle C^*EF^*=\sin \angle D^*EF^*.\] Now, again from law of sines in $\triangle F^*EC^*$ and $\triangle F^*ED^*$, \[\sin \angle C^*EF^*=\sin \angle D^*EF^*\Leftrightarrow \frac{\cos \delta}{F^*E\cdot F^*D^*} =\frac{\sin \angle F^*EC^*}{F^*C^*\cdot F^*D^*}=\frac{\cos \theta}{F^*E\cdot F^*C^*}\Leftrightarrow \frac{\cos \delta}{F^*D^*}=\frac{\cos \theta}{F^*C^*}.\] From law of sines in triangles $F^*D^*B^*$ and $F^*C^*A^*$, \[\frac{\cos \delta}{F^*D^*}=\frac{\cos \theta}{F^*C^*}\Leftrightarrow \frac{\cos \delta}{\cos \theta}=\frac{F^*D^*}{F^*C^*}=\frac{F^*D^*}{B^*D^*}\cdot \frac{A^*C^*}{F^*C^*}\cdot \frac{B^*D^*}{A^*C^*}=\frac{\sin (\alpha +\beta)}{\sin \angle B^*F^*D^*}\cdot \frac{\sin \angle A^*F^*C^*}{\sin (\alpha +\beta)} \cdot \frac{B^*D^*}{A^*C^*}=\frac{B^*D^*}{A^*C^*}.\] Now from law of sines in triangle $\triangle EC^*D^*$, \[\frac{B^*D^*}{A^*C^*}=\frac{\cos \delta}{\cos \theta}= \frac{ED^*}{EC^*}.\] Now from law of sines in triangles $A^*EC^*$ and $B^*ED^*$, \[\frac{B^*D^*}{A^*C^*}= \frac{ED^*}{EC^*}\Leftrightarrow \frac{\cos (\theta - \alpha)}{\sin \alpha}=\frac{\sin (90^{\circ}+\alpha-\theta)}{\sin \alpha}=\frac{A^*C^*}{EC^*}=\frac{B^*D^*}{ED^*}=\frac{\sin (90^{\circ}+\beta - \delta)}{\sin \beta}=\frac{\cos (\delta - \beta)}{\sin \beta}.\]So we need to prove the condition: \[\cos(\theta -\alpha)\sin \beta = \cos(\delta - \beta)\sin \alpha.\]As $\alpha,\beta, \delta, \theta <90^{\circ}$, we can also prove the square of the condition, i.e \[\cos(\theta -\alpha)\sin \beta = \cos(\delta - \beta)\sin \alpha\Leftrightarrow \cos^2(\theta -\alpha)\sin^2 \beta = \cos^2(\delta - \beta)\sin^2 \alpha.\] From tangent chord theorem, $\angle O_1EA^*=90^{\circ}-\alpha \implies \angle A^*EC^* =90^{\circ}+\alpha - \theta$. From law of sines in triangle $A^*EC^*$, \[\frac{A^*E}{\cos(\theta -2\alpha)}=\frac{A^*E}{\sin (90^{\circ}+2\alpha -\theta)}=\frac{C^*E}{\sin \alpha}=\frac{2EO_2\cos \theta}{\sin \alpha}.\] $\Rightarrow \frac{A^*E\sin \alpha}{2EO_2}=\cos \theta \cos (\theta -2\alpha)=\frac{1}{2}\left(\cos (2\alpha)+\cos (2\theta -2\alpha)\right) \implies \frac{A^*E\sin \alpha}{EO_2}=\cos (2\alpha)+\cos (2\theta -2\alpha)$. $\Rightarrow \frac{A^*E\sin \alpha}{EO_2}=2\cos^2(\theta - \alpha)-1+\cos (2\alpha) = 2\cos^2(\theta - \alpha) - 2\sin ^2(\alpha) \implies \frac{A^*E}{2EO_2\sin \alpha} + 1 =\frac{\cos (\theta - \alpha)}{\sin ^2 \alpha}.$ Similarly we get $\frac{B^*E}{2EO_2\sin \beta} + 1 =\frac{\cos (\delta - \beta)}{\sin ^2 \beta}$. We need to prove \[\cos^2(\theta -\alpha)\sin^2 \beta = \cos^2(\delta - \beta)\sin^2 \alpha \Leftrightarrow \frac{\cos^2(\theta -\alpha)}{\sin^2 \alpha}=\frac{\cos^2 (\delta -\beta)}{\sin^2 \beta}.\] \[LHS = \frac{\cos^2(\theta -\alpha)}{\sin^2 \alpha} = \frac{A^*E}{2EO_2\sin \alpha} + 1\]and \[RHS = \frac{\cos ^2(\delta -\beta)}{\sin^2 \beta}= \frac{B^*E}{2EO_2\sin \beta} + 1.\] So we need to show \[\frac{A^*E}{2EO_2\sin \alpha} + 1=\frac{B^*E}{2EO_2\sin \beta} + 1\Leftrightarrow \frac{A^*E}{\sin \alpha}=\frac{B^*E}{\sin \beta}\]which is true from the law of sines in $\triangle A^*EB^*$. Proved.

it is helpful to observe that the point where AC and BF concur lies on (ECD) and so does the point where AF and BD concur. The two mentioned points along with A and B are concyclic. This is sufficient to come to the desired result with some angle tracing.

Let $P$ and $Q$ be the intersection of $\odot{(ABE)}$ with lines $AC$ and $BD$ respectively. Also let $R$ be the intersection of $AC$ and $BD$. Notice that $$\angle{PQD}=\angle{PAB}=\angle{PCD}.$$So the quadrilateral $PQCD$ is cyclic. Let $X$ be some point on $AB$ beyond $A$. Observe that \begin{align*} \angle{QRC} &=\angle{RAB}+\angle{RBA} \\ &=\angle{XAC}+\angle{QEA} \\ &=\angle{AEC}+\angle{QEA} \\ &=\angle{QEC}. \end{align*}Thus the quadrilateral $CQER$ is cyclic. Similarly the quadrilateral $DPER$ is also cyclic. Thus we have $\angle{REQ}=\angle{RCQ}=\angle{RDP}=\angle{REP}.$ This means that $$\angle{PBQ}=\angle{PEQ}=2\angle{PCQ}=2\angle{PDQ}.$$So $\triangle{PBD}$ is isosceles. Similarly $\triangle{QAC}$ is also isosceles. Let $F'$ be the circumcenter of the cyclic quadrilateral $PQCD$. Then $\angle{PF'Q}=2\angle{PCQ}=\angle{PAQ}$, so $F'\in\odot{(ABE)}$. Observe that \begin{align*} \angle{CF'D} &=2\angle{CPD} \\ &=2\angle{APB}+2\angle{BPD} \\ &=\angle{APB}+\angle{APB}+\angle{PBR} \\ &=\angle{APB}+\angle{ARB} \\ &=(180^{\circ}-\angle{AEB})+(180^{\circ}-\angle{RAB}-\angle{RBA}) \\ &=360^{\circ}-\angle{CEA}-\angle{AEB}-\angle{BED} \\ &=\angle{CED}. \end{align*}Thus we have $F'\in \odot{(CED)}.$ Therefore we have $F'=F\implies CF=DF$. This means that $F$ bisects the arc $CD$ of $\odot (CDE)$ as desired.

WizardMath wrote: Invert at $E$ and use Pascal and Sawayama's lemma (or its extraversion). Any reference for the above theorem's?

TheDarkPrince wrote: Consider circles $\tau_1$ and $\tau_2$ centered at $O_1$ and $O_2$ tangent to each other at $E$. Points $A^*$ and $B^*$ lie on $\tau_1$. Tangents at $A^*$ and $B^*$ at $\tau_1$ intersect $\tau_2$ at $C^*$ and $D^*$ respectively. $A^*B^*$ and $C^*D^*$ intersect at $F^*$. Prove that \[\frac{C^*F^*}{EC^*}=\frac{D^*F^*}{ED^*}.\] Pretty much the same as this problem here.