$ A,D,F$ and $ B,D,E$ are collinear.let $ EF$ intersect $ t$ and $ AB$ at $ P,Q$ in order. we know point $ D$ is on polar of $ Q$ and $ t$ is perpendicular to $ AB$ hence $ t$ is polar of $ Q$ so $ O(PQ,EF)=-1$ and cause $ AB\parallel O_{1}O_{2}$ hence $ OP$ bisect $ O_{1}O_{2}$ therefore $ P,O_{1},A$ and $ P,O_{2},B$ are collinear.

Amir.S wrote: $ OP$ bisect $ O_{1}O_{2}$ therefore $ P,O_{1},A$ and $ P,O_{2},B$ are collinear. I don't understand why ? I would finish the problem as follows : let $ G$ denote the point of intersection of $ AE$ with $ O_{1}O_{2}$, then $ O_{1}$ is the midpoint of segment $ DG$, so $ O(AD, AG ; AB, AO_{1}) =-1$. On the other hand, since P lies on the polar of Q, we have $ O(AF, AE ; AQ, AP) =-1$. Therefore $ O(AD, AE ; AB, AO_{1}) = O(AD, AE ; AB, AP)$, which means, because of unicity of the cross ratio, that $ A$, $ O_{1}$ and $ P$ are colinear, qed. In fact this problem is well known, it was a problem from Bulgaria in 1996 (see Kedlaya's Geometry Unbound, problem 7.2.9) Benjamin

tanx for mentioning. The rules of mathlinks don't let me to edit my post(I don't know why) so I send it in a new topic. let $ X$ be a point on $ OE$ that $ EX=\frac{r_{1}+r_{2}}{2}$ and M,N be midpoints of $ O_{1}O_{2}$ and $ OO_{1}$ in order. hence we have $ MN=XN=\frac{R-r_{2}}{2}$ so $ XM\parallel EF$ hence $ \frac{PM}{PO}=\frac{r_{1}+r_{2}}{2R}$ now we are done.

From $ O_{1}D\parallel OB$ and because of $ \frac{EO_{1}}{EO}= \frac{O_{1}D}{OB},$ we conclude that the points $ E,$ $ D,$ $ B,$ are collinear and similarly the points $ F,$ $ D,$ $ A,$are also collinear. We consider now, the triads of collinear points $ O_{1},$ $ D,$ $ O_{2}$ and $ B,$ $ O,$ $ A$ and based on the Pappus theorem, we conclude that the points $ E\equiv O_{1}O\cap BD,$ $ P\equiv O_{1}A\cap BO_{2}$ and $ F\equiv DA\cap OO_{2},$ are collinear. That is, the point $ P,$ lies on the segment line $ EF$ and it is enough to prove that also lies on the line $ (t).$ $ \bullet$ Because of the segment $ AB$ as a diameter of $ (O),$ we have that $ AE\perp BE,$ $ AF\perp BF$ and so, we conclude that the point $ D,$ is the orthocenter of the triangle $ \bigtriangleup QAB,$ where $ Q\equiv AE\cap BF.$ That is we have that $ QT\equiv (t),$ where $ T\equiv AB\cap QD$ and we will prove that the point $ P\equiv AO_{1}\cap BO_{2},$ lies on the segment line $ QT.$ It is enough to prove that $ \frac{O_{1}D}{O_{2}D}= \frac{AT}{BT}.$ $ ,(1)$ We denote as $ E',$ $ F',$ the intersection points of the circles $ (\omega_{1}),$ $ (\omega_{2})$ respectively, from the segment line $ O_{1}O_{2}$ and it is easy to show that $ EE'\equiv AQ$ $ ($ $ EE'\perp ED,$ $ AE\perp ED$ $ )$ and similarly $ FF'\equiv BQ.$ So, in the triangle $ \bigtriangleup QAB,$ from $ E'F'\parallel AB$ $ \Longrightarrow$ $ \frac{E'D}{F'D}= \frac{AT}{BT}$ $ \Longrightarrow$ $ (1).$ Hence, we have that the point $ P\equiv AO_{1}\cap BO_{2},$ lies on the segment line $ EF$ and also on the segment line $ QT\equiv (t).$ We conclude now, that the segment lines $ QT\equiv (t),$ $ EF,$ $ AO_{1},$ $ BO_{2},$ are concurrent at one point and the proof is completed. Kostas Vittas.

This problem was proposed by Onofre Campos, from Brazil.

let $ EF$ and $ AB$ intersect $ t$ at $ P$ and $ T$ respectively, and the radius of $ w$ ,$ w_1$ and $ w_2$ be $ R$ ,$ r_1$ and $ r_2$respectively. since $ O_{1}D\parallel OB$,and $ \frac{O_1D}{OB}=\frac{EO_1}{EO}=\frac{r_1}{R}$, $ E,D,B$ are collinear and similarly the points $ F, D, A$,are also collinear. let $ \angle DAB=\alpha, \angle DBA= \beta$, since $ E,A,B,F$ are concyclic , $ \triangle EDF \sim \triangle ADB$, $ \angle FED= \angle DAB=\alpha , \angle EFD= \angle DBA=\beta$, and since $ O_{1}O_2\parallel AB$, $ \angle DFO_2=\angle FDO_2=\angle O_1DA=\angle DAT=\alpha$, $ \angle DEO_1=\angle O_1DE= \angle O_2DB=\angle DBT=\beta$, consider ,$ \triangle OO_1O_2$, by sine law, $ \frac{OO_1}{\sin \angle O_1O_2O}=\frac{OO_2}{\sin \angle O_2O_1O}=\frac{O_2O_1}{\sin \angle O_1OO_2}$, $ \iff \frac{R-r_1}{\sin 2\alpha}=\frac{R-r_2}{\sin 2\beta}=\frac{r_1+r_2}{\sin (\pi-2\beta-2\alpha)}$, which implies, $ r_1=\frac{R\tan \beta}{ \tan ( \alpha+ \beta)}.............(1),$ and $ r_2=\frac{R\tan \alpha}{ \tan ( \alpha+ \beta)}..............(2),$ consider $ \triangle DTA, \triangle DTB$ and $ \triangle ABD,$ we have $ \frac{AT}{DT}=\frac{1}{ \tan \alpha}, \frac{BT}{DT}=\frac{1}{\tan \beta},and AT+TB=AB=2R$, $ \therefore DT=\frac{2R \tan \alpha \tan \beta}{\tan \alpha+\tan \beta}, AT=\frac{2R \tan \beta}{\tan \alpha+\tan \beta}...........(3), BT=\frac{2R \tan \alpha }{\tan \alpha+\tan \beta}.........(4)$. $ \therefore \frac{O_1D}{AT-O_1D}=\frac{r_1}{\frac{2R \tan \beta}{\tan \alpha+\tan \beta}-r_1}=\frac{1}{\frac{2 \tan \beta}{\tan \alpha+\tan \beta} (\frac{R}{r_1})-1} =\frac{1-\tan \alpha \tan \beta}{1+\tan \alpha \tan \beta}=\frac{\cos(\alpha +\beta)}{\cos(\alpha-\beta)}............(5)$ similarly, $ \frac{DO_2}{TB-DO_2}=\frac{\cos(\alpha +\beta)}{\cos(\alpha-\beta)}......(6)$ since $ \angle EFO=\angle FEO=\alpha+\beta$, $ EF=2R\cos (\alpha+\beta)$, since $ \triangle EDF \sim \triangle ADB$, $ \cos(\alpha+\beta)=\frac{2R \cos(\alpha+\beta) }{2R}=\frac{EF}{AB}=\frac{ED}{AD}=\frac{DF}{DB}$, consider$ \triangle EPD, \angle EDP =\pi -\angle PED- \angle EDP=\frac{\pi}{2}-(\alpha-\beta)$, $ \therefore \frac{PD}{ED}=\frac{\sin \angle PED}{\sin \angle EPD}=\frac{\sin \alpha}{\cos (\alpha-\beta)}$, $ \frac{PD}{PT-PD}=\frac{PD}{DT}=\frac{PD}{ED} \frac{ED}{DA} \frac{DA}{DT}=\frac{\sin \alpha}{\cos (\alpha-\beta)} \frac{\cos(\alpha+\beta) }{1} \frac{1}{\sin \alpha} =\frac{\cos(\alpha +\beta)}{\cos(\alpha-\beta)}............(7)$ form (6), (7),$ \frac{DO_1}{AT}=\frac{PD}{PT}$. since $ O_{1}D\parallel AT$ and $ \frac{DO_1}{AT}=\frac{PD}{PT}$, $ P,O_1 , A$ are collinear and similarly $ P,O_2 , B$ are collinear. $ \therefore$ lines $ AO_{1}, BO_{2}, EF$and $ t$are concurrent.

Lemma 1: In $\triangle ABC$, let point $D$ lie on the interior of segment $BC$. Then $\frac{BD}{DC} = \frac{AB \sin \angle BAD}{AC \sin \angle DAC}$. Proof: $\frac{BD}{\sin \angle BAD} = \frac{AB}{\sin \angle ADB}$ and $\frac{DC}{\sin \angle DAC} = \frac{AC}{\sin \angle CDA} = \frac{AC}{\sin \angle CDA}$. Dividing these equations gives the desired result. Lemma 2: Let $\Gamma$ be a circle, let $\omega$ be internally tangent to $\Gamma$ at $P$, let $AB$ be a chord of $\Gamma$ tangent to $\omega$ at $Q$, and let $R$ be the midpoint of the arc $\stackrel \frown{BC}$ not containing $P$. Then line $PQ$ and the perpendicular bisector of $AB$ intersect on $\Gamma$. Proof: A homothety centered at $P$ mapping $\omega$ to $\Gamma$ maps segment $AB$ to the line tangent to $\Gamma$ parallel to $AB$, which is clearly tangent to $\Gamma$ at $R$. Since a homothety centered at $P$ maps $Q$ to $R$, $P$, $Q$, and $R$ are collinear, as desired. By lemma 2, $A$, $D$, and $F$ are collinear, and $E$, $D$, and $B$ are collinear. Let $R$ be the radius of $\omega$, $\angle DAB = \alpha$, and $\angle DBA = \beta$. Simple angle chasing gives us all the angles labeled in the diagram. Furthermore, it is easily seen that $AF = 2R \cos \alpha$, $EB = 2R \cos \beta$, $AE = 2 R \sin \beta$, and $FB = 2 R \sin \alpha$. Let $t$ hit $EF$ at $T$, let $AO_1$ hit $EF$ at $T_1$, and let $BO_2$ hit $EF$ at $T_2$. By the trigonometric form of Ceva's theorem on $\triangle DEA$, $\frac{\sin \angle DAO_1}{\sin \angle O_1 AE} = \frac{1}{\frac{\sin(90 - \beta)}{\sin \beta} \cdot \frac{\sin \beta}{\sin \alpha}} = \frac{\sin \alpha}{\cos \beta}$. By lemma 1, $\frac{FT_1}{T_1 E} = \frac{AF \sin \angle FAT_1}{AE \sin \angle T_1 AE} = \frac{2R \cos \alpha}{2 R \sin \beta} \cdot \frac{\sin \alpha}{\cos \beta} = \frac{\sin \alpha \cos \alpha}{\sin \beta \cos \beta}$. By symmetry (e.g., by swapping points $A$ and $B$ and points $E$ and $F$), we find that $\frac{FT_2}{T_2E} = \frac{\sin \alpha \cos \alpha}{\sin \beta \cos \beta}$ as well. Finally, by lemma 1 and the law of sines, $\frac{FT}{TE} = \frac{DF \cos \alpha}{DE \cos \beta} = \frac{\sin \alpha \cos \alpha}{\sin \beta \cos \beta}$. Hence, $\frac{FT}{TE} = \frac{FT_1}{T_1E} = \frac{FT_2}{T_2E} = \frac{\sin \alpha \cos \alpha}{\sin \beta \cos \beta}$, whence $T = T_1 = T_2$, i.e., lines $t$, $AO_1$, $BO_2$, and $EF$ concur.

Trazamos las tangentes a E y F por w claramente estas dos rectas concurren con t(por eje radical) llamemos al punto de concurrencia S ,llamemos M al punto de corte de AB con t.luego es facil ver que O1=X centro de W1 y O2=Y son colineales con D de hecho la recta que los une es perpendicular a t => XY // AB..Denotemos a U al punto de corte de AE con BF , facilmente se prueba que E,D,B y A,D,F son colineales =>D es el ortocentro de AUB luego U está sobre t, ademas SD=SE =>US=SF=SD=SE por lo que hay una circunferancia R de centro S y radio US, finalmente XS y AU son perpendiculares a ED por lo tanto XS//AU y de modo similar se tiene que SY//BU así que los triangulos XSY y AUB son homotéticos =>AX , BY y US=t son concurrentes llamemos a este punto K, ahora basta con probar queE,KyF son colineales para esto es suficiente con probar que K está en el eje radical de R y de J donde J es la circunferencia que pasa por S,E,M,O,F estos puntos son conciclicos ya que S,E,O Y F son conciclicos y ttambien lo son S,E,M,O pues <OED=90 y <SMO=90. Entonces como dijimos es suficiente con que (UK)(KD)=(SK)(SM) , ahora si DM =c, SK =b y KD=a =>UT=a+b , tenemos KX/XA=a/c=b/a+b=>a(a+b)=bc a(a+2b)=bc+ab=b(a+c) que justamente es (UK)(KD)=(KS)(KM) por lo tanto E,K,F son colineales ,esto completa la solución. Disculpen que este en ESPAÑOL, luego lo coloco en INGLES.

We draw tangents to E and F by w clearly these two lines together with t (by radical axis) call concurrency S point, call M cut-off point of AB with t.luego is easy to see that O1 = X W1 and O2 = Center and are collinear with D in fact the line that links is perpendicular to t = > XY / / AB...We denote to or AE with BF, cut-off point easily proves that E, D, B and A, D, F are collinear = > D is the orthocenter of AUB then U is on t, besides SD = SE = > US = SF = SD =SE there foulbrood is a circunferancia R, S Center and RADIUS US, finally XS and AU are perpendicular to ED therefore XS / / AU and similarly you have to SY / / BU so XSY and AUB triangles are homotéticos = > AX and BY, US = t are concurrent call this point K, now just try queE, KyF are collinear for this is sufficient to prove that K is the radical axis R and J where J is the circle which passes through S, E, M, O, F these points are conciclicos as S,E , O and F conciclicos and ttambien are S, E, M,O or because < OED = 90 and < SMO = 90. Then as we said is sufficient to (UK) (KD) = (SK) (SM), now if DM = c, SK = b and KD = a= > UT = a + b, have KX/XA = a/c = b/a + b = > (a + b) =bc a(a+2b) = bc + ab = b (a + c) which is precisely (UK) (KD) = (KS) (KM) by therefore E, K, F are collinear, this complete solution.

Another end to vittasko's solution. Using his notation, by Pappus like he said we have that $P$ lies on $EF$. Consider the homothety centred around $P$ bringing $O_1$ to $A$, $O_2$ to $B$, $w_1$ and $w_2$ to two tangent circles centred at $A$ and $B$. Let the tangent point of the two circles lying on $AB$ be $T'$. Since homothety preserves ratio of the radii of the circles, we have that $\frac{AT'}{T'B} = \frac{O_1D}{O_2D}$. However, $\frac{AT}{TB} = \frac{O_1D}{O_2D}$, so $T = T'$. Therefore, since $T, P,$ and $D$ are collinear, $T$ lies on $t$.

As previously posted, one can quickly establish that $E,D,B$ are collinear. Let $T = t\cap AB$. Let S be the reflection of $T$ in $AE$. Then $\triangle SAT$ and $\triangle EO_1D$ are isosceles, with $AT \parallel O_1D$, and $ST \parallel ED$, (both are perpendicular to $AE$). So $\triangle SAT$ and $\triangle EO_1D$ are similar with parallel sides, which means that they are homothetic, and $SE$, $AO_1$ and $DT$ are concurrent. If $\angle FAB = \theta$, then since $DEAT$ is cyclic, $\angle SEA = \angle AET =\angle ADT = 90-\theta$. Now $\angle AEF = \angle AEB + \angle FEB = 90 + \theta$, so $S,E,F$ are collinear. Thus $EF$, $AO_1$ and $t$ are concurrent. A similar argument shows that $EF$, $BO_2$ and $t$ are also concurrent.

Let $AB\cap t=K$. By homothety, we know that $E,O_1,K$,$F,O_2,K$,$A,D,F$, and $B,D,E$ are sets of collinear points. Furthermore, $\angle BEA=\angle AEF=90$ and $t$ is perpendicular to $AB$ thus $t,AE,BF$ are concurrent at a point $T$ and $D$ is the orthocenter of $\triangle ABT$. Let $AO_1\cap t=J, O_1D\cap AE=N$ and $BO_2\cap t=J', DO_2\cap BF=N$. Note that since $DN,DM$ are diameters and $O_1DO_2 || AB$, we have \[\frac{JD}{JK}=\frac{O_1D}{AK}=\frac{O_1N}{AK}=\frac{EO_1}{EK}=\frac{O_1D}{EK}\] and similarly $\frac{J'D}{J'K}=\frac{O_2D}{FK}$. Note that we know from harmonic divisions that $\angle O_1KD=\angle DKO_2$ and thus by the law of sines \[\frac{O_1K}{O_1D}=\frac{O_2K}{O_2D}=\frac{EK-O_1D}{O_1D}=\frac{FK-O_2D}{O_2D}\rightarrow\frac{O_1D}{EK}=\frac{O_2D}{FK}\] But this means that $\frac{JD}{JK}=\frac{J'D}{J'K}$, and we have $J=J'$. Thus, $AO_1,BO_2,t$ are concurrent. Now, we let $BE\cap FK=L$. Then $\frac{LO_2}{LK}=\frac{DO_2}{BK}=\frac{FO_2}{FK}$, thus $(FO_2LK)$ is a harmonic divsion. Let $BF\cap EF=X$. Consider pencil $B$ and line $FE$. Then by a lemma of harmonic divisions, $BF\cap EF,BO_2\cap EF, BL\cap EF, BK\cap EF=F,BO_2\cap EF,E,X$ is harmonic. But then $t\cap EF=BO_2\cap EF$ since by a lemma of harmonic divisions. Thus, doing this for $AO_1$ too, we have that $AO_1,BO_2,EF$ are concurrent. Putting these together, we have the desired concurrency.

I couldn't resist posting another solution. Let $v$ be the excircle at $O$ of $\triangle OO_1O_2$. A little arithmetic shows that $E,D,F$ are the points of tangency of $v$ with the sides of $\triangle OO_1O_2$. Let $K$ be the polar of $AB$ with respect to $v$. Then clearly $K$ lies on $t$ and also on the polars of $A,B,O$. The polar of $O$ is simply $EF$. Also, since $w,v$ are clearly orthogonal, the inverse of $B$, $B'$, lies on $w$. So $A$ lies on the polar of $B$ (because $AB' \perp BB'$), as does $O_1$ (since $B$ lies on the polar of $O_1$). Similarly $BO_2$ is the polar of $A$.

Proof: Let $R$ be the point of intersection of the tangent $t$ and the the line $AB$. Since $AB$ is a diameter, we easily obtain that quadrilaters $EDRA$ and $FDRB$ are cyclic. Thus, $BF$,$AE$ and $t$ concur at the radical center of the circles of the quadrilaters $EDRA$, $FDRB$ and $ABFE$. Let $t$ intersect $EF$ at point $S$. It is well-known(it is just a simple property regarding homothety) that $E$, $D$ and $B$ are collinear. Let $FO_2$ intersect $BS$ at $X$ and denote by $M$ the midpoint of side $AB$. Since $(T,D;S,R)=-1=B(F,D;X,M)=D(F,B;X,M)=D(A,B;X,M)$, we get that $DX$ must be parallel to $AB$, so $X$ is $O_2$. Since this is what the problem asks for, we are done.$\blacksquare$

As a starting observation note that $E,D,B$ and $F,D,A$ are collinear.We will start be letting $\angle{EAB}=y,\angle{FBA}=x$.Let $J=t \cap AB$.Then due to the concyclicities of $EDJA$ and $FDJB$ we get $\frac{AJ}{BJ}=\frac{tanx}{tany}=\frac{AF}{BF} \cdot \frac{AE}{EB}=\frac{AF}{BE} \cdot \frac{ED}{FD}=\frac{ED}{BE} \cdot \frac{AF}{FD}=\frac{r_1}{R} \cdot \frac{R}{r_2}=\frac{r_1}{r_2}=\frac{O_1D}{O_2D}$ Consequently $AO_1,BO_2,t$ concur. Also let $EF \cap t=K$.Note that $r_1=R \cdot \frac{ED}{EB}=R \cdot \frac{-AEcot(x+y)}{EB}=R \cdot \frac{-cot(x+y)}{tany}$ $AJ=ADsinx=\frac{2Rcosysinx}{sin(x+y)}$ So we get $\frac{O_1D}{AJ}=\frac{r_1}{AD}=\frac{-cos(x+y)}{2sinxsiny}$ By applying sine rule in $\triangle{EKD},\triangle{ADJ}$ we get $KD=\frac{EDcosx}{cos(x-y)}$ $DJ=ADcosx$ Thus $\frac{KD}{DJ}=\frac{ED}{ADcos(x-y)}=\frac{-cos(x+y)}{cos(x-y)}$ $\frac{KD}{KJ}=\frac{-cos(x+y)}{cos(x-y)-cos(x+y)}=\frac{-cos(x+y)}{2sinxsiny}$ Finally $\frac{KD}{KJ}=\frac{O_1D}{AJ} \implies A,O_1,K$ are collinear. Combining all these we get that $AO_1,BO_2,t,EF$ concur.

First, be $AO_1 \cap EF = K_1$, $BO_2 \cap EF = K_2$, $M$ midpoint of $AB$, $t \cap AB = R$, $EF\cap AB = Q$, $ME \cap AF = U$ and $MF \cap BE = V$. With some homotheties, it is easy to see that $A, D, F$ are collinear, such as $B, D, E$. Also, it is easy to see that $AE, t, BF$ are concurrent since $AEDR, AEFB, DRBF$ are cyclics, thus $AE \cap t \cap BF = T$. See that $D$ is on polar of $Q$ to $w$, and $D \in t \perp AB$, so $t$ is the polar of $Q$ to $w$. Now, we have $(E, F; K_1, Q) = (E, U; O_1, M) = (V, F; O_2, M) = (E, F; K_2, Q)$, so $K_1 = K_2$. Let $K_1 = K_2 = K$ be. Let $N = O_1O_2 \cap MK$ be a point. Since $O_1O_2 // AB$ and $MA = MB$, we have that $N$ is the midpoint of $O_1O_2$. Finally, $(E, F; K, Q) = (O_1, O_2; N, P_{?})$, where $P_{?} = AB \cap O_1O_2$, but $AB // O_1O_2$, and since $N$ is the midpoint of $O_1O_2$, it follows that $(O_1, O_2; N, P_{?}) = -1$, hence $(E, F; K, Q) = -1$, thus $K$ is on polar of $Q$ to $w$, in other words, $K \in t$, qed.

Quote: Circles $\omega_1$ and $\omega_2$ with centres $O_1$ and $O_2$ are externally tangent at point $D$ and internally tangent to a circle $\omega$ at points $E$ and $F$ respectively. Line $t$ is the common tangent of $\omega_1$ and $\omega_2$ at $D$. Let $AB$ be the diameter of $\omega$ perpendicular to $t$, so that $A, E, O_1$ are on the same side of $t$. Prove that lines $AO_1$, $BO_2$, $EF$ and $t$ are concurrent.

Circles are overrated, I was quite dismayed to see solutions involve so many circles. Here's a pure bash solution. Observe that since $OF$ and $OE$ are equal to the semiperimeter of $OO_1O_2$, that they are the extouch points. Hence, we can mark point $I_O$, which denotes the $O$ excenter of $OO_1O_2$. The upshot of all this is that $t = I_OD$. Now use barycentric coordinates on $OO_1O_2$. Let $O = (1 : 0 : 0)$, $O_1 = (0 : 1 : 0)$, and $O_2 = (0 : 0 : 1)$. Since $D$ is the $O$-extouch point $D = (0 : a + c - b : a + b - c)$, $I_O = (-a : b : c)$, $F = (b-s : 0 : s)$, $E = (c-s : s : 0)$. Since $OB = s$, observe that $B = (a : -s : s)$, $A = (a : s : -s)$. Since $O_2B$ and $O_1A$ are cevians in $OO_1O_2$, we have that $O_2B \cap O_1A = (-a : s : s)$. We now show this is collinear with $DI_O$ and $EF$, which will complete the question. For $EF$ the verification is incredibly simple. Its just \[ 0 = \det \begin{pmatrix} b-s & 0 & s \\ c-s & s & 0 \\ -a & s & s \end{pmatrix}. \] This is true because the sum of the first two rows is the third one, done. For $DI_O$ its \begin{align*} \det \begin{pmatrix} 0 & a + c - b & a + b - c \\ -a & b & c \\ -a & s & s \end{pmatrix} &= \det \begin{pmatrix} 0 & a + c - b & a + b - c \\ 0 & b-s & c-s \\ -a & s & s \end{pmatrix} \\ &= \det \begin{pmatrix} a + c - b & a + b - c \\ b-s & c-s \end{pmatrix} \\ &= \det \begin{pmatrix} b - s & c - s \\ b-s & c-s \end{pmatrix} = 0 \end{align*} so the intersection indeed lies on both the other lines, as desired.

Let $C=AE\cap BF$, $T=EF\cap AB$, $G=AB\cap t$, and $O$ be the center of $\omega$. Lemma: $D=AF\cap BE$. Proof: the homothety with center $E$ sending $\omega_1\mapsto \omega$ sends $D\mapsto B$ since $B$ is the arc midpoint defined by $t$, so $B, D, E$ are collinear. Similarly, $A, D, F$ are collinear and the result follows. $\blacksquare$ Since $\angle AEB=\angle AFB=90^{\circ}$, $D$ is the orthocenter of $\triangle CAB$. In particular, $CD\perp AB$. But since $t\perp AB$ and $D\in t$, then $CD=t$ so $t$ is the $C$-altitude of $\triangle CAB$. Let $X=EF\cap CD$; we want to prove $X$ lies on both $AO_1$ and $BO_2$. Let $\Gamma$ denote the circle with diameter $CD$. Let $O_1'$ be the intersection of the tangents at $D$ and $E$ to $\Gamma$, and let $O_2'$ be the intersection of the tangents at $D$ and $F$ to $\Gamma$. We know $\triangle ABX$ is a self-polar triangle w.r.t $\Gamma$ by Brokard. By La Hire, $B\in \text{polar}(O_1')\implies O_1'\in \text{polar}(B)=AX$ so $A, O_1', X$ are collinear. Similarly, $B, O_2', X$ are collinear. It remains to show that $O_1'=O_1$ and $O_2'=O_2$, or in other words $O, O_1', E$ and $O, O_2', F$ are both collinear. But this is true from the fact that the tangents at $E$ and $F$ of $\Gamma$ intersect at the midpoint of $AB$, which is just $O$, so we're done. $\Box$

Let $D'$ be the reflection of $D$ over $M$, and $D''$ be the reflection of $D$ over $I$. Note that a homothety centered at $A$ taking the incircle of $\triangle ABC$ to the $A$-excircle of $\triangle ABC$ maps $D''$ to $D'$ because the line parallel to $BC$ through $D''$ is tangent to $\triangle ABC$'s incircle, just like how $Dl$ is the point of tangency on $BC$ with respect to the $A$-excircle of $\triangle ABC$. Since $I$ is the midpoint of $DD''$ and $M$ is the midpoint of $DD'$, then $MI$ and $D'D''$ are parallel. Now, let $A'$ be where $AD'$ hits the circumcircle of $\triangle ABC$ again. Also note that $XA$ is the radical axis of the circumcircles of $\triangle ABC$ and $\triangle AEF$. Since $BF$ and $CE$ concur at $A$, then a spiral similarity centered at $X$ takes $FIE$ to $BMC$. This means that $\angle XA'A = \angle XBA = \angle XMI$ so since, from above, $MI$ and $D'D''$ are parallel and $M$ lies on $A'X$. Now let $AM$ and $CD$ hit at $P$. The Butterfly Theorem implies that assuming $MB=MC$ and $B$, $C$, $X$, $A$, and $A'$ are concyclic, then $MD=MD'$ if and only if $P$ lies on the same circle containing the other five points. Thus, since $MD=MD'$ in this case, then $P$ lies on the circumcircle of $\triangle ABC$ and we are done.

By a homothety centered at $F$ taking $\omega_2$ to $\omega$, we have $F,D,A$ are collinear. Similarly, $E,D,B$ are collinear. Let $T = l \cap AB$ and let $C = O_1D \cap \omega_1$. Since $EDTA$ and $FDTB$ are cyclic, by radical axis theorem on $(EDTA) , (FDTB) , (AEFB)$, we have $AE , BF$ and $DT$ are concurrent, call it $Z$. Let $X = EF \cap l$, we will prove that $X$ is the desired concurrency point. Let $K = EF \cap AB$. By a well known lemma, we know $(E,F;X,K) = -1$. We also have $-1 = (C,D;O_1,CD_{\infty}) \stackrel{A} = (E,F;AO_1\cap EF , K)$. So that means $AO_1 \cap EF = X$, implying that $A , O_1 , X$ are collinear. Similarly we can show $B,O_2,X$ are collinear. Thus we are done. $\square$

Let $C=\overline{AB}\cap t,X=\overline{EF}\cap t,$ and $J=\overline{EE}\cap\overline{FF}.$ Notice that $A,D,$ and $F$ are collinear since $\triangle FO_2D\sim\triangle FOA.$ Also, $J$ lies on $\overline{CD}$ by Radical Axis, $JE=JD=JF,$ and $\angle JFO=90$ so $J$ is the $O$-excenter of triangle $OO_1O_2.$ Let $D'$ be the antipode of $D$ with respect to $(DEF)$ and note that $A,E,$ and $D'$ are collinear as $\triangle D'DE\sim\triangle D'AC.$ By Pascal on $EEFDDD'$ we see that $X$ lies on $\overline{AO_1}.$ $\square$

Let $O$ be the center of $\omega$. We have $O_1O_2$ passes through $D$, and furthermore, it is perpendicular to $t$. Since $AB$ is also perpendicular, $O_1O_2\parallel AB$. Note that a homothety from $E$ takes $\omega_1$ to $\omega$. Since $O_1D\parallel OB$, it takes $D$ to $B$ which implies that $E$, $D$, $B$ are collinear. Similarly, $F$, $D$, $A$ are collinear. Let $C$ be the orthocenter of $\triangle ADB$, which must lie on $AE$ and $BF$. We must also have $CD\perp AB$ so $C$ lies on $t$. Let $H$ be the perpendicular from $C$ to $AB$. Clearly, $D$ is also the orthocenter of $\triangle ABC$. Let $M$ be the midpoint of $CD$. Note that $M$ is also the center of the circle $(CEDF)$. Thus, $ME$ and $MF$ are tangent to $\omega_1$ and $\omega_2$, respectively. This implies that $O_1M \parallel AC$ and similarly for the other sides. We have $O_1O_2M$ is a homothety of $ABC$, so $CM$, $AO_1$ and $BO_2$ are concurrent. Let $P$ be the intersection of $AO_1$ and $BO_2$. By Pappus's theorem on $A,O,B$ and $O_2,D,O_1$, the points $E$, $P$, $F$ are collinear. $C$, $P$, $M$ are also concurrent. We are done.

Kinda surprised at how quickly this fell, but hey, i'll take an easier problem as a break :3 Define $O_3$ as the center of $\omega$. All polars spoken about this problem are with respect to $(DEF)$. Note that $(DEF)$ is the excircle of $\triangle O_1O_2O_3$. Observe that $\triangle EO_1D$ and $\triangle EO_3B$ are homothetic (tangent circles and parallel lines, say), and hence $B$ lies on line $DE$. Similarly we have $A$ lying on line $DF$. Now note that $(DEF)$ is tangent to the lines through the centers of the circles, and it is also orthogonal to $\omega$. Since $B$ lies on the polar of $O_1$, we have $O_1$ on the polar of $B$. Since we have also have $(DEF)$ orthogonal to $\omega$, $A$ is also on the polar of $B$, so $AO_1$ is the polar of $B$. Similar story for $BO_2$. Let $AO_1$ and $BO_2$ meet at $X$. Since $X$ lies on the polar of $A$ and $B$, it must hold that $AB$ is the polar of $X$; indeed, it also must lie on $t$ as such since $t \perp AB$ and $t$ passes through the center of $(DEF)$. Now at last, since the polar of $X$ passes through $O_3$, the polar of $O_3$, which is $EF$ must pass through $X$, hence we have the four lines concurrent. $\textbf{Remark.}$ After noticing the collinearity it's just La Hire spam and well known facts about orthogonal circles.

most convoluted solution. Let $O$ be the center of $w$. By homothety note that $B,D,E$ and $A,D,F$ are collinear, so by converse of Pappus we find that $AO_1,BO_2,EF$ are concurrent. Now we just need to show that $AO_1,BO_2,t$ are concurrent. Note that $D$ is the $O$-extouch point of $\triangle OO_1O_2$. Consider the angle bisector of $\angle OO_2O_1$. I claim that this line is parallel to $AD$. To see this, simply let the bisector intersect $AB$ at $P$: \[AP=AO-PO=(DO_2+OO_2)-PO=DO_2\]so $APO_2D$ is a parallelogram. Let $I$ be the incenter of $OO_1O_2$ and note that $\triangle ADB\sim \triangle O_2IO_1$. Also let $H$ be the $O$-intouch point of $\triangle OO_1O_2$. In order to show that $AO_1$ and $BO_2$ intersect on $t$, let's draw the foot of the altitude from $D$ to $AB$. Call it $G$. It suffices to show that \[\frac{AG}{BG}=\frac{O_1D}{O_2D}\]but this is easy with our information now: \[\frac{AG}{BG}=\frac{O_2H}{O_1H}=\frac{O_1D}{O_2D}.\] done!

Really nice problem Let $C$ be the center of $(DEF)$; notice that $C = \overline{EE} \cap \overline{FF}$. Also, $B, D, E$ and $A, D, F$ are collinear by homothety. Let $I = \overline{EF} \cap \overline{DD}$. Now let's do some magic! $(O_3)$ and $(C)$ are orthogonal, so it follows that $\overline{EF}$ is the polar of $O_3$ with respect to $(C)$. Thus $I$ lies on the polar of $O_3$, and it follows by perpendicularity that the polar of $I$ is precisely $\overline{AB}$. Let $X = \overline{AE} \cap \overline{DD}$. By an easy angle chase $\angle EAB + \angle EFD = 90^\circ$, $X$ lies on $(DEF)$, so it is actually the $D$-antipode and also lies on $\overline{BF}$. Then by Brokard on $EDFX$, $ABI$ is self-polar and hence $B$ lies on the polar of $A$. But by previous discussion, $A$ lies on the polar of $O_2$ so $O_2$ lies on the polar of $A$. It follows that $I$ lies on $\overline{BO_2}$, and similarly, $I$ lies on $\overline{AO_1}$. Done! Next, observe that $A$ lies on the polar of $O_2$, so $O_2$ lies on the polar of $A$.

Denote the center of $\omega$ as $O$. Clearly, $\overline{O_1O_2}$ passes through point $D$, and is perpendicular to $t$. This means that $\overline{O_1O_2} \parallel \overline{AB}$. Consider the homothety centered at $E$ sending $\omega_1$ to $\omega$. This homothety maps $D$ to $B$, so points $E$, $D$, $B$ are collinear. Analogously, points $F$, $D$, $A$ are collinear. Now, let the orthocenter of $\triangle ADC$ be point $H$. Due to our aforementioned collinearities, $H$ is the intersection of $\overline{AE}$, $\overline{BF}$, and $t$. Denote $C = \overline{AB} \cap \overline{DH}$. Point $H$ is the orthocenter of $\triangle ABD$, implying that $D$ is the orthocenter of $\triangle ABH$. Denote the midpoint of $\overline{DH}$ as $M$. Point $M$ is obviously the center of $(CEHF)$. Thus, $MD=ME=MF$, which is equivalent to $ME$ being tangent to $\omega_1$ and $MF$ being tangent to $\omega_2$. Hence, $\overline{O_1M} \parallel \overline{AH}$ and $\overline{O_2M} \parallel \overline{BH}$. This implies that $\triangle O_1O_2M \sim \triangle ABH$, meaning that lines $AO_1$, $BO_2$, and $MH \equiv t$ are concurrent at the center of the homothety sending one to another. Pappus's Theorem on $\overline{AOB}$ and $\overline{O_1DO_2}$ shows our last desired concurrent line. $\square$

Let $O$ be the center of $\omega$. We make the following initial observations: Homothety at $E$, $F$ tells us $EO_1D$, $FO_2O$, $EDB$, $FDA$ are collinear. Pappus on $O_1DO_2$ and $AOB$ tells us $AO_1$, $BO_2$, $EF$ concur at a point, say $M$. Let the exsimilicenter of $\omega_1$ and $\omega_2$ be $N$, which is $EF \cap O_1O_2$ by Monge's. Ratios tell us \[-1 = (O_1O_2; DN) \overset{M}{=} (AB; MD \cap AB, EF \cap AB).\] Let $X = AE \cap BF$. Note that $D$ is the orthocenter of $\triangle XAB$, so $MD \cap AB$ must be the foot of the altitude from $X$ to $AB$, as desired. $\blacksquare$

Let $D_1$ and $D_2$ be the antipodal points of $D$ in $\omega_1$ and $\omega_2$ respectively. By a homothety about $E$ sending $\omega_1$ to $\omega$, $E,D,B$ are collinear and $E,D_1,A$ are collinear. Similarly, $F,D,A$ are collinear and $F,D_2,B$ are collinear. Extend $AE$ and $BF$ to meet at $C$. Since $\angle AEB = \angle AFB = 90^{\circ}$, $D$ is the orthocenter of $\Delta ABC$. It follows that $CD \perp AB$, so $C$ lies on $t$. Let $H$ be the foot of the altitude from $C$ to $AB$, $X$ be the intersection of $EF$ with $t$. It is well-known that $(D,C;X,H)=-1$. Now $-1=(D,D_1;O_1,\infty_{AB}) \stackrel{A}{=} (D,C;AO_1 \cap t, H)$, so $A,O_1,X$ are collinear. Similarly, $B,O_2,X$ are collinear, so we are done. $\square$

Focus on $\triangle OO_1O_2$. Let $O_1O_2=a$, $OO_2=b$, and $OO_1=c$. Note that since $$r_1+r_2=a,c+r_1=b+r_2,$$we can solve for $$r_1=s-c,r_2=s-b$$which means that $D$ is the excircle touchpoint of $\triangle OO_1O_2$ to $O_1O_2$. Let $K$ be the intersection of the perpendicular at $D$ to $O_1O_2$ (equivalently, $I_aD$) with $AB$. We will first show that $AO_1,BO_2,KD$ concur. It of course suffices to show that $$\frac{AK}{O_1D}=\frac{AB}{O_1O_2}$$ We have $AK=s+s-c-c\cos\beta=a+b-c\cos\beta$, $O_1D=s-c$, $AB=a+b+c$, and $O_1O_2=a$. Thus, this becomes $$\frac{a+b-c\cos\beta}{s-c}=\frac{a+b+c}{a}$$$$2a^2+2ab-2ac\cos\beta=a^2+b^2-c^2+2ab$$$$\cos\beta=\frac{a^2-b^2+c^2}{2ac},$$which is clearly true. Let the concurrency point be $T$. Now, we wish to show that $T$ lies on $EF$. Note that, however, $EF$ is the polar of $O$ with respect to the excircle. Thus, it suffices to show that $O$ lies on the polar of $T$. Since the polar of $T$ is parallel to $AB$, and $K$ is the foot from $O$ to $AB$, this is equivalent to $$I_AT\cdot I_AK=r_a^2$$. Letting $h_a$ being the height from $A$ to $BC$, we can expand this to $$(r_a-DT)(r_a+h_a)=r_a^2$$$$DT=\frac{r_ah_a}{r_a+h_a}.$$However, we have $$\frac{DT}{KT}=\frac{O_1O_2}{AB}$$$$\frac{DT}{DT+h_a}=\frac{a}{a+b+c}$$$$DT=\frac{ah_a}{b+c}.$$ Thus, it suffices to show that $$\frac{ah_a}{b+c}=\frac{r_ah_a}{r_a+h_a}$$$$ar_a+ah_a=(b+c)r_a$$$$r_a=\frac{2A}{-a+b+c}=\frac{A}{s-a},$$which is clearly true. remark (alternative for the part about showing $ETF$ collinear) that I didn't see: $BDE$ and $ADF$ are collinear since $D$ and $B$ are the "rightmost points" of the two circles that are homothetic at $E$. after this, Pappus on $ABO$ and $O_2O_1D$ gives $ETF$ collinear, oops i need to improve on seeing pappus

Let $C = AE \cap BF$. From homothety, $E, D, B$ are collinear and $A, D, F$ are collinear, so $D$ is the orthocenter of $\triangle ABC$. Then $O_1D$ and $O_1E$ are tangents to $(CEDF)$, so from Pascal on $EEFDDC$ we get $A$, $O_1$, $EF \cap t$ are collinear. Similarly, $B$, $O_2$, $EF \cap t$ are collinear. Therefore, $AO_1$, $BO_2$, $EF$, and $t$ are concurrent, as desired.

oops.

By homothety, we have $\overline{A-D-F}$ and $\overline{B-D-E}$. Let $t$ intersect $AB$ at $X$, $EF$ at $T$ and $O$ be the midpoint of $AB$. It suffices to prove that: $$\frac{\sin(\angle EAO_1)}{\sin(\angle DAO_1)}=\frac{\sin(\angle EAT)}{\sin(\angle DAT)}$$ We compute both ratios using trig ceva: \begin{align*} \frac{\sin(\angle EAO_1)}{\sin(\angle DAO_1)}&=\frac{\sin(\angle AEO_1)}{\sin(\angle O_1ED)}\cdot \frac{\sin(\angle EDO_1)}{\sin(\angle O_1DA)}=\frac{\sin(\angle AEO_1)}{\sin(\angle O_1DA)}\\ &=\frac{\cos(\angle ABD)}{\cos(\angle ADX)}=\frac{DA}{DB}\cdot \frac{XB}{XD} \end{align*} \begin{align*} \frac{\sin(\angle EAT)}{\sin(\angle DAT)}&=\frac{\sin(\angle AET)}{\sin(\angle TED)}\cdot\frac{\sin(\angle TDE)}{\sin(\angle TDA)}\\ &=\frac{\sin(\angle ABF)}{\sin(\angle FAB)}\cdot\frac{\sin(\angle BDX)}{\sin(\angle ADX)}\\ &=\frac{AF}{FB}\cdot\frac{XB}{XA}\cdot\frac{DA}{DB} \end{align*} The conclusion follows by POP on $(DXBF)$. $\blacksquare$