I noticed that this problem has a configuration issue . Also, Coordinate-Bash works! It seems that when the point $C$ lies on the side of the line $T_1T_2$ not containing the circles $\omega_1$ and $\omega_2$, this configuration issue pops up. The flaw can be demonstrated from the following picture. The nagelian of $\triangle ABC$ from $C$ is coloured in blue and four different instances of the nagelian are shown in which each of the cases, the point $C$ lies on the side of $T_1T_2$ not containing $\omega_1$ and $\omega_2$. However! The issue is fixed when we take point $C$ on the same side of $T_1T_2$ as of the circles. Here is the figure for the case when $C$ lies on the same side of $T_1T_2$ as that of $\omega_1$ and $\omega_2$. The nagelian of $\triangle ABC$ from $C$ is coloured in blue and four different instances of the nagelian are shown again. Now we Coordinate Bash this one! $\textbf{LEMMA: }$If $\omega_1,\omega_2$ are two circles with centres $O_1,O_2$ and radius $r_1, r_2$ respectively, where the coordinates of $O_1 = (x_1, y_1)$ and $O_2 = (x_2, y_2)$. Then the coordinates of the insimilicenter $X$ of the two cirlces is given by $$X = \left( \dfrac{r_1 \cdot x_2 + r_2 \cdot x_1}{r_1 + r_2}, \dfrac{r_1 \cdot y_2 + r_2 \cdot y_1}{r_1 + r_2}\right).$$ $\textbf{PROOF: }$ Follows directly from Homothety and Section-Formula.

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@below you just read my mind lol check this image

@above $A, B$ need to be sufficiently far away for it to work, otherwise it just becomes a Gergonne cevian, interestingly. And yeah, I don't see a way to do this without using Cartesian Coordinates.

In $\triangle ABC$, draw the incircle $\omega$. Let it hit $\overline{AB}$ at $D$, and let the opposite point of $D$ on $\omega$ be $D'$. Let $D_1$ be the reflection of $D$ across the midpoint of $AB$. It is well known that $\overline{D'D_1}$ is the nagelian of $\triangle ABC$ (because $\overline{CD_1}$ is the nagelian and the homothety from $C$ sending $\omega$ to the $C$-excircle sends $D'$ to $D_1$). Let $P_1$ be the opposite point of $T_1$ on $\omega_1$ and let $P_2$ be the opposite point of $T_2$ on $\omega_2$. The homothety from $A$ sending $\omega_1$ to $\omega$ sends $P_1$ to $D'$, so $A-P_1-D'$. Similarly $B-P_2-D'$. We can show that $X=\overline{P_2T_1}\cap\overline{P_1T_2}$ lies on $\overline{D'D_1}$, and since $X$ is fixed we will be done. Let $P_{\infty}$ be the point at infinity perpendicular to $\overline{AB}$. Then by DDIT on $D'$ and the complete quadrilateral $(\overline{P_1T_1},\overline{P_1T_2},\overline{P_2T_1},\overline{P_2T_2})$, we have that $(\overline{D'P_1},\overline{D'P_2}),(\overline{D'T_1},\overline{D'T_2}),(\overline{D'P_{\infty}},\overline{D'X})$ are pairs under an involution. Projecting this involution from $D'$ to $\overline{AB}$, we get that $(A,B),(T_1,T_2),(D,\overline{D'X}\cap\overline{AB})$ are pairs under an involution. Since $(A,B)$ and $(T_1,T_2)$ are pairs under this involution, it must be the reflection across $M$, so $\overline{D'X}\cap\overline{AB}=D_1$, as desired.

Here is a not entirely synthetic solution using some (not too much) trig bash: Let $O_1$ and $O_2$ be the centers of $\omega_1$ and $\omega_2$ respectively and let $M$ be the midpoint of $T_1T_2$. Then $O_1T_1T_2O_2$ is clearly a right-angled trapezoid. Also let $O_1T_1=r_1$ and $O_2T_2=r_2$. Let $W$ be the point on $O_1O_2$ such that $\frac{O_1W}{WO_2}=\frac{r_1}{r_2}$. Lemma 1. $T_1W\parallel O_2M$ and $T_2W\parallel O_1M$. Proof. Let $O_1'$ be the point on the extention on $O_1T_1$ after $T_1$ such that $T_1O_1'=O_2T_2$. Similarly, let $O_2'$ be the point on the extention of $O_2T_2$ after $T_2$ such that $O_2'T_2=O_1T_1$. Then clearly $O_1T_2O_2'T_1$ and $O_1'T_2O_2T_1$ are parallelograms. Also $\frac{O_1T_1}{T_1O_1'}=\frac{r_1}{r_2}=\frac{O_1W}{WO_2}$ so $T_1W\parallel O_1'O_2$ and $T_2W\parallel O_1O_2'$. We will prove the nagelian through $C$ always passes through the fixed point $W$. Let $J$ be the touching point of the ex-circle of $ABC$ from $C$ and let $H$ be the foot of the perpendicular from $C$ to $AB$. Also let $K$ be the foot of the perpendicular from $W$ to $AB$ and $T_1T_2=a$ and $AT_1=BT_2=x$. Firstly, $\frac{r_1}{r_2}=\frac{O_1W}{O_2W}=\frac{T_1K}{KT_2}$ so $T_2K=\frac{ar_2}{r_1+r_2}$. Let $I$ be the incenter of $ABC$ and $r$ be the radius of the incircle of $ABC$ and $D$ be the touching point of the incircle with $AB$. Then $\frac{r_1}{r}=\frac{x}{x+T_1D}$ and $\frac{r_2}{r}=\frac{x}{x+a-T_1D}$. By dividing the to equalities above we get $T_1D=\frac{r_2x-r_1x+r_2a}{r_1+r_2}$. It is well-known $AD=BJ$, so $x+T_1D=x+JT_2$, so $JT_2=\frac{r_2x-r_1x+r_2a}{r_1+r_2}$. So $\boxed{KJ=KT_2-JT_2=\frac{ar_2}{r_1+r_2}-\frac{r_2x-r_1x+r_2a}{r_1+r_2}=\frac{(r_1-r_2)x}{r_1+r_2}}$. Also, $S_{O_1WK}+S_{O_1T_1K}+S_{O_2WK}+S_{O_2T_2K}=S_{O_1O_2T_2T_1}$, so $\frac{T_1K\cdot WK}{2}+\frac{T_2K\cdot WK}{2}+\frac{T_1K\cdot O_1T_1}{2}+\frac{T_2K\cdot O_2T_2}{2}=\frac{(r_1+r_2)a}{2}$. From here we find $\boxed{WK=\frac{2r_1r_2}{r_1+r_2}}$. Let $\angle BAC=\alpha$, $\angle ABC=\beta$. Then $\tan\frac{\alpha}{2}=\frac{r_1}{x}$ and $\tan\frac{\beta}{2}=\frac{r_2}{x}$. So $\tan\alpha=\frac{2\tan\frac{\alpha}{2}}{1-\tan^2\frac{\alpha}{2}}=\frac{2r_1x}{x^2-r_1^2}$ and similarly $\tan\beta=\frac{2r_2x}{x^2-r_2^2}$. Then $\frac{CH}{AH}=\tan{\alpha}$ and $\frac{BH}{CH}=\frac{1}{\tan\beta}$. Multiplying these two we get $\frac{AH}{BH}=\frac{\tan\beta}{\tan\alpha}$, so because $AH+BH=2x+a$, $BH=\frac{\tan\alpha}{\tan\alpha+\tan\beta}(2x+a)$ and $HJ=BH-(JT_2+T_2B)=\frac{\tan\alpha}{\tan\alpha+\tan\beta}(2x+a)-(x+\frac{r_2x-r_1x+r_2a}{r_1+r_2})$. So $\boxed{HJ=\frac{\tan\alpha}{\tan\alpha+\tan\beta}(2x+a)-(x+\frac{r_2x-r_1x+r_2a}{r_1+r_2})}$ Also, $\boxed{CH=BH\tan\beta=\frac{\tan\alpha\tan\beta}{\tan\alpha+\tan\beta}(2x+a)}$. Now we only have to do the following computations: \[HJ=\frac{(2x+a)\frac{2r_1x}{x^2-r_1^2}}{\frac{2r_1x}{x^2-r_1^2}+\frac{2r_2x}{x^2-r_2^2}}=\frac{(2x+a)x^2(r_1-r_2)}{(r_1+r_2)(x^2-r_1r_2)}\]\[CH=\frac{\tan\alpha\tan\beta}{\tan\alpha+\tan\beta}=\frac{2x+a}{\frac{1}{\tan\alpha}+\frac{1}{\tan\beta}}=\frac{2x+a}{\frac{x^2-r_1^2}{2r_1x}+\frac{x^2-r_2^2}{2r_2x}}=\frac{(2x+a)2xr_1r_2}{(x^2-r_1r_2)(r_1+r_2)}\]So \[\frac{CH}{HJ}=\frac{2r_1r_2}{x(r_1-r_2)}\]Similarly, \[\frac{WK}{KJ}=\frac{2r_1r_2}{r_1+r_2}\frac{r_1+r_2}{(r_1-r_2)x}=\frac{2r_1r_2}{x(r_1-r_2)}\]So \[\frac{WK}{KJ}=\frac{CH}{HJ}\]and by Thales' theorem it follows $C, W$ and $J$ are collinear. $\Box$

Problem 17 sharygin (9–11) A common external tangent to circles $\omega_1$ and $\omega_2$ touches them at points $T_1, T_2$respectively. Let $ A $ be an arbitrary point on the extension of $(T_1 T_2]$ beyond $T_1$, and $B$ be a point on the extension of $[T1T2) $ beyond $T_2$ such that $AT_1 = B_T2$ , The tangents from $A$ to $\omega_1$ and from $B$ to $\omega_2$ distinct from $ T_1T_2$ meet at point C (on the same side of $\omega_1$ wrt $ T_1T_2$). Prove that all nagelians of triangles $ABC$ from $C$ have a common point. Solution: Let $(I),(I_a) $ the incircle and the $A$-excircle ;$I_1,I_2$ the centers of $\omega_1$ and $\omega_2$; $T,E$ the tangency points of $(I),(I_a)$ with $BC$ ; $U $ the antipode of $T$ ; $S,S’$ the in-similicenter of $( \omega_1),( \omega_2)$ and $( \omega_2),(I)$. Claim:$U,S,E$ collinear i.e. the nagelians pass through $S$ Proof: Monge Alembert’s leads $A,S,S’$ $ collinear$ plus by homothety we get $T,S,U$ collinear. Using Menelaus’s in $AT_2S’$ it suffices to show that $\frac{ET_2}{EA}=\frac{UT_2}{US’}.\frac{SA}{SS’}$. By Menelaus’s in $SS’I_2 $ we get $ \frac{AS}{AS’}=\frac{I_1S}{I1I_2}.\frac{II_2}{IS’}= \frac{r_1 }{r_1+r_2}.\frac{r+r_2}{r } $ thus $ \frac{SA}{SS’} =\frac{r_1(r+r_2)}{r_2(r-r_1)}$. More by symmetry in the midpoint of $BC$ we get $\frac{ET_2}{EA}= \frac{TT_1}{TB}=\frac{TT_1}{TA}.\frac{TA}{AT_1}.\frac{AT_1}{BT_2}.\frac{BT_2}{TB}$ since $\frac{AT_1}{BT_2}=1$ hence $\frac{ET_2}{EA}= \frac{ r-r_1}{r }.\frac{ r}{r_1 }.\frac{ r_2}{r }=\frac{r_2(r-r_1) }{ r.r_1} $. Further $\frac{UT_2}{US’}= \frac{ r+r_2}{r} $therefore multiplying we get $\frac{ET_2}{EA}=\frac{UT_2}{US’}.\frac{SA}{SS’}$ which ends the proof. Best regards RH HAS

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