Lemma.- $ \alpha$ is the circumcircle of $ ABC$,Let $ D$ be an arbitrary point on the side $ BC$ of a triangle $ ABC$ . $ \beta$ is a circle tangent to $ \alpha$,$ AD$ and $ BC$, in $ D_0$,$ X$ and $ A_1$ respectively. Prove that $ I \in XA_1$where $ I$: Incenter of $ ABC$. Dem: (WLG-$ A_1$ between $ D$ and $ C$.) $ IA \cap \alpha = M$, know that by the theorem Catalan $ IM=MB=MC=R$. Let $ \phi$ cirumcircle of $ BIC$ with center $ M$,Know that $ \alpha$ and $ \beta$ They are homothetic with Homothetic transformation centre in D. Then, $ M,A_1,D_0$ are collinear , $ AD \cap \beta =T$, $ D_0,T,A$ are collinear, moreover $ TA_1$//$ AM$. If $ XA_1 \cap MA=L$, $ <MAD_0=<A_1TD_0=<A_1XD_0$, Then the circle $ \theta$ passes through points $ X,L,D_0,A$. Investing on the circumference $ \phi$ . -$ \alpha$ Invested in the line $ BC$,then $ \beta$ is invested in one circle tangent to the line $ BC$ and $ \alpha$,But $ M,A_1,D_0$, are collinear, then $ \beta$ invested in $ \beta$. if $ XM \cap \beta=T_1$, $ A_0=AM \cap BC$, $ A,X$ is invested in $ A_0,T_1$,respectively. Then, $ R^2=MT_1(MX)=ML(MA)$, then $ <T_1A_0A=<T_1XD=<XA_1T_1$, then the circle $ \mu$ passes through points $ T_1,A_0,A_1,L$. Then $ \theta$ (circumcircle of $ XAD_0$), is invested in $ \mu$ (circumcircle of $ T_1A_0A_1$), Then $ L$ is invested in $ L$, hence $ L \in \phi$, but $ \phi \cap AM=I$, then $ I=L$. Then $ I \in XA_1$. $ Q.E.D.$ In the problem. Let $ A'$ be an arbitrary point on the side $ BC$ of a triangle $ ABC$. Denote by $ T^b_A$,$ T^c_A$ the circles simultanously tangent to $ AA'$,$ A'B$,$ \gamma$ and $ AA'$,$ A'C$ ,$ \gamma$ , respectively, where $ \gamma$ is the circumcircle of $ ABC$. *) If $ T^b_A$,$ T^c_A$ are congruent. $ BC \cap T^b_A=B_1$,$ T^b_A \cap AA'=X$ $ BC \cap T^c_A=C_1$,$ T^c_A \cap AA'=Y$ Know by Lemma.($ I$:incenter of $ ABC$) $ I \in B_1X$, $ I \in C_1Y$, Then , $ B_1X \cap C_1Y =I$ moreover, $ <YC_1A'=<C_1YA'$ => $ <XA'B_1=2<C_1YA'$, then $ , then $ Now, $ M$ is mind point of $ XY$, then $ <MIY=<MYI=<A'YC_1=<YC_1A'$, then $ IM$//$ BC$, If $ O_b$ and $ O_c$ are center of $ T^b_A$, $ T^c_A$ respectively.Know that $ O_a$,$ O_b$ and $ M$ are collinear, moreover, $ O_aO_b$//$ BC$(because $ O_bB_1=O_cC_1$ and $ O_bB_1$//$ O_cC_1$) then $ I \ O_bO_c$, $ K$ is the point of tangency of the circle inscribed to $ ABC$ with $ CB$. then, $ IK=O_cC_1=O_bB_1$, moreover, $ <B_1IK=<O_cIC_1=<A'O_cC_1=\alpha$ (because $ <OB_1C_1=<B_1IC_1=90$)Then $ B_1K=(IK)Tang( \alpha)$, $ A'C_1=(O_cC_1)Tang( \alpha)$, (0< $ \alpha$<90) Then $ B_1K=A'C_1$, moreover $ BB_1=C_1C$( by simetry), Then $ A'$ is the points of tangency of the excircles of the triangle relative to $ BC$, hence $ AA'$, passes through the Nagel point of triangle $ ABC$. $ Q.E.D.$ **)If $ AA'$, passes through the Nagel point of triangle $ ABC$ Suppose the opposite,$ T^b_A$,$ T^c_A$ aren't congruent. Then Exist $ X \in BC$,$ X$ different $ A'$,such that the circles tangents to $ XC,AX, \gamma$, $ BX,AX, \gamma$, are congruent,It is true, By the continuity of $ A'$ in $ BC$, (WGL-$ AB \leq AC$), then if $ A'$ is approaching $ B$,radius of $ T^b_A$ is $ r_b$ and radius of $ T^c_A$ is $ r_c$, then $ r_b \leq r_c$, but if $ A'$ is approaching $ C$, $ r_c \leq r_b$. Then by (*), $ AX$ passes through the Nagel point of triangle $ ABC$.(-><-) next, If $ AA'$, passes through the Nagel point of triangle $ ABC$, then $ T^b_A$,$ T^c_A$ are congruent. $ Q.E.D.$

1.this come easy by trilinear or baricentric coordinates follows further easily by invertion

Let $ T_A^b$ touches $ \Gamma,\ BC,\ AA'$ at the points $ K,\ X,\ P$ respectively and $ T_A^c$ touches $ \Gamma,\ BC,\ AA'$ at the points $ L,\ Y,\ S$ respectively Let $ M$ the midpoint of the small arc $ BC$ . Then the homothety with center $ K$ which sends $ T_A^b$ to $ \Gamma$ sends $ X$ to $ M$ so $ K,X,M$ are collinear and in a similar way we can see that $ L,Y,M$ are also collinear . From Casey's theorem to the 4-tuple of circles $ (A,T_A^b,B,C)$ we have $ AP\cdot BC + BX\cdot AC = AB\cdot CX \\ \Leftrightarrow (AA' - A'X)BC + BX\cdot AC = AB(BC - BX) \\ \Leftrightarrow AA'\cdot BC - (BA' - BX)BC + BX\cdot AC = AB\cdot BC - BX\cdot AB \\ \Leftrightarrow BX(AB + BC + CA) = AB\cdot BC + BA'\cdot BC - AA'\cdot BC$ (1) In a similar way we find that $ CY(AB + BC + CA) = AC\cdot BC + CA'\cdot BC - AA'\cdot BC$ (2) We will prove first that if $ AA'$ passes through Nagel point then the circles have equal radius .Then since $ BA' = p - c$ and $ CA' = p - b$ (where $ p$ is the semiperimeter) the relation (1) and (2) become $ BX(AB + BC + CA) = pBC - AA'\cdot BC$ and $ CY(AB + BC + CA) = pBC - AA'\cdot BC$ so $ BX = CY$ . From this we obtain that $ XY$ and $ BC$ share the same midpoint and $ M$ is the midpoint of arc $ BC$ so if $ O$ is the circumcenter of $ \Gamma$ then $ OM$ is the midperpedicular of $ XY$ so $ MXY$ is an isosceles triangle and so $ \angle{KXB} = \angle{LYC}$ But also $ \angle{YLC} = \angle{XKB}$ so the triangles $ KBX,\ LYC$ are equal so $ KX = LY$ (3) . Also $ \angle{LYC} = \frac {\angle{YO_2L}}{2}$ and $ \angle{KXB} = \frac {\angle{KO_1X}}{2}$ so ${ \angle{YO_2L}} = \angle{KO_1X}$ (4) . From (3),(4) and the law of sines we get $ R_1 = R_2$ Now we will prove that if the circles have equal radius then $ AA'$ passes through Nagel point. Since the radius are equal we have that $ KX = LY$ and also $ \angle{KXB} = \angle{LYC}$ and also $ \angle{BKM} = \angle{MLC}$ so the triangles $ LYC,\ KXB$ are equal so $ BX = YC$ and from (1) and (2) we find that $ AC\cdot BC + CA'\cdot BC = AB\cdot BC + BA'\cdot BC$ and $ BA' = BC - CA'$ and so $ CA' = p - b$ so $ A'$ is the $ A$ excircle touching point , the desired. ***Note From the Thebault's theorem we obtain that the points $ O_1,\ I,\ O_2$ are collinear so $ R_1=R_2=r$ where $ r$ is the radius of the incircle !