Raja Oktovin wrote: Given triangle $ ABC$ with $ AB > AC$. $ l$ is tangent line of the circumcircle of triangle $ ABC$ at $ A$. A circle with center $ A$ and radius $ AC$, intersect $ AB$ at $ D$ and $ l$ at $ E$ and $ F$. Prove that the lines $ DE$ and $ DF$ pass through the incenter and excenter of triangle $ ABC$. Let I be an intersection of ED and the bisector of angle C. Since $ EA=AD, \angle EAD=180^o-C$ we have $ \angle AED=\frac{\angle C}{2}=\angle ACI$ Then $ AECI$ is cyclic we get $ \angle AIC=\angle AIE+\angle EIC=\angle ACE+\angle EAC$ $ =\angle ABC+\frac{180^o-\angle EAC}{2}=90^o+\frac{\angle ABC}{2}$ So I is the incenter of triangle ABC Let J is an intersection of FD and the external bisector of angle C We have $ \angle ICJ=\angle IDJ=90^o$ hence $ CIDJ$ is cyclic We obtain $ \angle DJI=\angle ICD (1)$ On the other hand, $ \angle ADE=1/2C= \angle ICB$ then $ CIDB$ is cyclic So $ \angle ICD=\angle IBD (2)$ From (1) and (2) we have $ \angle IJD=\angle IBD$ Then $ IDBJ$ is cyclic we get $ \angle IBJ=90^o$ Therefore J is the excenter of triangle $ ABC$

Raja Oktovin wrote: Let $ ABC$ be a triangle with $ c>b$ , the incenter $ I$ and the $ A$ - exincenter $ I_a$ . Denote $ D\in (AB)$ for which $ AD=b$ . Consider $ E$ , $ F$ which belong to the tangent in $ A$ at the circumcircle of $ \triangle ABC$ so that $ AE=AF$ and $ AB$ separates $ I$ , $ F$ . Prove that $ E\in DI$ and $ F\in DI_a$ . Proof. $ \odot\ AD=AE=AF\ \implies\ DE\perp DF$ . Observe that $ \left\|\begin{array}{ccc} CD=CF & \implies & m(\widehat {ADF})=90^{\circ}-\frac C2\\\\ AD=AC & \implies & m(\widehat{ADI})=m(\widehat {ACI})=\frac C2\end{array}\right\|\ \bigoplus\ \implies$ $ \ m(\widehat {FDI})=$ $ m(\widehat{ADF})+m(\widehat {ADI})=90^{\circ}\ \implies\ DI\perp DF$ . Thus, $ DE\perp DF\ \wedge\ DI\perp DF\ \implies\ E\in DI$ . $ \odot\ SD=SC=SB=SI=SI_a\ \implies$ the point $ D$ belongs to the circle with diameter $ II_a\ \implies\ DI_a\perp DI\ \stackrel{(DI\perp DF)}{\ \implies\ }\ F\in DI_a$ . Remark. Prove easily that $ \frac {ID}{IE}=\frac {a}{b+c}$ and $ \frac {I_aD}{I_aF}=\frac ap$ .