The A-mixtilinear incircle is defined to be the circle tangent to segments [math]AB[/math] and [math]AC[/math] of triangle [math]ABC[/math] and internally tangent to the circumcircle of [math]ABC[/math]. [math]E[/math] and [math]F[/math] denote the tangency points of the mixtilinear incircle with [math]AB[/math] and [math]AC[/math] respectively.[br][br]This is a (relatively) simple proof that points [math]E[/math],[math]I[/math], and [math]F[/math] are collinear without the need for root-bc inversions or other transformations. Those are apparently more instructive though so you probably miss a few cool things.[br][br]Denote [math]G[/math] to be the center of the mixtilinear incircle, and [math]O[/math] to be the center of the [math]\left(ABC\right)[/math]. [br][math]M[/math],[math]G[/math], and [math]O[/math] are collinear because both circles are tangent to eachother (tangent line at [math]M[/math] is perpendicular to both [math]GM[/math] and [math]OM[/math]). [br][br]Consider the dilation centered at [math]M[/math] that sends [math]G[/math] to [math]O[/math]. This will take any point [math]X[/math] on the mixtilinear incircle to a point on [math]\left(ABC\right)[/math] [math]X'[/math] because [math]\frac{XG}{GM}=1[/math], and ratios of lengths are preserved so [math]\frac{X'O}{OM}=1[/math] so [math]X'[/math] lies on [math]\left(ABC\right)[/math]. [br]Let [math]M_c[/math] be the point that [math]E[/math] goes to. The line [math]GE[/math] is perpendicular to [math]AB[/math] before the transformation because [math]E[/math] is a tangency point of the mixtilinear incircle. [math]OM_c[/math] is parallel to [math]GE[/math], so is also perpendicular to [math]AB[/math]. This means that [math]M_c[/math] is the midpoint of arc [math]AB[/math] because [math]OM_c[/math] will be the perpendicular bisector of [math]AB[/math]. That also means [math]M_c[/math] is the intersection of [math]CI[/math] with [math]\left(ABC\right)[/math]. And [math]M_b[/math], defined similarly is the intersection of [math]BI[/math] with [math]\left(ABC\right)[/math].[br][br]Now at this point we could just invoke pascal's theorem and call it done. But three of our points are determined by the other three, so it seems unnecessary that we should have to use pascal's theorem.[br][br]Since [math]MEF[/math] is similar to [math]MM_cM_b[/math], [math]\angle MEF=\angle MM_cM_b=\angle MBM_b=\angle MBI[/math], so we get that if [math]I_b[/math] is the intersection of line [math]BI[/math] with [math]EF[/math], then [math]I_bEBM[/math] is cyclic. By a similar argument we get that if [math]I_c[/math] is the intersection of line [math]CI[/math] with [math]EF[/math], then [math]I_cFCM[/math] is cyclic.[br][br]Now observe that [math]M[/math] can be seen as a miquel point of triangle [math]AEF[/math]. This means that [math]\left(EBM\right)[/math] and [math]\left(FCM\right)[/math] intersect [math]EF[/math] at the same point ([math]\angle MI_bF=180-\angle MI_bE=\angle MBE=\angle MBA=180-\angle MCA=\angle180-MCF[/math], so [math]FCMI_b[/math] is cyclic)[br][br]So [math]I_b=I_c[/math][br]But we said that [math]I_b[/math] lies on line [math]BI[/math], and [math]I_c[/math] lies on line [math]CI[/math], and the only way for this to happen is if [math]I_b=I_c=I[/math].[br]As a corollary we also get that [math]EBMI[/math] and [math]FCMI[/math] are cyclic.