Re: A circle tangent to the incircle
> >> Let ABC be a triangle, and A'B'C', A"B"C" the pedal triangles[Francisco Javier]
> >> of H,I, resp. The circle (A,AB"=AC") intersects AA' at A*
> >> between A,A'. Denote A1 := BC /\ IA*. The circle with
> >> diameter AA1 touches the incircle (I) at Ta. Similarly Tb, Tc.
> >> [extraversion]:
> >> Let ABC be a triangle, A'B'C', A"aA"bA"c the pedal
> >> triangles of H, Ia, resp.
> >> The circle (A, A"b = A"c) intersects AA' at A**, with A'
> >> between A and A**.
> >> Denote A2 := BC /\ IaA**. The circle with diameter AA2
> >> touches externally the excircle (Ia) at Sa.
> >> Similarly Sb, Sc.
> > The triangles TaTbTc is perspective with ABC at X1336.[APH]
> > If XYZ is the cevian triangle of X1336, then Sa and Ta are
> > harmonic conjugates with respect A,X; thus we have that SaSbSc
> > is perspective with ABC at X1336 also.
> So we have one theorem instead of two!Now, let T1T2T3 be the triangle bounded by the tangents to Incircle
> TaTbTc, SaSbSc are perspective at X(1336)
(I) at Ta,Tb,Tc, and S1S2S3 the triangle bounded by the tangents
to excircles (Ia),(Ib),(Ic) at Sa, Sb, Sc, resp.
Is the Orthic triangle A'B'C' perspective with T1T2T3 and