Dear friends,

[1] Given a triangle ABC right in A, its
incircle(I), its circumcircle(O), and its excircles (Ia), (Ib), (Ic) with their
centers I, O, Ia, Ib, Ic, denote:

S, Q, R= the contact points of (I) with BC,
CA, AB respectively,

Aa, Ba, Ca=the contact points of (Ia) with
BC, CA, AB respectively,

Ab, Bb, Cb=the contact points of (Ib) with
BC, CA, AB respectively,

Ac, Bc, Cc=the contact points of (Ic) with
BC, CA, AB respectively,

B’, C’=the midpoints of CA, AB
respectively,

B’’, C’’=the intersections of B’C’ with (O)
respectively.

8 points B’’, Q, R, C’’, Ac, Ca, Ba, and Ab are on a circle
(1-circle).

8 points B’’, Bb, Aa, S, Cc, C’’, Bc, and Cb are on another circle
(2-circle).

See the attached pictures.

This is an extension of Le Probleme de
Toshio Seimiya in Quelques Theoremes Oublies, vol. 1 (2007), Geometrie by
Jean-Louis Ayme.

[2] Given a triangle ABC right in A, its excircles
(Ia), (Ib), (Ic) with their centers Ia, Ib, Ic, and its angle bisectors
(A-bsc), (B-bsc), (C-bsc), denote:

B’, C’=the midpoints of CA, AB
respectively,

Ka, Kb, Kc=the intersections farthest from
the sides of ABC of (A-bsc), (B-bsc), (C-bsc) with (Ia), (Ib), (Ic)
respectibvely,

Jb, Jc=the intersections nearest to the
sides of ABC of (B-bsc), (C-bsc) with (Ib), (Ic),

Mb, Mc=the intersections farthest from BC
of (Ia) with the angle bisector of CBCa, BCBa respectively,

Lb, Lc=the intersections nearest to BC of
(Ia) with the angle bisector of CBCa, BCBa respectively.

Kb, B’, Lb are on a line (1-line),

Kc, C’, Lc are on a line (2-line),

Mc, C’, Jc are on a line (3-line),

Mb, B’, Jb are on a line (4-line),

P= the intersection of 1-line and 2-line,

Q=the intersection of 3-line and 4-line.

5.P, Q, Ka are on a line (5-line).

See the attached pictures.

This is the continuation of Anopolis message #817.

Best regards,

Seiichi.