Angle bisector in touching circles:
This result could also be proved by considering the second point of intersection K of TP with the larger circle. By considering the scaling about T taking the smaller to the larger circle, The tangent to the larger circle at K is parallel to AB, so K is the midpoint of the arc AB, and so TK bisects ATB.
The generalisation
can actually be extended even further. Given two circles that cross at S and T, let a chord AB of one intersect the other in P and Q. Then ASP = AST - PST = (π - ABT) - (π - PQT) = QTB.

Three parallels in a triangle:
This can be proved in a slightly simpler manner by considering the symmetry of the figure.
EK is the reflection of BC in the bisector of angle A, so the point X where these lines meet is the intersection of BC with that bisector: It'satisfies XB:XC = c:b = BD:CL, so XD:XL = (BD - XB):(CL - XC) = c:b = XE:XK, so DE and KL are parallel.

A property of the line IO:
This has a neat proof by transformation geometry. D is the reflection of A in BI, so DI = AI and AID = 2 AIB = 2(π - A/2 - B/2) = 2π - (π - C) = π + C. So letting f be the operation of rotation about I by π + C, D = f(A). Similarly, B = f(E). Let g be the operation of rotation about O by -2C. Then g(B) = A, so fgf(E) = D. But fgf rotates the plane by a total angle of (&pi + C) -2C + (π + C) = 2π, so fgf is a translation by the vector D - E. Let E' = f^-1(O), and let D' = E' + D - E = fgf(E') = fg(O) = f(O). Then DE is parallel and equal in length to D'E', which is perpendicular to IO and of length |2 IO sin(π + C)| = 2 IO sin(C) = IO when C = 30 degrees.
This also throws some light on the related theorem mentioned elsewhere. Let O' be the reflection of O in CI. Then OO' is at right angles to the bisector CI of ACB, so to the internal bisector of NOM, so it is the external bisector of that angle. So O' is the midpoint of the longer arc MN and MIO' = O'IN = MIN/2 = (2π - 2C)/2 = π - C. So O'IM = NIO' = π + C, and so M and N are the reflections of D' and E' in CI. Thus MN is equal and antiparallel to D'E', which is equal and parallel to DE. Thus MN, NO and OM are antiparallel to ED, DC and CE respectively, so MNO and EDC are similar. As also MN = ED, they are congruent.

A sangaku: Two unrelated circles:
Another approach is to start from the smaller circle. Take coordinates such that it is the unit circle, with Q on the x axis. Let G = (g, h), W = (u, v). Then:
QV has equation xg + yh = 1.
Q is at (1/g, 0).
O is at (1/g, v/ug) = W/ug.
V satisfies xg + yh = 1 and (x - 1/g)^2 + (y - v/gu)^2 = (1 - 1/ug)^2
So y^2 - 2ygv/u + 2g/u - g^2 - 1 = 0
So QR = -y = -gv/u - sqrt(g^2/u^2 - 2g/u + 1) = -gv/u + 1 - g/u = 1 - g(1+v)/u
and PQ = 1 - 1/ug + v/ug = 1 - (1-v)/ug.
So (PQ - 1)(QR - 1) = (1-v)(1+v)/u^2 = 1.
Equivalently, 1/PQ + 1/QR = 1 = 1/q.

A median in touching circles:
Apply Menelaus with respect to triangle ABC.

Dissection of a cyclic quadrilateral
"for any P, points M, N, Q, R always exist on the side lines of ABCD such the quadrilaterals AQPR, DNPQ, CMPN, and BRPM are cyclic." Just let M, N, Q, R be the bases of the perpendiculars from P to the side lines.
"And for which P the set of N's that answers that question is not empty?"
The rays PM, PN, PQ and PR are at fixed angles to one another, and must lie inside the angles CPB, DPC, APD and BPA respectively. Say AD and BC intersect at X, and AB and CD intersect at Y. Without loss of generality, A lies between X and D and B lies between A and Y. We must have BPA <= MPQ = π - AXB, so P lies outside the circumcircle of ABX. Similarly P lies outside the circumcircle of BCY. Suppose we are given any P lying outside these two circles.
For any choices of M on BC and R on AB, let Q be the point on DA with MPQ = π - AXB, and N the unique point on CD with NPR = π - BYC. We say a choice of M is suitable if both of M and Q lie in the interiors of the respective sides of the quadrilateral ABCD; the definition of suitability for R is similar. Now choose M' and R' as close to B as possible so that both choices are suitable. At most one of M', R' is not B; without loss of generality M' Then M'PR' <= M'PQ' - BPD <= π - CXD - XCD = CDA. On the other hand, choose M'' and R'' as far from B as possible so that both choices are suitable. We have either M'' = C or Q'' = A, and also R'' = A or N'' = C. This gives 4 possible values for M''PR'': CPA, M''PQ'', N''PR'' or M''PQ'' + N''PR'' - CPA, and in all of these cases M''PR'' >= CDA (The last case is least obvious. We have M''PQ'' + N''PR'' - CPA >= (π - BXD) + (π - XBY) - (π - XBY) = XDY = CDA). So we may choose M between M' and M'' and R between R' and R'' so that MPR = CDA, and then M, N, Q and R meet all of the conditions in the problem.
To summarise, a necessary and sufficient condition is that P should lie outside the circumcircles of ABX and BXY.

Projections on internal and external angle bisectors:
P, N and M are collinear since they lie on the Simson line of A with respect to BII_C.

Thankyou

sfwc
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Thank you on the whole; but where you've been all this time? There is so much to do now.

Alex

Thankyou

sfwc
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