CONCURRENT CIRCLES -- EULER LINE

Theorem:

Let ABC be a triangle and A',B',C' three points. If the circumcircles of A'BC, B'CA, C'AB are concurrent, then also the circumcircles of AB'C', BC'A', CA'B' are concurrent.

Corollary:

Let ABC be a triangle and P a point. If A',B',C' are arbitrary points on the circumcircles of PBC,PCA,PAB, resp. then the circumcircles of AB'C', BC'A', CA'B' are concurrent.

Applications:

Let P,Q be two points and PaPbPc, Q1Q2Q3 the antipedal, pedal triangles of P,Q, resp. The orthogonal projections A',B',C' of Pa,Pb,Pc on PQ1,PQ2,PQ3, resp. lie on the circumcircles of PBC,PCA,PAB, resp.

The circumcircles of AB'C', BC'A', CA'B' concur at a point D.

1. For P = O, Q = H:

The point D lies on the Euler line of ABC

2. For P = Q = I:

The point D lies on the Euler line of ABC

3. For P = Q = H:

The Euler lines L,L1,L2,L3 of ABC, DBC,DCA,DAB are concurrent at a point D' on the Neuberg cubic.

The parallels to L1,L2,L3 through A,B,C, resp. are concurrent at a point D"

Coordinates of the points D's, D' and D"?

More pairs (P,Q) such that the circumcircles concur on the Euler line?

Antreas P. Hatzipolakis, 11 April 2014

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It seems that the general case for D has truly appalling barycentrics.

But, for the cases you mention we have:

1) PQ = OH, D= X(186)

2) PQ = II, D = X(1325)

3) PQ = HH, D = X(1157)

Also PQ = OI, D = X(36)

Peter Moses, 14 April 2014