54-A generalization Colling theorem

P. Yiu, Introduction to the Geometry of the Triangle, Florida Atlantic University Lecture Notes, 2001; with corrections, 2013, available athttp://math.fau.edu/Yiu/Geometry.html at 2.4.3 More on reflections

(1) (Colling) The reflections of a line L in the side lines of triangle ABC are concurrent if and only if L passes through the orthocenter. In this case, the intersection is a point on the circumcircle.

I generalization this result following:

Let  be a triangle,  is the orthocenter of the triangle . are projection of  to . Let  lie on such that: . Let  be any point on the plain.  are reflection of  on . Show that:  are concurrent. If  we have Colling's theorem above

Proof by Telv Cohl https://www.facebook.com/telv.cohl?fref=ufi:

Lemma:
Given triangle ABC and a point P
D is the projection of A on BC 
E is the projection of B on CA 
F is the projection of C on AB 
Pa Pb Pc is the reflection of P wrt BC CA AB
prove that PaD PbE PcF are concurrent

Proof lemma:
Since PaD is the reflection of PD wrt AD so PaD is the isogonal line of PD wrt angle FDE similarity. PbE is the isogonal line of PE wrt angle DEF. PcF is the isogonal line of PF wrt angle EFD. Hence PaD PbE PcF are concur at the isogonal conjugate of P wrt triangle DEF done

Proof theorem:

Let H be the orthocenter of triangle ABC
S1 be the intersection of A1Da and B1Db
S2 be the intersection of B1Db and C1Dc
D' be the isogonal conjugate of D wrt triangle HaHbHc
K be the pole of the simson line which is parallel to DH

We move A1 on AH and move B1 on BH accordingly so as to preserve the condition given in hypothesis the map A1 --- > B1 is homography

so the map pencil{DaA1} --- > pencil{DbB1} is homography Hence the locus of S1 is a conic through Da and Db ... (1)

When A1=A2=H, S1 coincide with H ... (2)

When A1=infinity point of the direction perpendicular to BC then B1=infinity point of the direction perpendicular to CA S1 coincide with P ... (3)

When A1=Ha B1=Hb from lemma we deduce that S1 coincide with D' ... (4)

When A1=the intersection of AH and the circumcircle of triangle ABC
then B1=the intersection of BH and the circumcircle of triangle ABC
S1 concide with K ... (5)

From (1) (2) (3) (4) (5)
we get the locus of S1 is the conic through Da Db H P D' K ... (6)
similarity we get the locus of S2 is the conic through Db Dc H P D' K ... (7)

From (6) (7) we get the locus of S1 is coincide with the locus of S2
ie. S1 is coincide with S2 so DaA1 DbB1 DcC1 concur at one point
Q.E.D