Point secAsec(B-C) :: (in trilinears)

Let ABC be a triangle and H its orthocenter.

Define

Ab : the intersection of the parallel from C to AH and the parallel from A to CH (ie AHCAb is parallelogram)

Ac : the intersection of the parallel from B to AH and the parallel from A to BH (ie AHBAc is parallelogram)

The circle (B, BAc) intersects AB at Ac1 (between A,B) and Ac2 and the circle (C, CAb) intersects AC at Ab1 (between A,C) and Ab2.

The lines Ab1Ac2 and Ac1Ab2 intersect at A1.

Similarly the points B1, C1.

The triangles ABC, A1B1C1 are perspective (ie the lines AA1,BB1,CC1 are concurrent).

Proof

We have:

sin(A1AB) / sin(A1AC) = (AAb1 / AAc1).(AAb2 / AAc2).(Ac1Ac2 / Ab1Ab2)
(See THIS)

CAb1 = CAb2 = BAc1 = BAc2 = AH = 2RcosA

AAb1 = AC - CAb1 = 2R(sinB - cosA)

AAc1 = AB - BAc1 = 2R(sinC - cosA)

AAb2 = AC + CAb2 = 2R(sinB + cosA)

AAc2 = AB + BAc2 = 2R(sinC + cosA)

Ac1Ac2 = Ab1Ab2 [= 4RcosA]

==>


sin(A1AB)/sin(A1AC) = [(sinB - cosA)/(sinC - cosA)].[(sinB + cosA)/(sinC + cosA)] =

= ((sinB)^2 - (cosA)^2)/((sinC)^2 - (cosA)^2) =

= (1 - [(cosB)^2 + (cosA)^2]) / (1 - [(cosC)^2 + (cosA)^2])

Cyclically:

sin(B1BC) / sin(B1BA) = ...

sin(C1BA) / sin(C1CB) = ...

By multiplying them we get 1, therefore the lines AA1,BB1,CC1 are concurrent.

The trilinears of the point of concurrence (perspector of the triangles) are:

( 1 / (1 - [(cosB)^2 + (cosC)^2]) :: = 1 / (cos2B + cos2C) :: =

= 1 / cosAcos(B-C) :: = secAsec(B-C) ::

Note: We used figure of acute triangle. The case of non-acute triangle is left to the reader.

CORRECTION:

The trilinears of the point of concurrence (perspector of the triangles) are:

((1 - [(cosB)^2 + (cosC)^2]) :: = (cos2B + cos2C) :: =

= cosAcos(B-C) ::

See the Hyacinthos discussion HERE