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A midpoint on the Euler line

(4)
[APH]: Let ABC be a triangle and A'B'C' the pedal triangle of O. Denote: Ab, Ac = the midpoints of AB', AC', resp. Bc, Ba = the midpoints of BC', BA', resp.
Antreas Hatzipolakis
4
posts
Oct 19

NPC, Orthologic, Euler line

(3)
[APH]: Let ABC be a triangle and A'B'C' the pedal triangle of O.. Denote: (Na), (Nb), (Nc) = the NPCs of IBC, ICA, IAB, resp. The perpendicular bisector of BC
Antreas Hatzipolakis
3
posts
Oct 18

Simson Lines

(14)
[APH]: Let ABC be a triangle, P a point, A'B'C' be the circumcevian triangle of P, A"B"C" the triangle bounded by the Simson lines of A', B', C' and O", R" the
Antreas Hatzipolakis
14
posts
Oct 17

N, Orthologic

(5)
[APH]: Let ABC be a triangle. Denote: Na, Nb, Nc = the NPC centers of NBC, NCA, NAB, resp. The perpendicular from Na to BC intersects AB, AC at Ab, Ac, resp.
Antreas Hatzipolakis
5
posts
Oct 17

Re: Euler lines, parallelogic

(5)
Loci (Antreas Hatzipolakis): Let's take the Euler line as a fixed line intersecting BC, CA, AB, resp. at A1, B1, C1, resp. Let L be a line passing through a
Antreas Hatzipolakis
5
posts
Oct 14

Concurrent perpendiculars

(9)
[APH]: Let ABC be a triangle and P,Q two isogonal conjugate points. Denote: (Na), (Nb), (Nc) = the NPCs of PBC, PCA, PAB, resp. (N1), (N2), (N3) = the NPCs of
Antreas Hatzipolakis
9
posts
Oct 13

X(110)

(3)
[APH]: Let ABC be a triangle. The Euler line intersects BC, CA, AB at A', B', C', resp. The perpendicular to BC at A' intersects AB, AC at Ab, Ac, resp. The
Antreas Hatzipolakis
3
posts
Oct 6

A centroid on the Euler line of the orthic triangle

(3)
[APH]: Orthic version: Let ABC be a triangle and A'B'C', A"B"C" the cevian triangles of H,O, resp. Denote: Ma, Mb, Mc = the midpoints of HA", HB", HC", resp.
Antreas Hatzipolakis
3
posts
Oct 5

Reflections, midpoints, perspective, orthocenter on OI line

(3)
[APH]: Let ABC be a triangle and IaIbIc the antipedal triangle of I (excentral triangle). Denote: Bc, Cb = the reflections of B, C in CI, BI, resp. Ca, Ac =
Antreas Hatzipolakis
3
posts
Oct 3

Circles with diameters BC,CA,AB

(3)
[APH]: Let ABC be a triangle and (Oa), (Ob), (Oc) the circles with diameters BC, CA, AB, resp. Questions: 1. Which are the centers of the circles touching
Antreas Hatzipolakis
3
posts
Sep 29

Medialtial

(4)
[APH] Hyacinthos 253 Definition: Mi : the midpoints of sides BC,CA,AB, resp. Hi :
Antreas Hatzipolakis
4
posts
Sep 24

Rectangles erected on sides of a triangle

(3)
[APH]: Let ABC be a triangle. Let BCCaBa, CAAbCb, ABBcAc be three arbitrary rectangles erected on the sides BC, CA, AB, resp. outwardly ABC. The perpendicular
Antreas Hatzipolakis
3
posts
Sep 22

Bicentric points

(4)
[APH] Let ABC be a triangle. The perpendicular to AB at A intersects BC at A1 The perpendicular to BC at B intersects CA at B1 The perpendicular to CA at C
Antreas Hatzipolakis
4
posts
Sep 12

I, Reflected Euler lines

(7)
[APH]: Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P. Denote: La, Lb, Lc = the Euler lines of PBC, PCA, PAB, resp. L1, L2, L3 = the
Antreas Hatzipolakis
7
posts
Sep 12

N, Reflected Euler lines

(5)
[APH]: Let ABC be a triangle, P a point and A'B'C' the pedal triangle of P. Denote: La, Lb, Lc = the Euler lines of AB'C', BC'A', CA'B', resp. L1, L2, L3 = the
Antreas Hatzipolakis
5
posts
Sep 9
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