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### 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

5:48 AM

### 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

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

Oct 17

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### 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

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

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

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

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

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

Sep 29

Antreas Hatzipolakis

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

Sep 22

### Collinear points, Coaxial circles

(3)

[APH]: Let ABC be a triangle, P a point and A1B1C1 the cevian triangle of P. Let K be the point of concurrence of the circumcircles of AB1C1, BC1A1, CA1B1,
Antreas Hatzipolakis

Sep 20

### 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

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

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

Sep 9

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## Trending Topics

See All- NPC, Orthologic, Euler line 3 Posts
- Simson Lines 14 Posts
- N, Orthologic 5 Posts
- Re: Euler lines, parallelogic 5 Posts
- Concurrent perpendiculars 9 Posts

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