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### Midway

(3)

Let ABC be a triangle and P a point Denote: Na, Nb, Nc = the NPC centers of PBC, PCA, PAB, resp. NaaNabNac = the midway triangle of Na (ie Naa, Mab, Nac = the
Antreas Hatzipolakis

2:15 PM

### A theorem, well known I guess

(3)

Q( X(1), X(4614) ) = Infinity point of line X(1)X(4822): This point is now X(15309) in ETC Q( X(1), X(8851) ) = Infinity point of line X(44)X(573): This point
Antreas Hatzipolakis

10:25 AM

### N, Euler lines

(5)

In general : Let ABC be a triangle and L1, L2, L3 three arbitrary lines passing through O Denote: Aa, Ab, Ac = the orthogonal projections of A on L1, L2, L3,
Antreas Hatzipolakis

Nov 16

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### Euler lines, Parallelogic

(3)

[APH]: ORTHIC VERSION Let ABC be a triangle, A'B'C' the orthic triangle and A"B"C" the orthic triangle of A'B'C'. Denote: Aa, Ab, Ac = the orthogonal
Antreas Hatzipolakis

Nov 16

### Perpendicular bisectors, loci

(3)

[APH]: Let ABC be a triangle, P a point and A'B'C' the cevian triangle of P. Denote: A", B", C" = the reflections of P in the midpoints Ma, Mb, Mc of AA', BB',
Antreas Hatzipolakis

Nov 15

### O, NPC, circumcyclologic

(4)

[APH]: Let ABC be a triangle and A'B'C', A"B"C" the antimedial, medial triangle, resp. Denote: Na, Nb, Nc = the NPC centers of HB'C', HC'A', HA'B', resp. (H of
Antreas Hatzipolakis

Nov 13

### H, Orthologic

(14)

Let ABC be a triangle and HaHbHc, AaBbCc the pedal triangles of H,O, resp. Denote: Ab, Ac = the orthogonal projections of Ha on HB, HC, resp. Bc, Ba = the
Antreas Hatzipolakis

Nov 12

### X(15222) - X(15225)

(3)

[César Lozada]: Assuming a<=b<=c or a>=b>=c, It is possible to obtain some interesting properties for centers X(15222) - X(15225) by using the following well
Antreas Hatzipolakis

Nov 10

### NPC, Orthologic

(7)

[APH]: Let ABC be a triangle and P a point. Denote: Na, Nb, Nc = the NPC centers of PBC, PCA, PAB, resp. Naa, Nab, Nac = the orthogonal projections of Na on
Antreas Hatzipolakis

Nov 9

### A variant of Gossard perspector

(4)

[APH] Gossard Theorem: Let ABC be a triangle. The Euler line L of ABC intersects AB, AC at Ab, Ac, resp. The Euler line La of AAbAc is parallel to BC.
Antreas Hatzipolakis

Nov 8

### I, Euler lines, orthologic

(4)

Continued from Hyacinthos 26753 [César Lozada]: Sorry. I forgot to complete the
Antreas Hatzipolakis

Nov 6

### H, Euler lines, Parallelogic

(3)

[APH] Let ABC be a triangle and A'B'C' the pedal triangle of H. Denote: MaMbMc= the midheight triangle (ie Ma, Mb, Mc = the midpoints of AA', BB', CC', resp.)
Antreas Hatzipolakis

Nov 3

### Another construction of X(110) (with Brocard axes)

(3)

[Seiichi Kirikami]: Dear friends, Given a triangle ABC, its X(13) and X(14), let La, Lb and Lc be the Brocard axes of AX(13)X(14), BX(13)X(14) and CX(13)X(14).
Antreas Hatzipolakis

Oct 31

## Trending Topics

See All- Midway 3 Posts
- A theorem, well known I guess 3 Posts
- N, Euler lines 5 Posts
- Euler lines, Parallelogic 3 Posts
- Perpendicular bisectors, loci 3 Posts

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