Editing Orthocentric system (section)

==Euler lines and homothetic orthocentric systems==

[[File:Orthocentric system and their circumcenters.PNG|thumb|'''Orthocentric system'''. Where {{math|''O''{{sub|1}}, ''O''{{sub|2}}, ''O''{{sub|3}}, ''O''{{sub|4}}}} are the circumcenters of the four possible triangles formed from the orthocentric points {{math|''A''{{sub|1}}, ''A''{{sub|2}}, ''A''{{sub|3}}, ''A''{{sub|4}}}}.]]

Let [[vector (geometry)|vectors]] {{math|'''a''', '''b''', '''c''', '''h'''}} determine the position of each of the four orthocentric points and let {{math|1='''n''' = ('''a''' + '''b''' + '''c''' + '''h''') / 4}} be the position vector of {{mvar|N}}, the common nine-point center. Join each of the four orthocentric points to their common nine-point center and extend them into four lines. These four lines now represent the Euler lines of the four possible triangles where the extended line {{mvar|HN}} is the Euler line of triangle {{math|△''ABC''}} and the extended line {{mvar|AN}} is the [[Euler line]] of triangle {{math|△''BCH''}} etc. If a point {{mvar|P}} is chosen on the Euler line {{mvar|HN}} of the reference triangle {{math|△''ABC''}} with a position vector {{math|'''p'''}} such that {{math|1='''p''' = '''n''' + α('''h''' – '''n''')}} where {{math|α}} is a pure constant independent of the positioning of the four orthocentric points and three more points {{mvar|P{{sub|A}}, P{{sub|B}}, P{{sub|C}}}} such that {{math|1='''p{{sub|a}}''' = '''n''' + α('''a''' – '''n''')}} etc., then {{mvar|P, P{{sub|A}}, P{{sub|B}}, P{{sub|C}}}} form an orthocentric system. This generated orthocentric system is always [[Homothetic transformation|homothetic]] to the original system of four points with the common nine-point center as the homothetic center and α the ratio of [[:wikt:similitude|similitude]].

When {{mvar|P}} is chosen as the centroid {{mvar|G}}, then {{math|1=α = –⅓}}. When {{mvar|P}} is chosen as the [[circumcenter]] {{mvar|O}}, then {{math|1=α = –1}} and the generated orthocentric system is [[congruence (geometry)|congruent]] to the original system as well as being a reflection of it about the nine-point center. In this configuration {{mvar|P{{sub|A}}, P{{sub|B}}, P{{sub|C}}}} form a [[Johnson circles|Johnson triangle]] of the original reference triangle {{math|△''ABC''}}. Consequently the [[circumcircle]]s of the four triangles {{math|△''ABC'', △''ABH'', △''ACH'', △''BCH''}} are all equal and form a set of [[Johnson circles]] as shown in the diagram adjacent.