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Difference between revisions of "Steiner line"

(Created page with "==Steiner line== Let <math>ABC</math> be a triangle with orthocenter <math>H. S</math> is a point on the circumcircle <math>\Omega</math> of <math>\triangle ABC.</math> Then,...")
 
(→‎Collings Clime)
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Let <math>H_A, H_B,</math> and <math>H_C</math> be the points symmetric to <math>H</math> with respect <math>BC, AC,</math> and <math>AB,</math> respectively.
 
Let <math>H_A, H_B,</math> and <math>H_C</math> be the points symmetric to <math>H</math> with respect <math>BC, AC,</math> and <math>AB,</math> respectively.
  
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Therefore <math>H_A \in l_A, H_B \in l_B, H_C \in l_C, AH = AH_B = AH_C, BH = BH_A = BH_C, CH = CH_A = CH_B \implies</math>  
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Therefore <math>H_A \in l_A, H_B \in l_B, H_C \in l_C,</math>
 +
<cmath>AH = AH_B = AH_C, BH = BH_A = BH_C, CH = CH_A = CH_B \implies</cmath>  
 
<cmath>\angle HH_BE = \angle EHH_B = \angle BHD = \angle BH_CD.</cmath>
 
<cmath>\angle HH_BE = \angle EHH_B = \angle BHD = \angle BH_CD.</cmath>
  
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Let <math>P</math> be the crosspoint of  <math>l_B</math> and <math>l_C  \implies BH_CH_BP</math> is cyclic <math>\implies P \in \omega.</math>
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Let <math>P</math> be the crosspoint of  <math>l_B</math> and <math>l_C  \implies BH_CH_BP</math> is cyclic <math>\implies P \in \Omega.</math>
  
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Similarly <math>\angle CH_BE = \angle CHE = \angle CH_A \implies CH_BH_AP</math> is cyclic <math>\implies P \in \omega \implies</math> the crosspoint of  <math>l_B</math> and <math>l_A</math> is point <math>P.</math>
+
Similarly <math>\angle CH_BE = \angle CHE = \angle CH_A \implies CH_BH_AP</math> is cyclic <math>\implies P \in \Omega \implies</math> the crosspoint of  <math>l_B</math> and <math>l_A</math> is point <math>P.</math>
  
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''

Revision as of 01:40, 7 December 2022

Steiner line

Let $ABC$ be a triangle with orthocenter $H. S$ is a point on the circumcircle $\Omega$ of $\triangle ABC.$ Then, the reflections of $S$ in three edges $BC, CA, AB$ and point $H$ lie on a line $s$ which is known as the Steiner line of point $S$ with respect to $\triangle ABC.$

Collings Clime

Steiner H line.png

Let triangle $ABC$ be the triangle with the orthocenter $H$ and circumcircle $\Omega.$ Denote $H–line$ any line containing point $H.$

Let $l_A, l_B,$ and $l_C$ be the reflections of $H-line$ in the edges $BC, AC,$ and $AB,$ respectively.

Prove that lines $l_A, l_B,$ and $l_C$ are concurrent and the point of concurrence lies on $\Omega.$

Proof

Let $D, E,$ and $F$ be the crosspoints of $H–line$ with $AB, AC,$ and $BC,$ respectively.

WLOG $D \in AB, E \in AC.$ Let $H_A, H_B,$ and $H_C$ be the points symmetric to $H$ with respect $BC, AC,$ and $AB,$ respectively.

Therefore $H_A \in l_A, H_B \in l_B, H_C \in l_C,$ \[AH = AH_B = AH_C, BH = BH_A = BH_C, CH = CH_A = CH_B \implies\] \[\angle HH_BE = \angle EHH_B = \angle BHD = \angle BH_CD.\]

Let $P$ be the crosspoint of $l_B$ and $l_C \implies BH_CH_BP$ is cyclic $\implies P \in \Omega.$

Similarly $\angle CH_BE = \angle CHE = \angle CH_A \implies CH_BH_AP$ is cyclic $\implies P \in \Omega \implies$ the crosspoint of $l_B$ and $l_A$ is point $P.$

vladimir.shelomovskii@gmail.com, vvsss

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