Difference between revisions of "Steiner line"

Revision as of 10:34, 12 December 2022

Steiner line

Let $ABC$ be a triangle with orthocenter $H. P$ is a point on the circumcircle $\Omega$ of $\triangle ABC.$

Let $P_A, P_B,$ and $P_C$ be the reflections of $P$ in three lines which contains edges $BC, AC,$ and $AB,$ respectively.

Prove that $P_A, P_B, P_C,$ and $H$ are collinear. Respective line is known as the Steiner line of point $P$ with respect to $\triangle ABC.$

Proof

Let $D, E,$ and $F$ be the foots of the perpendiculars dropped from $P$ to lines $AB, AC,$ and $BC,$ respectively.

WLOG, Steiner line cross $AB$ at $Y$ and $AC$ at $Z.$

The line $DEF$ is Simson line of point $P$ with respect of $\triangle ABC.$

$D$ is midpoint of segment $PP_C \implies$ homothety centered at $P$ with ratio $2$ sends point $D$ to a point $P_C.$

Similarly, this homothety sends point $E$ to a point $P_B$ , point $F$ to a point $P_A,$ therefore this homothety send Simson line to line $P_AP_BP_C.$

Let $\angle ABC = \beta, \angle BFD = \varphi \implies \angle BDF = \beta – \varphi.$ $P_CP_A||DF \implies \angle P_CYB = \beta – \varphi.$ $P$ is simmetric to $P_C \implies \angle PYD = \beta – \varphi.$

Quadrungle $BDPF$ is cyclic $\implies \angle BPD = \varphi \implies \angle BPY = 90^\circ – \angle BYP – \angle BPD = 90^\circ – \beta.$

$\angle BCH = \angle BPY \implies PY \cap CH$ at point $H_C \in \Omega.$ Similarly, line $BH \cap PZ$ at $H_B \in \Omega.$

According the Collins Claim $YZ$ is $H-line,$ therefore $H \in P_AP_B.$

Simson line

vladimir.shelomovskii@gmail.com, vvsss

Collings Clime

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 $P.$

Usually the point $P$ is called the anti-Steiner point of the $H-line$ with respect to $\triangle ABC.$

vladimir.shelomovskii@gmail.com, vvsss

Ortholine

Let four lines made four triangles of a complete quadrilateral.

In the diagram these are $\triangle ABC, \triangle ADE, \triangle CEF, \triangle BDF.$

Let points $H, H_A, H_B,$ and $H_C$ be the orthocenters of $\triangle ABC, \triangle ADE, \triangle BDF,$ and $\triangle CEF,$ respectively.

Prove that points $H, H_A, H_B,$ and $H_C$ are collinear.

Proof

Let $M$ be Miquel point of a complete quadrilateral.

Line $KLMN$ is the line which contain $4$ Simson lines of $4$ triangles.

Using homothety centered at $M$ with ratio $2$ we get $4$ coinciding Stainer lines which contain points $H, H_A, H_B,$ and $H_C$ .

Proof 2

$AH_A \perp DE, CH_C \perp EF \implies AH_A ||CH_C,$

$AH \perp BC, EH_C \perp CF \implies AH ||EH_C,$

$EH_A \perp AD, CH \perp AB \implies EH_A ||CH.$

Points $A, E,$ and $C$ are collinear.

According the Claim of parallel lines, points $H, H_A,$ and $H_C$ are collinear.

Similarly points $H, H_B,$ and $H_C$ are collinear as desired.

Claim of parallel lines

Let points $A, B,$ and $C$ be collinear.

Let points $D, E, F$ be such that $AF||CD, BF||CE, AE||BD.$

Prove that points $D, E,$ and $F$ are collinear.

Proof

Let $P = AE \cap CD, Q = AF \cap CE.$

$\angle CEP = \angle QEA, AQ||CP \implies \angle QAE = \angle CPE \implies$ $\triangle AEQ \sim \triangle PEC.$

$AP||BD \implies \frac {PD}{CD} = \frac {AB}{BC},$

$CQ||BF \implies \frac {AF}{QF} = \frac {AB}{BC} = \frac {PD}{CD}.$

The segments $EF$ and $ED$ are corresponding segments in similar triangles. Therefore $\angle CED = \angle QEF \implies D, E,$ and $F$ are collinear.

vladimir.shelomovskii@gmail.com, vvsss

@@ Line 95: / Line 95: @@
 <i><b>Claim of parallel lines</b></i>
-[[File:Pras 1 12.png|400px|right]]
-Let points <math>A, B,</math> and <math>C</math> are collinear.
+Let points <math>A, B,</math> and <math>C</math> be collinear.
-Let points <math>D, E, F</math> are such that <math>AF||CD, BF||CE, AE||BD.</math>
+Let points <math>D, E, F</math> be such that <math>AF||CD, BF||CE, AE||BD.</math>
 Prove that points <math>D, E,</math> and <math>F</math> are collinear.
 <i><b>Proof</b></i>
+[[File:Pras 1 12.png|400px|right]]
 Let <math>P = AE \cap CD, Q = AF \cap CE.</math>
-<math>\angle CEP = \angle QEA, AQ||CP \implies \angle QAE = \angle CPE \implies \triangle AEQ \sim \triangle PEC.</math>
+<cmath>\angle CEP = \angle QEA, AQ||CP \implies \angle QAE = \angle CPE \implies</cmath>
+<cmath>\triangle AEQ \sim \triangle PEC.</cmath>
-<math>AP||BD \implies \frac {PD}{CD} = \frac {AB}{BC},</math>
+<cmath>AP||BD \implies \frac {PD}{CD} = \frac {AB}{BC},</cmath>
-<math>CQ||BF \implies \frac {AF}{QF} = \frac {AB}{BC} = \frac {PD}{CD}.</math>
+<cmath>CQ||BF \implies \frac {AF}{QF} = \frac {AB}{BC} = \frac {PD}{CD}.</cmath>
-The segments <math>EF</math> and <math>ED</math> are corresponding segments in similar triangles <math>\implies \angle CED = \angle QEF \implies D, E,</math> and <math>F</math> are collinear.
+The segments <math>EF</math> and <math>ED</math> are corresponding segments in similar triangles.
+Therefore <math>\angle CED = \angle QEF \implies D, E,</math> and <math>F</math> are collinear.
 '''vladimir.shelomovskii@gmail.com, vvsss'''

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Difference between revisions of "Steiner line"

Revision as of 10:34, 12 December 2022

Steiner line

Collings Clime

Ortholine