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Difference between revisions of "Simson line"
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==Simson line (main)==
 
==Simson line (main)==
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[[File:Simson line.png|300px|right]]
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[[File:Simson line.png|270px|right]]
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[[File:Simson line inverse.png|300px|right]]
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[[File:Simson line inverse.png|270px|right]]
 
Let a triangle <math>\triangle ABC</math> and a point <math>P</math> be given.
 
Let a triangle <math>\triangle ABC</math> and a point <math>P</math> be given.
  
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Let the points  <math>D, E,</math> and <math>F</math> be collinear.  
 
Let the points  <math>D, E,</math> and <math>F</math> be collinear.  
  
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<math>AEPD</math> is cyclic <math>\implies \angle APE = \angle ADE, \angle APE = \angle BAC.</math>  
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<math>AEPD</math> is cyclic <math>\implies \angle APE = \angle ADE, \angle DPE = \angle BAC.</math>  
  
 
<math>BFDP</math> is cyclic <math>\implies \angle BPF = \angle BDF, \angle DPF = \angle ABC.</math>
 
<math>BFDP</math> is cyclic <math>\implies \angle BPF = \angle BDF, \angle DPF = \angle ABC.</math>
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<math>ACBP</math> is cyclis as desired.
 
<math>ACBP</math> is cyclis as desired.
  
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'''vladimir.shelomovskii@gmail.com, vvsss'''
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==Simson line of a complete quadrilateral==
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[[File:Simson complite.png|430px|right]]
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Let four lines made four triangles of a complete quadrilateral. In the diagram these are <math>\triangle ABC, \triangle ADE, \triangle CEF, \triangle BDF.</math>
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Let <math>M</math> be the Miquel point of a complete quadrilateral.
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Let <math>K, L, N,</math> and <math>G</math> be the foots of the perpendiculars dropped from <math>M</math> to lines <math>AB, AC, EF,</math> and <math>BC,</math> respectively.
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Prove that points <math>K,L, N,</math> and <math>G</math> are collinear.
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<i><b>Proof</b></i>
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Let <math>\Omega</math> be the circumcircle of <math>\triangle ABC, \omega</math> be the circumcircle of <math>\triangle CEF.</math> Then <math>M = \Omega \cap \omega.</math>
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Points <math>K, L,</math> and <math>G</math> are collinear as Simson line of <math>\triangle ABC.</math>
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Points <math>L, N,</math> and <math>G</math> are collinear as Simson line of <math>\triangle CEF.</math>
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Therefore points <math>K, L, N,</math> and <math>G</math> are collinear, as desired.
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*[[Miquel's point]]
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*[[Steiner line]]
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
  
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<math>PD \perp OO_0, PE \perp OO_1, PF \perp O_0O_1 \implies</math>
 
<math>PD \perp OO_0, PE \perp OO_1, PF \perp O_0O_1 \implies</math>
 
<math>DEF</math> is Simson line of <math>\triangle OO_0O_1 \implies P</math> lies on circumcircle of <math>\triangle OO_0O_1</math> as desired.  
 
<math>DEF</math> is Simson line of <math>\triangle OO_0O_1 \implies P</math> lies on circumcircle of <math>\triangle OO_0O_1</math> as desired.  
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*[[Euler line]]
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''

Latest revision as of 08:28, 5 August 2024

In geometry, given a triangle ABC and a point P on its circumcircle, the three closest points to P on lines AB, AC, and BC are collinear. Simsonline.png

Simson line (main)

Simson line.png
Simson line inverse.png

Let a triangle $\triangle ABC$ and a point $P$ be given.

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

Then points $D, E,$ and $F$ are collinear iff the point $P$ lies on circumcircle of $\triangle ABC.$

Proof

Let the point $P$ be on the circumcircle of $\triangle ABC.$

$\angle BFP = \angle BDP = 90^\circ \implies$

$BPDF$ is cyclic $\implies \angle PDF = 180^\circ – \angle CBP.$

$\angle ADP = \angle AEP = 90^\circ \implies$

$AEPD$ is cyclic $\implies \angle PDE = \angle PAE.$

$ACBP$ is cyclic $\implies \angle PBC = \angle PAE \implies \angle PDF + \angle PDE = 180^\circ$

$\implies D, E,$ and $F$ are collinear as desired.

Proof

Let the points $D, E,$ and $F$ be collinear.

$AEPD$ is cyclic $\implies \angle APE = \angle ADE, \angle DPE = \angle BAC.$

$BFDP$ is cyclic $\implies \angle BPF = \angle BDF, \angle DPF = \angle ABC.$

$\angle ADE = \angle BDF \implies \angle BPA = \angle EPF$

$= \angle BAC + \angle ABC = 180^\circ – \angle ACB \implies$

$ACBP$ is cyclis as desired.

vladimir.shelomovskii@gmail.com, vvsss

Simson line of a complete quadrilateral

Simson complite.png

Let four lines made four triangles of a complete quadrilateral. In the diagram these are $\triangle ABC, \triangle ADE, \triangle CEF, \triangle BDF.$

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

Let $K, L, N,$ and $G$ be the foots of the perpendiculars dropped from $M$ to lines $AB, AC, EF,$ and $BC,$ respectively.

Prove that points $K,L, N,$ and $G$ are collinear.

Proof

Let $\Omega$ be the circumcircle of $\triangle ABC, \omega$ be the circumcircle of $\triangle CEF.$ Then $M = \Omega \cap \omega.$

Points $K, L,$ and $G$ are collinear as Simson line of $\triangle ABC.$

Points $L, N,$ and $G$ are collinear as Simson line of $\triangle CEF.$

Therefore points $K, L, N,$ and $G$ are collinear, as desired.

vladimir.shelomovskii@gmail.com, vvsss

Problem

Problem on Simson line.png

Let the points $A, B,$ and $C$ be collinear and the point $P \notin AB.$

Let $O,O_0,$ and $O_1$ be the circumcenters of triangles $\triangle ABP, \triangle ACP,$ and $\triangle BCP.$

Prove that $P$ lies on circumcircle of $\triangle OO_0O_1.$

Proof

Let $D, E,$ and $F$ be the midpoints of segments $AB, AC,$ and $BC,$ respectively.

Then points $D, E,$ and $F$ are collinear $(DE||AB, EF||DC).$

$PD \perp OO_0, PE \perp OO_1, PF \perp O_0O_1 \implies$ $DEF$ is Simson line of $\triangle OO_0O_1 \implies P$ lies on circumcircle of $\triangle OO_0O_1$ as desired.

vladimir.shelomovskii@gmail.com, vvsss

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