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Difference between revisions of "Simson line"

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[[File:Simsonline.png]]
 
[[File:Simsonline.png]]
  
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== Proof ==
 
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In the shown diagram, we draw additional lines <math>AP</math> and <math>BP</math>. Then, we have cyclic quadrilaterals <math>ACBP</math>, <math>PC_1A_1B</math>, and <math>PB_1AC_1</math>. (more will be added)
 
 
==Simson line (main)==
 
==Simson line (main)==
 
[[File:Simson line.png|300px|right]]
 
[[File:Simson line.png|300px|right]]

Revision as of 15:58, 30 November 2022

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 $\implies \angle APE = \angle ADE, \angle APE = \angle BAC.$

$BFDP$ 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

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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