Difference between revisions of "Simson line"
(→Simson line (main)) |
Sharonwang (talk | contribs) m |
||
(13 intermediate revisions by one other user not shown) | |||
Line 2: | Line 2: | ||
[[File:Simsonline.png]] | [[File:Simsonline.png]] | ||
− | |||
− | |||
==Simson line (main)== | ==Simson line (main)== | ||
− | [[File:Simson line.png| | + | [[File:Simson line.png|270px|right]] |
− | Let a triangle <math>\triangle ABC</math> and a point <math>P</math> be given. Let <math>D, E,</math> and <math>F</math> be the foots of the perpendiculars dropped from P to lines AB, AC, and BC, respectively. | + | [[File:Simson line inverse.png|270px|right]] |
+ | Let a triangle <math>\triangle ABC</math> and a point <math>P</math> be given. | ||
+ | |||
+ | Let <math>D, E,</math> and <math>F</math> be the foots of the perpendiculars dropped from P to lines AB, AC, and BC, respectively. | ||
Then points <math>D, E,</math> and <math>F</math> are collinear iff the point <math>P</math> lies on circumcircle of <math>\triangle ABC.</math> | Then points <math>D, E,</math> and <math>F</math> are collinear iff the point <math>P</math> lies on circumcircle of <math>\triangle ABC.</math> | ||
Line 13: | Line 14: | ||
Let the point <math>P</math> be on the circumcircle of <math>\triangle ABC.</math> | Let the point <math>P</math> be on the circumcircle of <math>\triangle ABC.</math> | ||
− | <math>\angle BFP = \angle BDP = 90^\circ \implies BPDF</math> is cyclic <math>\implies \angle PDF = 180^\circ – \angle CBP.</math> | + | |
− | <math>\angle ADP = \angle AEP = 90^\circ \implies AEPD</math> is cyclic <math>\implies \angle PDE = \angle PAE.</math> | + | <math>\angle BFP = \angle BDP = 90^\circ \implies</math> |
+ | |||
+ | <math>BPDF</math> is cyclic <math>\implies \angle PDF = 180^\circ – \angle CBP.</math> | ||
+ | |||
+ | <math>\angle ADP = \angle AEP = 90^\circ \implies</math> | ||
+ | |||
+ | <math>AEPD</math> is cyclic <math>\implies \angle PDE = \angle PAE.</math> | ||
<math>ACBP</math> is cyclic <math>\implies \angle PBC = \angle PAE \implies \angle PDF + \angle PDE = 180^\circ</math> | <math>ACBP</math> is cyclic <math>\implies \angle PBC = \angle PAE \implies \angle PDF + \angle PDE = 180^\circ</math> | ||
+ | |||
<math>\implies D, E,</math> and <math>F</math> are collinear as desired. | <math>\implies D, E,</math> and <math>F</math> are collinear as desired. | ||
Line 22: | Line 30: | ||
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. | ||
− | + | ||
− | <math>AEPD</math> is cyclic <math>\implies \angle APE = \angle ADE, \angle | + | <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> | ||
<math>\angle ADE = \angle BDF \implies \angle BPA = \angle EPF</math> | <math>\angle ADE = \angle BDF \implies \angle BPA = \angle EPF</math> | ||
− | |||
+ | <math>= \angle BAC + \angle ABC = 180^\circ – \angle ACB \implies</math> | ||
+ | |||
+ | <math>ACBP</math> is cyclis as desired. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Simson line of a complete quadrilateral== | ||
+ | [[File:Simson complite.png|430px|right]] | ||
+ | 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> | ||
+ | |||
+ | Let <math>M</math> be the Miquel point of a complete quadrilateral. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | Prove that points <math>K,L, N,</math> and <math>G</math> are collinear. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | Points <math>K, L,</math> and <math>G</math> are collinear as Simson line of <math>\triangle ABC.</math> | ||
+ | |||
+ | Points <math>L, N,</math> and <math>G</math> are collinear as Simson line of <math>\triangle CEF.</math> | ||
+ | |||
+ | Therefore points <math>K, L, N,</math> and <math>G</math> are collinear, as desired. | ||
+ | |||
+ | *[[Miquel's point]] | ||
+ | *[[Steiner line]] | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Problem== | ||
+ | [[File:Problem on Simson line.png |400px|right]] | ||
+ | |||
+ | Let the points <math>A, B,</math> and <math>C</math> be collinear and the point <math>P \notin AB.</math> | ||
+ | |||
+ | Let <math>O,O_0,</math> and <math>O_1</math> be the circumcenters of triangles <math>\triangle ABP, \triangle ACP,</math> and <math>\triangle BCP.</math> | ||
+ | |||
+ | Prove that <math>P</math> lies on circumcircle of <math>\triangle OO_0O_1.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>D, E,</math> and <math>F</math> be the midpoints of segments <math>AB, AC,</math> and <math>BC,</math> respectively. | ||
+ | |||
+ | Then points <math>D, E,</math> and <math>F</math> are collinear <math>(DE||AB, EF||DC).</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. | ||
+ | *[[Euler line]] | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Latest revision as of 09:28, 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.
Simson line (main)
Let a triangle and a point
be given.
Let and
be the foots of the perpendiculars dropped from P to lines AB, AC, and BC, respectively.
Then points and
are collinear iff the point
lies on circumcircle of
Proof
Let the point be on the circumcircle of
is cyclic
is cyclic
is cyclic
and
are collinear as desired.
Proof
Let the points and
be collinear.
is cyclic
is cyclic
is cyclis as desired.
vladimir.shelomovskii@gmail.com, vvsss
Simson line of a complete quadrilateral
Let four lines made four triangles of a complete quadrilateral. In the diagram these are
Let be the Miquel point of a complete quadrilateral.
Let and
be the foots of the perpendiculars dropped from
to lines
and
respectively.
Prove that points and
are collinear.
Proof
Let be the circumcircle of
be the circumcircle of
Then
Points and
are collinear as Simson line of
Points and
are collinear as Simson line of
Therefore points and
are collinear, as desired.
vladimir.shelomovskii@gmail.com, vvsss
Problem
Let the points and
be collinear and the point
Let and
be the circumcenters of triangles
and
Prove that lies on circumcircle of
Proof
Let and
be the midpoints of segments
and
respectively.
Then points and
are collinear
is Simson line of
lies on circumcircle of
as desired.
vladimir.shelomovskii@gmail.com, vvsss