Difference between revisions of "Simson line"

Revision as of 14:31, 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.

Proof

In the shown diagram, we draw additional lines $AP$ and $BP$ . Then, we have cyclic quadrilaterals $ACBP$ , $PC_1A_1B$ , and $PB_1AC_1$ . (more will be added)

Simson line (main)

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.

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.$

@@ Line 4: / Line 4: @@
 == Proof ==
 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)==
+[[File:Simson line.png|350px|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>
+<i><b>Proof</b></i>
+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>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.
+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 APE = \angle BAC.</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 BAC + \angle ABC = 180^\circ – \angle ACB \implies ACBP</math> is cyclis as desired.$

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Difference between revisions of "Simson line"

Revision as of 14:31, 30 November 2022

Proof

Simson line (main)