Difference between revisions of "Miquel's point"

Revision as of 09:10, 5 December 2022

Miquel and Steiner's quadrilateral theorem

500px|right Let four lines made four triangles of a complete quadrilateral. In the diagram these are $\triangle ABC, \triangle ADE, \triangle CEF, \triangle BDF.$

Prove that the circumcircles of all four triangles meet at a single point.

Proof

Let circumcircle of $\triangle ABC$ circle $\Omega$ cross the circumcircle of $\triangle CEF \omega$ at point $M.$

Let $AM$ cross $\omega$ second time in the point $G.$

$CMGF$ is cyclic $\implies \angle BCM = \angle MGF.$

$AMCB$ is cyclic $\implies \angle BCM + \angle BAM = 180^\circ \implies \angle BAG + \angle AGF = 180^\circ \implies AB||GF.$

$CMGF$ is cyclic $\implies \angle AME = \angle EFG.$

$AD||GF \implies \angle ADE + \angle DFG = 180^\circ \implies \angle ADE + \angle AME = 180^\circ \implies ADEM$ is cyclic and circumcircle of $\triangle ADE$ contain the point $M.$

Similarly circumcircle of $\triangle BDF$ contain the point $M$ as desired.

vladimir.shelomovskii@gmail.com, vvsss

Circle of circumcenters

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

Prove that the circumcenters of all four triangles and point $M$ are concyclic.

Proof

Let $\Omega, \omega, \Omega',$ and $\omega'$ be the circumcircles of $\triangle ABC, \triangle CEF, \triangle BDF,$ and $\triangle ADE,$ respectively.

In $\Omega' \angle MDF = \angle MBF.$

In $\omega' \angle MDE = \frac {\overset{\Large\frown} {ME}} {2}.$

$ME$ is the common chord of $\omega$ and $\omega' \implies \angle MOE = \overset{\Large\frown} {ME} \implies$

$\angle MO'o' = \frac {\overset{\Large\frown} {ME}} {2} = \angle MDE.$

Similarly, $MF$ is the common chord of $\omega$ and $\Omega' \implies \angle MDF = \angle Moo' = \angle MO'o'.$

Similarly, $MC$ is the common chord of $\Omega$ and $\omega' \implies \angle MBC = \angle MOo' \implies$

$\angle MOo' = \angle MO'o' \implies$ points $M, O, O', o,$ and $o'$ are concyclic as desired.

vladimir.shelomovskii@gmail.com, vvsss

@@ Line 20: / Line 20: @@
 Similarly circumcircle of <math>\triangle BDF</math> contain the point <math>M</math> as desired.
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Circle of circumcenters==
+[[File:Miquel point.png|450px|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>
+Prove that the circumcenters of all four triangles and point <math>M</math> are concyclic.
+<i><b>Proof</b></i>
+Let <math>\Omega, \omega, \Omega',</math> and <math>\omega'</math> be the circumcircles of <math>\triangle ABC, \triangle CEF, \triangle BDF,</math> and <math>\triangle ADE,</math> respectively.
+In <math>\Omega' \angle MDF = \angle MBF.</math>
+In <math>\omega' \angle MDE = \frac {\overset{\Large\frown} {ME}} {2}.</math>
+<math>ME</math> is the common chord of <math>\omega</math> and <math>\omega' \implies \angle MOE = \overset{\Large\frown} {ME} \implies</math>
+<cmath>\angle MO'o' = \frac {\overset{\Large\frown} {ME}} {2} =  \angle MDE.</cmath>
+Similarly, <math>MF</math> is the common chord of <math>\omega</math> and <math>\Omega' \implies  \angle MDF = \angle Moo' = \angle MO'o'.</math>
+Similarly, <math>MC</math> is the common chord of <math>\Omega</math> and <math>\omega' \implies  \angle MBC = \angle MOo' \implies</math>
+<math>\angle MOo' = \angle MO'o' \implies</math> points <math>M, O, O', o,</math> and <math>o'</math> are concyclic as desired.
 '''vladimir.shelomovskii@gmail.com, vvsss'''

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Difference between revisions of "Miquel's point"

Revision as of 09:10, 5 December 2022

Miquel and Steiner's quadrilateral theorem

Circle of circumcenters