Miquel's point

Miquel and Steiner's quadrilateral theorem

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 $\triangle CEF$ circle $\omega$ 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

Triangle of circumcenters

Let four lines made four triangles of a complete quadrilateral.

In the diagram these are $\triangle ABC, \triangle ADE, \triangle CEF, \triangle BDF.$

Let points $O,O_A, O_B,$ and $O_C$ be the circumcenters of $\triangle ABC, \triangle ADE, \triangle BDF,$ and $\triangle CEF,$ respectively.

Prove that $\triangle O_AO_BO_C \sim \triangle ABC,$ and perspector of these triangles point $X$ is the second (different from $M$ ) point of intersection $\Omega \cap \Theta,$ where $\Omega$ is circumcircle of $\triangle ABC$ and $\Theta$ is circumcircle of $\triangle O_AO_BO_C.$

Proof

Quadrungle $MECF$ is cyclic $\implies \angle AEM = \angle BFM \implies$ $\angle AO_AM = 2\angle AEM = 2 \angle BFM = \angle BO_BM.$ $\angle CO_CM = 2\angle CFM = 2 \angle BFM = \angle BO_BM.$ $AO_A = MO_A, BO_B = MO_B, CO_C = MO_C \implies \triangle AO_AM \sim \triangle BO_BM \sim \triangle CO_CM.$

Spiral similarity sentered at point $M$ with rotation angle $\angle AMO_A = \angle BMO_B = CMO_C$ and the coefficient of homothety $\frac {AM}{MO_A} = \frac {BM}{MO_B} =\frac {CM}{MO_C}$ mapping $A$ to $O_A$ , $B$ to $O_B$ , $C$ to $O_C \implies \triangle O_AO_BO_C \sim \triangle ABC.$

$\triangle AO_AM, \triangle BO_BM, \triangle CO_CM$ are triangles in double perspective at point $M \implies$

These triangles are in triple perspective $\implies AO_A, BO_B, CO_C$ are concurrent at the point $X.$

The rotation angle $\triangle AO_AM$ to $\triangle BO_BM$ is $\angle O_AMO_B$ for sides $O_AM$ and $O_BM$ or angle between $AO_A$ and $BO_B$ which is $\angle AXB \implies M O_AO_BX$ is cyclic $\implies M O_AO_BXO_C$ is cyclic.

Therefore $\angle O_AXO_B = \angle O_AO_CO_B = \angle ACB \implies ABCX$ is cyclic as desired.

Similarly, one can prove that $\triangle ADE \sim \triangle OO_BO_C, \triangle BDF \sim \triangle OO_AO_C, \triangle CEF \sim \triangle OO_AO_B.$

Double perspective triangles

vladimir.shelomovskii@gmail.com, vvsss

Analogue of Miquel's point

Let inscribed quadrilateral $ABB'A'$ and

points $C \in AB', C' \in A'B', D \in A'B$ be given.

$\theta = \odot CC'B', \Theta = \odot BDD', M = \theta \cap \Theta,$ $E = C'D \cap \odot B'CC', D' = AB \cap CE, F = CC' \cap DD'.$ Prove that points $A, B, B',$ and $M$ are concyclic.

Proof

$\angle BMB' = \angle EMB' - \angle EMB = \angle ECB' - \angle ED'B = \angle BAB' \blacksquare$

Corollary

The points $F, C, D',$ and $M$ are concyclic.

The points $F, C', D,$ and $M$ are concyclic.

vladimir.shelomovskii@gmail.com, vvsss

Six circles crossing point

Let $\triangle ABC,$ point $P \in BC,$ point $X \in \Omega = \odot ABC$ be given.

Denote $Y = PX \cap \Omega, D = AX \cap BC, E = AY \cap BC,$ $\omega = \odot ADE, \theta = \odot PEY, \Theta = \odot PDX,$ $\sigma = \odot BP$ tangent to $AB, \Sigma = \odot CP$ tangent to $AC.$

Prove that the circles $\omega, \Omega, \theta, \Theta, \sigma,$ and $\Sigma$ have the common point.

Proof

Let $M = \omega \cap \Omega \ne A.$ $\angle YCM = \angle YXM = \angle YAM = \angle EAM = \angle EDM \implies$ $\angle PDM = \angle PXM \implies$ points $M,P,D,$ and $X$ are concyclic, $M \in \Theta.$ Similarly $M \in \theta, M$ is the Miquel point of quadrungle $EDXY.$ $\angle CMY = 180^\circ - \angle CAB - \angle BAY.$ $\angle PMY = \angle PEY = \angle ABC - \angle BAY.$ $\angle PMC = \angle CMY - \angle PMY = (180^\circ - \angle CAB - \angle BAY) - (\angle ABC - \angle BAY) = \angle ACB.$ $\angle PMC = \angle ACB \implies AC$ is tangent to $\Sigma.$ Similarly, $AB$ is tangent to $\sigma.$