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 circumcircles of $\triangle ABC$ and $\triangle O_AO_BO_C.$ Proof Quadrungle $MECF$ is concyclic $\implies \angle AEM = \angle BFM \implies \angle AO_AB = 2\angle AEM = 2 \angle BFM = \angle BO_BM.$ $\angle CO_CM = 2\angle CFM = 2 \angle BFM = \angle BO_BM.$ $AO_A = MO_A, BO_B = MO_B, CO_C = MO_C \implies \triangle AO_AM \sim \triangle BO_BM \sim \triangle CO_CM.$ 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.$$ (Error compiling LaTeX. Unknown error_msg)\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 $is$ 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 $is cyclic$ \implies \angle O_AXO_B = \angle O_AO_CO_B = \angle ACB \implies ABCX$ is cyclic as desired.

vladimir.shelomovskii@gmail.com, vvsss

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Miquel's point

Miquel and Steiner's quadrilateral theorem

Circle of circumcenters

Triangle of circumcenters