Miquel Point for Complete Quadrilateral

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Miquel Point for Complete Quadrilateral

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#1 h Apr 17, 2014, 1:22 PM • 2 Y

Y by Adventure10, Mango247

Could someone provide a proof of Miquel's theorem for complete quadrilaterals? I know it's well known, but couldn't find a proof.

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#2 h Apr 17, 2014, 4:38 PM • 2 Y

Y by Adventure10, Mango247

Suppose ABCD is a quad, AB hits CD at E and AD hits BC at F. If XYZ is a triangle, let (XYZ) denote its circumcircle (this isn't actually notation - I
'm just lazy).
We want to show (ADE), (FCD), (FAB), (ECB) concur.
Ok suppose (FCD) and (EAD) concur at a point P. I wts PFAB is cyclic (it's clear that this is symmetric with showing PECB is as well).
Because PFCD is cyclic, angle PFC equals angle PDE. Because PEAD is cyclic, angle PDE equals angle PEA.
Thus, angle PFC equals angle PAE, and we're done.

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pi37

#3 h Apr 17, 2014, 4:43 PM • 2 Y

Y by Adventure10, Mango247

Unnecessary solution shamelessly taken from a blog post 3 years ago:

[geogebra]cbccce0dad99363a30f6cc1b797fad92dbfe7195[/geogebra]
Let the points and lines be labeled as shown, and let $(XYZ)$ represent the circumcircle of triangle $XYZ$ . Also, let the orthogonal projection of point $X$ onto line $j$ be $G_j$ . We wish to show that $(BEF)$ , $(ADE)$ , $(CDF)$ , and $(ABC)$ are concurrent. Let $G=(BEF)\cap (ADE)$ , where $G\neq E$ . Since $G$ is on $(BEF)$ , $G_a$ , $G_b$ , and $G_d$ are collinear on the Simson line of $G$ with respect to $BEF$ . $G$ is also on $(ADE)$ , so $G_c$ is also on this same line. Since $G_a$ , $G_b$ , $G_c$ are collinear, and $G_b$ , $G_c$ , $G_d$ are collinear, it follows that $G$ is also on $(ABC)$ and $(CDF)$ , so the circles all concur at $G$ .

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jayme

#4 h Apr 18, 2014, 2:17 PM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,
you can see : http://perso.orange.fr/jl.ayme vol. 13 Miquel... p. 12-14.
and more about the results of Auguste Miquel.
Sincerely
Jean-Louis

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ABCDE

#5 h Apr 18, 2014, 2:49 PM • 1 Y

Y by Adventure10

Alternatively, spiral similarity makes this pretty easy.

Basically, two circles intersect at the center of spiral similarity between two sides. But this spiral similarity also maps another side to another other side. Thus, the other two circles pass through the center of spiral similarity so we're done.

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