AB,CD,XY concurrent or //, cyclic, circumcircle (Kharkiv City XI 2017- Ukr)

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AB,CD,XY concurrent or //, cyclic, circumcircle (Kharkiv City XI 2017- Ukr)

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#1 h Mar 19, 2020, 9:32 PM • 1 Y

Y by TaDucDuy

The quadrilateral $ABCD$ is inscribed in the circle $\omega$ . Lines $AD$ and $BC$ intersect at point $E$ . Points $M$ and $N$ are selected on segments $AD$ and $BC$ , respectively, so that $AM: MD = BN: NC$ . The circumscribed circle of the triangle $EMN$ intersects the circle $\omega$ at points $X$ and $Y$ . Prove that the lines $AB, CD$ and $XY$ intersect at the same point or are parallel.

This post has been edited 3 times. Last edited by parmenides51, Mar 19, 2020, 9:34 PM

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#2 h Mar 23, 2020, 8:55 AM

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Solution:

https://i.imgur.com/kceQ8pF.jpg

Let $F=AB \cap CD$ and T be the intersection of EF and $\odot (EAB)$ .As you know,T is the Miquel point of complete quadrilateral ABCD.EF(which means that T is the common point of four circles $\odot (EAB)$ , $\odot (EDC)$ , $\odot (FAD)$ and $\odot (FBC)$ ).Notice 2 cyclic quadrilaterals TEBC and TECD so it's easy to show that $\triangle TAD\sim\triangle TBC$ (a-a).On the other hand,points M,N divide the segment lines AD,BC in the same ratio respectively so we get $\triangle TAM\sim\triangle TBN\Longrightarrow\angle TME=\angle TMA=\angle TNB=\angle TNE\Longrightarrow$ TENM is the cyclic quadrilateral(or T lies on the circle (EMN)) and hence we get $P_{F/ \odot (EMN)}=\overline{FT}\cdot\overline{FE}=\overline{FD}\cdot\overline{FC}=P_{F/ \odot (ABCD)}$ .Therefore F lies on the radical axis of $\odot (EMN)$ and $\odot (ABCD)$ so F,X,Y are collinear (which means that AB,XY and CD are concurrent at a point F).We done!

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