Geometry and circles

Let A,B,C and D be four points on a lime, in that order. The circles with diameters AC and BD intersect at X and Y. The line XY meets BC at Z. Let P be a point on the line XY other than Z. The line CP intersects the circle with diameter AC at C and M, and the line BP intersects the circle with diameter BD at B and N. Prive that the lines AM, DN and XY are concurrent.

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mathprince2000

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mathprince2000

#2 h May 12, 2016, 9:44 AM • 1 Y

Y by Adventure10

Let $K$ be the point of intersection of $AM$ and $DN$ . We can easily see that $XY$ is the radical axis of the two circles. Thus, since $P$ lies on $XY$ , $\overline{PB}.\overline{PN}=\overline{PC}.\overline{PM}$ . Hence, $BCMN$ is a cyclic quadrilateral, so $\widehat{CMN}=\widehat{CBN}$ , or $\widehat{KMN}=\widehat{KDA}$ . Therefore, $AMND$ is a cyclic quadrilateral $\Rightarrow \overline{KM}.\overline{KA}=\overline{KN}.\overline{KD}$ . This means $K$ lies on $XY$ , so $AM$ , $DN$ and $XY$ are concurrent.

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This post has been edited 1 time. Last edited by mathprince2000, May 12, 2016, 9:46 AM
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