Let $\triangle ABC$ be an acute-angled triangle with $AB<AC$ and let $(c)$ be its circumcircle with center $O$. Let $M$ be the midpoint of $BC$. The line $AM$ meets the circle $(c)$ again at the point $D$. The circumcircle $(c_1)$ of triangle $\triangle MDC$ intersects the line $AC$ at the points $C$ and $I$, and the circumcircle $(c_2)$ of $\triangle AMI$ intersects the line $AB$ at the points $A$ and $Z$. If $N$ is the foot of the perpendicular from $B$ on $AC$, and $P$ is the second point of intersection of $ZN$ with $(c_2)$, prove that the quadrilateral with vertices the points $N, P, I$ and $M$ is a parallelogram.
Problem
Source: Cyprus 2022 TST-1 Problem 3
Tags: geometry, parallelogram, circumcircle
Y.J
27.03.2022 11:07
Easy to verify $BZ\bot AC$, which implies $N,Z,C,B$ in a circle.
User768799
02.04.2022 18:19
Could you tell how to verify that BZ is perpendicular to AC,thanks so much
gnoka
04.04.2022 19:23
User768799 wrote: Could you tell how to verify that BZ is perpendicular to AC,thanks so much Since $\angle CZM=\angle AIM=\angle BDM=\angle BCA$ we have $ZM=MC$ but $BN \perp NC, BM=MC$, so we have $MN=BM=MC=MZ$ $\therefore B,N,Z,C$ concyclic. btw: There is something different between the posts by @Demetres and @Y.J, I hope you can overcome it.
User768799
05.04.2022 03:57
No problem ,I can learn it ,thanks