SAUDI ARABIAN IMO Booklet 2023

Since triangles $AEB$ and $CAB$ are similar, we have $\frac{AB}{EB}=\frac{CB}{AB}$. Note that $AB=BZ$, thus $$\frac{BZ}{EB}=\frac{CB}{AB},$$ from which it follows that triangles $ZBE$ and $CBZ$ are also similar. Since $FEBZ$ is cyclic, then $\angle BEZ=\angle BFZ$. So by the similarity of triangles $ZBE$ and $CBZ$, we get $$\angle BFZ=\angle BEZ=\angle BZC=\angle BZF$$ and thus $BFZ$ is isosceles. Since $BF=BZ=AB$, triangle $AFZ$ is right with $\angle AFZ=90^{\circ }$. It follows that points $A,E,F,C$ are concyclic. Since $A,P,E,N$ are also concyclic, then $$\angle ENP=\angle EAP=\angle EAF=\angle BCZ=\angle BZE,$$ by using the similarity of triangles $ZBE$ and $CBZ$. Since $N,B,E,Z$ are concyclic, then $\angle ENP=\angle BZE=\angle ENB$, which implies that $N,B,P$ are collinear. $\square$