This problem was proposed by Cyprus

Nice config! Although this problem is not as hard as it looks Here's my solution :

This one is harder to draw than to actually solve, due to so many points given it get's kind of confusing to get a good grip of what's happening, it's actually better to "erase" or "ignore" one one of the circles when dealing with angles. Claim: $EACF$ is cyclic. Proof: $$\angle EAC = \angle EBA = \angle EFZ.$$So $\angle AEC = \angle AFC = 90^\circ \implies AF \perp ZC$. Also since $B$ is midpoint of side $ZA$ in $\triangle AZF \implies BA = BF = BZ$. $$\angle EFA = \angle ECA = \angle EAB$$$$\implies BFA - \angle EFA = \angle BAF - \angle BAE \implies \angle BFE = \angle EAF.$$Thus $$\angle BNE = \angle BFE = \angle EAF$$also we know that $$\angle PNE = \angle PAE = \angle FAE$$So $\angle BNE = \angle PNE \implies \overline{N-B-P}$ are collinear.

BZ^2 = BA^2 = BE.BC ---> circumcircle of AEC is tangent to BZ. CF.CZ = CE.CB = CA^2 ---> ∠CFA = 90 ---> CFEA is cyclic. ∠BNE = ∠BAE = ∠ZCE = ∠PAE = ∠PNE ---> P,N,B are collinear.

Here is a $\textbf{Really Cool}$ technique of $\textbf{Inversion}$ that I found back then . This was my first time ever using such an Inversion ! Also posting soln after almost a year after solving cuz IDK why . Moving onto the solution. We have a lot of "unecessary" circles and points. So let's fix the wording and make it a little better . Restated Problem wrote: Let $\triangle ABC$ be a right-angled triangle with $\angle BAC = 90^{\circ}$, and let $E$ be the foot of the perpendicular from $A$ to $BC$. Let $Z \neq A$ be a point on the line $AB$ with $AB = BZ$. Let $F = CZ \cap \odot(BEZ)$ and $P$ be an arbitrary point on line $AF$, and let $N = \odot(BEZ) \cap \odot(AEP)$. Prove that the points $\{ N, B, P \}$ are collinear. Much nicer to look at . Now we prepare our $\textbf{Cool Inversion}$ .

Before we start, note that, $$\measuredangle EFC = \measuredangle EFZ = \measuredangle EBZ = \measuredangle EBA = \measuredangle EAC \implies AEFC \text{ is cyclic.}$$ Also, $$BE \cdot BC = BA^2 = BZ^2 \implies \measuredangle EZB = \measuredangle ECZ. \qquad\ (\ddagger)$$ Now we move onto solving the problem . $\textbf{The Tool:}$ Let $\Psi$ denote the Inversion with center $E$ and radius $EA$, followed by a reflection across the line $EA$. How does this help? . There u go $\downarrow$ . Also, let $X^{*} = \Psi(X)$. Note that $\Psi$ maps the following: $$A \xleftrightarrow{} A$$$$B \xleftrightarrow{} C$$$$AB \xleftrightarrow{} \odot(EAC).$$ Now to the most important claim! Claim: $Z^{*} \equiv F$. Proof: Note that, $$Z \in AB \implies Z^{*} \in (AB)^{*} = \odot(EACF)$$and also, $$\measuredangle ECZ^{*} = \measuredangle EB^{*}Z^{*} \stackrel{\dagger}{=} \measuredangle EZB \stackrel{\ddagger}{=} \measuredangle ECZ \implies Z^{*} \in CZ \implies Z^{*} = CZ \cap \odot(EACF) = F.$$

And we are done with our claim . Thus we conclude that $\Psi$ maps the follwing too: $$AF \xleftrightarrow{} \odot(EAZ)$$$$CZ \xleftrightarrow{} \odot(EBF)$$$$\odot(EAP) \xleftrightarrow{} AP^{*}$$$$N = \odot(BEZ) \cap \odot(AEP) \xleftrightarrow{} N^{*} = CF \cap AP^{*}.$$ Also note that $CF\equiv CZ$. Now, since $P$ was given to be as an arbitrary point on line $AF$, we can basically redefine $P^{*}$ as an arbitrary point on $\odot(AEZ)$. Thus it finally remains to show that $EB^{*}N^{*}P^{*}$ is cyclic, that is equivalent to showing $ECN^{*}P^*$ is cyclic. Now we can restate our problem for one last final time . Restated Problem wrote: Let $\triangle ABC$ be a triangle with $\angle BAC = 90^{\circ}$, and let $E$ be the foot of the perpendicular from $A$ to $BC$. Let $Z \neq A$ be a point on the line $AB$ with $AB = BZ$. Let $P^*$ be an arbitrary point on $\odot(AEZ)$, and let $N^* = CZ \cap AP^*$. Prove that $ECN^*P^*$ is cyclic. Wayyy better isn't it?!

This is a lot easier . \[ \measuredangle ECN^* = \measuredangle ECZ = \measuredangle EZB = \measuredangle EZA = \measuredangle EP^*A = \measuredangle EP^*N^*. \] We are done !

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