[asy][asy] usepackage("tikz"); label("\begin{tikzpicture}[font=\small] \coordinate[label=below left:$C$] (C) at (0,0); \coordinate[label=above:$A$] (A) at (0,6); \coordinate[label=below:$B$] (B) at (5,0); \coordinate[label=above right:$O$] (O) at (2.5,3); \coordinate[label=below:$P$] (P) at (2.5,-2.08); \coordinate[label=right:$M$] (M) at (3.75,-1.04); \coordinate[label=above left:$N$] (N) at (3.09,-.86); \coordinate[label=above right:$E$] (E) at (4.06,1.13); \coordinate[label=below:$D$] (D) at (2.03,-.56); \draw[thick] (A)--(B)--(C)--cycle; \draw[thick] (C)--(P)--(B); \draw[thick] (D)--(E)--(P); \draw[dashed] (1.93,.94) circle (2.14); \draw[thick] (C)--(E); \draw[thick] (C)--(M); \draw[thick] (C)--(O)--(N); \draw[thick] (O) circle (3.91); \draw[thick] (B)--(N); \coordinate[label=below right:$N'$] (N') at (4.41,-1.22); \draw[thick] (P)--(N')--(B); \draw[thick] (M)--(N'); \draw[dotted] (2.5,.46) circle (2.54); \foreach \s in {A,B,C,O,E,D,M,N,P,N'}\filldraw (\s) circle (1.5pt); \end{tikzpicture}"); [/asy][/asy] Suppose $O$ is the centre of $(ABC)$, it is clear that $O$ is the midpoint of $AB$. Suppose $\angle CAB=\alpha$, then $\angle CBP=90^\circ-\angle CBA=\alpha$. Given the length $PB=PC$, then $\angle BCP=\alpha$ and $\angle CPB=180^\circ-2\alpha$. Suppose $N'$ is the reflection of point $M$ to point $N$. From PoP to $(ABC)$ holds $MP\cdot MB=MB^2=MN\cdot MC=MN'\cdot MC$, then $BN'PC$ is cyclic. Consider $BNPN'$ a parallelogram and $\angle COB = 2\angle CAB = 2\alpha$ and \[\angle CNE = \angle PNN' = \angle BN'N=\angle BN'C = \angle BPC= 180^\circ-2\alpha.\]Since $\angle BOE+\angle BNE=180^\circ$, then $CNEO$ is cyclic. From this, it is obtained \[\angle NCB = \frac{\angle NOB}{2}=\frac{\angle NOE}{2} =\frac{\angle NCE}{2}\implies \angle NCB=\angle ECB.\]Then \begin{align*} \angle CED &= 180^\circ - \angle ECD - \angle EDC\\\\ &= 180^\circ-2\angle MCB - \angle BMC\\\ &= \angle MBC - \angle MCB\\\ &= \angle PBC - \angle MCB\\\ &= \angle PCB - \angle MCB\\\ &= \angle PCN, \end{align*}which is proven by Alternate Segment Theorem. Remark. Looking at the midpoint will most likely pertain to PoP. After a few steps back, I needed to prove the CB bisect $\angle ECN$. Seeing the midpoint condition also motivated me to construct the parallelogram, which killed it :p

INAMO 2023/7 wrote: Given a triangle $ABC$ with $\angle ACB = 90^{\circ}$. Let $\omega$ be the circumcircle of $ABC$. The tangents to $\omega$ at points $B$ and $C$ meet at $P$. Let $M$ be the midpoint of $PB$. The line $CM$ intersects $\omega$ at $N$, and the line $PN$ intersects $AB$ at $E$. Point $D$ is on $CM$ such that $ED \parallel BM$. Prove that the circumcircle of triangle $CDE$ is tangent to $\omega$. Reflect $N$ with respect to $AB$ to get point $F$. Since $\angle ACB = 90^{\circ}$, $AB$ is the diameter of the circle and hence $AB \perp PB$. Since $FN \perp AB$ by definition, we obtain $FN \parallel PB$. Claim. $C, E, F$ are collinear. Proof. We'll do this by phantom point. Let $E' = CF \cap AB$. We'll prove that $P, N, E'$ are collinear. Note that by Power of Point, $MB^2 = MN \cdot MC$ and since $M$ is the midpoint of $BP$, we obtain $MP^2 = MN \cdot MC$, which implies $\triangle MPN \sim \triangle MCP$. This implies $\angle MCP = \angle MPN$. Therefore, we have \[ \angle FNE' = \angle E'FN = \angle CFN = \angle NCP = \angle MCP = \angle MPN \]However, as $NF \parallel MP$, we obtain the desired result. To finish the problem, note that we have $DE \parallel PB \parallel NF$, $C,D,N$ are collinear and $C,E,F$ are collinear. Thus, the homothety from point $C$ sends $(CNF) \equiv \omega$ to $(CDE)$ and hence $(CDE)$ is tangent to $\omega$ as desired. [asy][asy] real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.05, xmax = 0.79, ymin = -3.1, ymax = 7.1; /* image dimensions */ /* draw figures */ draw((-6.54,0.38)--(-0.42,0.36)); draw(circle((-3.48,0.37), 3.0600163398256552)); draw((-6.54,0.38)--(-5.420412291713705,2.736114988365182)); draw((-5.420412291713705,2.736114988365182)--(-0.42,0.36)); draw((-0.39877687499800984,6.854276250609016)--(-0.42,0.36)); draw((-0.39877687499800984,6.854276250609016)--(-5.420412291713705,2.736114988365182)); draw((-5.420412291713705,2.736114988365182)--(-0.4093884374990049,3.607138125304508)); draw((xmin, 1.7545276097405513*xmin + 7.5539412879190815)--(xmax, 1.7545276097405513*xmax + 7.5539412879190815)); /* line */ draw((-4.084888999721857,2.9682575047628763)--(-4.093373486703047,0.3720044885186374) ); draw(circle((-4.547444499860928,1.6716287523814384), 1.3766639794122426), gray); draw((-2.4707016232838974,-2.518774270648062)--(-2.4518425050199295,3.252115918126178), dotted + red); draw((-5.420412291713705,2.736114988365182)--(-2.4707016232838974,-2.518774270648062), dotted + red); /* dots and labels */ dot((-6.54,0.38),dotstyle); label("$A$", (-6.95,0.08), NE * labelscalefactor); dot((-0.42,0.36),dotstyle); label("$B$", (-0.33,0.56), NE * labelscalefactor); dot((-5.420412291713705,2.736114988365182),dotstyle); label("$C$", (-5.35,2.94), NE * labelscalefactor); dot((-0.39877687499800984,6.854276250609016), dotstyle); label("$P$", (-0.31,6.62), NE * labelscalefactor); dot((-0.4093884374990049,3.607138125304508), dotstyle); label("$M$", (-0.33,3.76), NE * labelscalefactor); dot((-2.4518425050199295,3.252115918126178), dotstyle); label("$N$", (-2.37,3.42), NE * labelscalefactor); dot((-4.093373486703047,0.3720044885186374), dotstyle); label("$E$", (-4.01,0.54), NE * labelscalefactor); dot((-4.084888999721857,2.9682575047628763), dotstyle); label("$D$", (-4.01,3.12), NE * labelscalefactor); dot((-2.4707016232838974,-2.518774270648062), dotstyle); label("$F$", (-2.39,-2.36), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

$PN$ intersects circle at $F$. $NF \cap BC=G$. $-1=(F,N;G,P)=(CF\cap BP,M; B,P) \implies CF \parallel BP \parallel DE$. If $CE \cap (ABC)=H$, then $FE=EC, FC \perp AB \implies HN \parallel FC \parallel ED$ which is enough.

We remove $A$; all we need from the information about $A$ is that $E \in PN$ and $BE$ is perpendicular to $BP$. Instead, we start with an isosceles triangle $BCP$ with $PB = PC$, and construct $\omega$ as the unique circle tangent to $BP$ and $B$ and $CP$ at $C$. (Note that $\omega$ only has to pass through $C$; the property $PB = PC$ would automatically ensure that $\omega$ is tangent to $CP$ at $C$.) We replace the condition $ED || BM$ with $ED \bot BE$, since they are indeed equivalent due to $BE \bot BP$. Now invert at $B$, and you get the following very nice picture. The description changes as follows. Let $BCP$ be an isosceles triangle with $BC = CP$. Let $\omega$ be the line parallel to $PB$ passing through $C$. Let $M$ be the point on the plane such that $P$ is the midpoint of $BM$. The line $\omega$ and the circumcircle of $BCM$ intersects at $N$. The circumcircle of $BPN$ intersects the line perpendicular to $BP$ passing through $B$ at $E$. (In short, $E$ is the antipodal point of $P$ with respect to the circumcircle of $BPN$.) Now let $D$ be point on the circumcircle of $BCM$ such that $BD \bot DE$ (i.e. it is the second intersection of the circumcircle of $BCM$ and the circle with diameter $BE$.) Show that the circumcircle of $CDE$ is tangent to the line $\omega$. Solution. Let $F$ be the second intersection of the line $\omega$ (or $NC$) with the circumcircle of $BPN$. It suffices to show that $\angle EDC = \angle ECF$. Now notice that \[ \angle EDC = \angle EDB - \angle CDB = \pi/2 - \angle CNB = \pi/2 - \angle FNB = \pi/2 - \angle FEB, \]with the second equation being true since the problem condition implies that $B, C, D, N, M$ lie in one circle. Since $\overline{BE} \bot \overline{BM} || \overline{CN} = \overline{CF}$, we have $\angle ECF = \pi/2 - \angle CEB$. So it remains to show that $\angle CEB = \angle FEB$. It now suffices to show that $F$ is actually the reflection of $C$ with respect to the line $BE$. Since $CNMB$ is cyclic and $\overline{CN}$ is parallel to $\overline{BM}$, $N$ is actually the reflection of $C$ with respect to the perpendicular bisector of $BM$, which in turn is the line through $P$ perpendicular to $BM$ since $P$ is the midpoint of $BM$. In particular, $PN = PC = CB$. With the same reasoning, since $FPBN$ is cyclic and $\overline{FN}$ is parallel to $\overline{BM}$, $F$ is the reflection of $N$ with respect to the perpendicular bisector of $BP$, which passes through $C$ since $BC = CP$. In summary, the reflection of the perpendicular bisector of $PB$ brings $P$ to $B$, $N$ to $F$, and $C$ to itself. Thus $CB = CP = PN = BF$. Since $\overline{BE}$ is perpendicular to $\overline{CF}$, the rest follows. For some reason, I've written too much for showing that $\overline{BE}$ is the perpendicular bisector of $CF$. Seriously, I think that the inverted diagram speaks for itself. When I tried to solve it, I noticed that maybe I could invert at $B$, but I was also looking towards inverting at $C$. The point $M$ was very annoying to me, maybe because I actually looked at $C$ instead of $B$ for a long time. Initially, I only actually try to invert at $B$ not because $M$ can be inverted nicely, but because I noticed that the circle $PCN$ is tangent to $PB$ at $P$ (after which I removed $M$.) After solving it this way, I realized that inverting $M$ at $B$ adds no complication at all...

Attachments:

For homothety reasons it suffices to show that if $N'$ is the reflection of $N$ over $\overline{AB}$ and $E'=\overline{AB} \cap \overline{CN'}$, then $E',N,P$ are collinear. This is straightforward by complex, with $A=-1$ and $B=1$, so $c$ lies on the unit circle. First, $$P=\frac{2c}{c+1} \implies M=\frac{3c+1}{2c+2}.$$Then we can calculate $$N=\frac{c-\frac{3c+1}{2c+2}}{\frac{c(c+3)}{2c+2}-1}=\frac{2c^2-c-1}{c^2+c-2}=\frac{2c+1}{c+2}.$$In general we should have $E'=\tfrac{cn+1}{c+n}$ by complex intersection. Plugging in our expression for $N$ yields $$E'=\frac{\frac{2c^2+c}{c+2}+1}{c+\frac{2c+1}{c+2}}=\frac{2c^2+2c+2}{c^2+4c+1}.$$Now we want to show that $$\frac{\frac{2c^2+2c+2}{c^2+4c+1}-\frac{2c+1}{c+2}}{\frac{2c^2+2c+2}{c^2+4c+1}-\frac{2c}{c+1}} \in \mathbb{R} \iff \frac{(c+1)((2c^2+2c+2)(c+2)-(2c+1)(c^2+4c+1)}{(c+2)((2c^2+2c+2)(c+1)-2c(c^2+4c+1))} \in \mathbb{R}.$$It turns out that this second fraction simplifies nicely to $$\frac{(c+1)(-3c^2+3)}{(c+2)(-4c^2+2c+2)}=\frac{3}{4}\cdot\frac{(c-1)(c+1)^2}{(c-1)(c+2)(2c+1)}=\frac{3}{8}\cdot \frac{c+\frac{1}{c}+2}{c+\frac{1}{c}+\frac{5}{2}},$$which is evidently real. $\blacksquare$

You want to prove that if $CE$ intersect $\omega$ at $F$, then $NF\parallel BP$. But this is the same as if $C'$ is the reflection of $C$ about $AB$ then we want $C', N, P$ colinear. But project from $C$, $CC'$ is parallel to $PB$, $CN$ bisects $BP$, and $C'C$ is tangent to $\omega$, so $(C', N; B, C)=-1$, so $P, N, C'$ colinear. QED

Good problem. I have different sol. Let $AN\cap BC=L$, and $E'$ be the foot of perpendicular through $L$ to $AB$. Since $MB$ is tangent to $(N C B)$ we have $MN*MC=MB^2=MP^2$, then $MP$ is tangent to $(C N P)$. Then $\angle NPM=\angle PCM=\angle NBC=\angle NBL$, since $(L E' B N)$ is cyclic, $\angle NPM=\angle NBL=\angle NE'L$ adding to this $E'L//PB$ (because $\angle LE'B=\angle PBE'=90$) gives that $P$,$N$,$E'$ are collinear which means that $E'=E$. Then we know that $( C A E L)$ is cyclic, then $\angle LEC=\angle LAC=\angle NAC=\angle PCM$ which means that $PN$ is also tangent to $(C D E)$, so done:).

Good problem. Here is my solution. [asy][asy] usepackage("tikz"); label("\begin{tikzpicture}[font=\small] \coordinate[label=below left:$C$] (C) at (0,0); \coordinate[label=above:$A$] (A) at (0,6); \coordinate[label=below:$B$] (B) at (5,0); \coordinate[label=above right:$O$] (O) at (2.5,3); \coordinate[label=below:$P$] (P) at (2.5,-2.08); \coordinate[label=right:$M$] (M) at (3.75,-1.04); \coordinate[label=above left:$N$] (N) at (3.09,-.86); \coordinate[label=above right:$E$] (E) at (4.06,1.13); \coordinate[label=below:$D$] (D) at (2.03,-.56); \draw[thick] (A)--(B)--(C)--cycle; \draw[thick] (C)--(P)--(B); \draw[thick] (D)--(E)--(P); \draw[dashed] (1.93,.94) circle (2.14); \draw[thick] (C)--(E); \draw[thick] (C)--(M); \draw[thick] (C)--(O)--(N); \draw[thick] (O) circle (3.91); \draw[thick] (B)--(N); \coordinate[label=below right:$N'$] (N') at (4.41,-1.22); \draw[thick] (P)--(N')--(B); \draw[thick] (M)--(N'); \draw[dotted] (2.5,.46) circle (2.54); \foreach \s in {A,B,C,O,E,D,M,N,P,N'}\filldraw (\s) circle (1.5pt); \end{tikzpicture}"); [/asy][/asy] Let $ED \cap CP=X$ and $O$ is the center of $\omega$. Claim. $ONXC$ is a kite Proof. Note that using Power of Point we have $MB^2=MN \cdot MC=MP^2$, therefore by Alternate Segment Theorem $MP$ tangent to $(CNP)$. So, $\angle{MPN}=\angle{DEN}=\angle{XEN}$. Thus, $\angle{XCN}=\angle{XEN} \implies$ $CENX$ cyclic. Furthermore, because $ED\parallel BM$ thus $\angle{DEO}=\angle{XEO}=90^\circ$. Because $\angle{XCO}=90^\circ$ we have $OEXC$ cyclic. Therefore, $ONEXC$ is cyclic quadrilateral. So $\angle{ONX}=\angle{OEX}=90^\circ$, and because $ON=OC$ we have $ONXC$ is a kite. $\square$ Therefore, we have $\angle{XON}=\angle{COX}=\angle{CEX}=\angle{CED}$ but we know that $\angle{XON}=\angle{XEN}=\angle{MPN}=\angle{PCD}$. By Alternate Segment Theorem, we have $(CDE)$ is tangent to $\omega$. $\blacksquare$