I chanced across the following problem recently while working on some geometry problems. I did not expect to solve it so easily; hence I am asking for help to verify if my proof is valid, or if any logical gaps are present.
Below is the problem, copied verbatim -
Let $\Gamma$ be the circumcircle of an acute scalene triangle $ABC$. Reflect $A$ in $BC$ to obtain $A'$, and let $A'C$ intersect $\Gamma$ a second time at $P$. Denote by $E$ the altitude from $B$ to $AC$ and denote by $N$ the altitude from $P$ to $BC$. Prove that the intersection of $BE$ and $PN$ lies on $\Gamma$.
For convenience's sake, I have attached a Geogebra-generated diagram below for easy reference.
Anyways, my proof proceeds as follows:
Proof:
Since $B, P$ and $C$ all lie on $\Gamma$, it suffices for us to prove that the points $B, P, C$ and $M$ are concyclic. By the Power of a Point Theorem, this is equivalent to showing that $BN \cdot CN = PN \cdot MN $.
We define points $Q, R, S$ as in the diagram, where $Q$ is the point of intersection of $\overline{BE}$ and $\overline{AA'}$, $R$ is the point of intersection of $\overline{AC}$ and $\overline{PM}$, and $S$ is the altitude from $A$ to $\overline{BC}$. We note that $R$ is also the point of reflection of $P$ over $\overline{BC}$ onto $\overline{AC}$; hence $\triangle CNR \equiv \triangle CNP$.
Now, we proceed by simple angle chasing. Since $\overline{AA'} \perp \overline{BC}$ and $\overline{PM} \perp \overline{BC}$, it is easy to see that $\overline{AA'} \parallel \overline{PM}$. We then have the following:
Let $\angle AQE = \theta \Rightarrow \angle QAE = 90^{\circ} - \theta$. Additionally, we have that $\angle RME =\theta$ and $\angle MRE = \angle NRC= \angle NPC= 90^{\circ} - \theta \Rightarrow \angle NCR = \angle NCP = \theta$. Now, $\angle RME = \angle NMB = \theta \Rightarrow NBM = 90^{\circ} - \theta $.
From the above, we essentially have that $\triangle BNM \sim \triangle PNC$. Thus, $\frac{BN}{NM}=\frac{PN}{NC} \Rightarrow BN \cdot NC = PN \cdot NM$, and we are done.
On a side note, why is the scalene triangle condition even necessary?
Another proof: Notice $ENCM$ cyclic. SO, $\angle BMP=\angle EMN=\angle ECN=\angle ACB=\angle BCA'$(reflection property)$=\angle BCP$
So,$BCMP$ cyclic.Hence the claim follows.