# Inscribed square in circle generates three collinear points.

Let $ABCD$ be a square inscribed in a circle $\mathcal{C}$ and P in $\mathcal{C}$ different from $A$. Assume that $PA$ intersects $BD$ at the point $E$. Let $M$ be the intersection of the line $PB$ with the line parallel to $AC$ that goes through $E$. Prove that $M$,$C$, and $D$ are collinear.

The idea is to prove this without a coordinate system (using properties of cyclic quadrilaterals, inscribed angles and such). I think I should do this in separate cases depending on the position of P relative to $A,B,C,D$, but I'm stuck in the first case (the first image) I'm trying to show that $\angle MCB=90°$, but I'm not sure how. I've observed that $DEMP$ is always cyclic, but I couldn't prove it either.

Here there is a Geogebra doc.

In the following solution we only use the fact that $ABCD$ is cyclic and $AB=BC$. We will assume that $P$ lies on the arc $CD$; the proof in other configurations is similar.
We have to prove that the lines $PB$, $CD$ and the line parallel to $AC$ through $E$ are concurrent. It will be convenient for us to consider the common point of the lines $CD$ and the line parallel to $AC$ through $E$, call this point $N$. We need to prove that $N$ lies on $PB$.
Note that $DENP$ is cyclic, because $$\angle EPD = \angle APD = \angle ACD = \angle END.$$ It follows that $$\angle NPE = \angle NDE = \angle CDB = \angle BPA = \angle BPE.$$ Therefore $B,N,P$ are collinear, or equivalently $N$ lies on $BP$.
I'll take configuration from 1.st picture. The others are similar. Redefine $M$ as to be point of intersection beetwen $CD$ and $PB$. Since $$\angle EPM = \angle APB = 45^{\circ} = \angle BDC = \angle EDM$$ we see that $DPME$ is cyclic. Since $$\angle DPM = \angle DPB = 90^{\circ}$$ we have $$\angle DEM = 90^{\circ}$$ so $$EM\bot BD \implies EM||AC$$