TITLE: Help Finishing Proof ?!
QUESTION [6 upvotes]: Let $AB$, $AC$, $AD$ be three chords in a circle. Denote $M, N, $ and $P$ the pairwise intersections of the circles of diameters $AB, AC$, and $AD$. Show that $M, N, P$ are collinear. 
I am attempting to approach the proof using angle chasing. 
So far I have: $DANP$ is a cyclic quadrilateral, so $\angle ADP + \angle ANP = 180^\circ$. Denote $\angle ADP = \theta$. Then $\angle ANP = 180^\circ - \theta$. If we can show that $\angle ANM = \theta = \angle ADP$, we are done.
We also have that $AMBN$ is a cyclic quadrilateral so $\angle ANM = \angle ABM = \frac{1}{2}\text{arc}AM = \beta$. Now we must show that $\beta = \theta$. 
$AMCP$ is cylic  
I am having trouble with chasing the rest of the angles necessary to finish the proof.

REPLY [2 votes]: Here's a slight variation on the proof you're looking for. $$\begin{align}
\angle APM & = \angle ACM \quad \textrm{as ACPM cyclic}\\ &= \angle ACB  \quad \textrm{as C,M,B collinear}\\ &= \angle ADB \quad \textrm{as ABCD cyclic}\\ &= \angle ADN \quad \textrm{as B,N,D collinear}\\ &= \angle APN \quad \textrm{as ANDP cyclic}\end{align}$$ so $M$, $N$, $P$ are collinear.  Note that $C$,$M$,$B$ are collinear because $\angle AMB = \angle AMC = 90^\circ$, since $AB$, $AC$ are diameters of their respective circles.