# Help Finishing Proof ?!

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. • I love this problem! But I wonder how the figure can look so fine despite $AD$ being so obviously NOT the diameter of its circle May 20 '17 at 0:36

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.