Problem 1

Let

A,B,C,D

be four distinct points on a line, in that order. The circles with diameters

AC

and

BD

intersect at

X

and

Y

. The line

XY

meets

BC

Z

. Let

P

be a point on the line

XY

other than

Z

. The line

CP

intersects the circle with diameter

AC

C

and

M

, and the line

BP

intersects the circle with diameter

BD

B

and

N

. Prove that the lines

AM

DN

, and

XY

are concurrent.

Step 1 of 3: Locate the intersection from

DN

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In plain words

Right angles from the diameter circles turn into similar triangles.

Q=DN\cap XY

Detailed analysis

Let $Q=DN\cap XY$ . Since $B,N,P$ are collinear and $D,N,Q$ are collinear, the right-angle cyclic relations give $\angle QDZ=90^\circ-\angle NBD=\angle BPZ$ . Thus triangles $QDZ$ and $BPZ$ are similar, so $QZ=\frac{BZ\cdot DZ}{PZ}$ .