Advanced High-School Mathematics

SECTION 1.5 Nine-Point Circle 45

(XZ) and (AC) are parallel as are (PX) and (BY′). Therefore,∠PXZ
is a right angle. By the theorem on page 35 we conclude that the
quadrilateral Y PXZ is cyclic and hence the corresponding points all
lie on a common circle. Likewise, the quadrilateralPXZZ′ is cyclic
forcing its vertices to lie on a common circle. As three non-collinear
points determine a unique circle (namely the circumscribed circle of
the corresponding triangle—see Exercise 8 on page 17) we have already
thatP, X, Y, Z, andZ′all lie on a common circle.
In an entirely analogous fashion we can show that the quadrilaterals
Y XQZandY XZRare cyclic and so we now have thatP, Q, R, X, Y, Z,
andZ′all lie on a common circle. Further analysis of cyclic quadrilat-
erals putsY′andZ′on this circle, and we’re done!

C

X

B

Q

Z' Z

A

R

O

P

Y'

Y

X'

Note, finally, that the nine-point circle of 4 ABC lies on this trian-
gle’s Euler line (see page 22).

Exercises.

1. Prove that the center of the nine-point circle is the circumcenter
   of 4 XY Z.


2. Referring to the above diagram, prove that the center of the nine-
   point circle lies at the midpoint of the segment [NO], whereN is
   the orthocenter of 4 ABC.