Given regular octagon RSTUVWXY with each side of

Figure 7.37 PERSPECTIVE ON APPLICATION The Nine-Point Circle In the study of geometry, there is a curiosity known as the Nine-Point Circle—a curiosity because its practical value consists of the reasoning needed to verify its plausibility. In , in Figure 7.35 we locate these points: M, N, and P, the midpoints of the sides of , D, E, and F, points on determined by its altitudes, and X, Y, and Z, the midpoints of the line segments determined by orthocenter O and the vertices of . Through these nine points, it is possible to draw or construct the circle shown in Figure 7.35. 䉭ABC 䉭ABC 䉭ABC 䉭ABC Although we do not provide the details, it can also be shown that quadrilateral XZPN of Figure 7.36 is a rectangle as well. Further, is also a diagonal of rectangle XZPN. Then we can choose the radius of the circumscribed circle for rectangle XZPN to have the length Because it has the same center G and the same length of radius r as the circle that was circumscribed about rectangle NMZY, we see that the same circle must contain points N, X, M, Z, P, and Y Finally, we need to show that the circle in Figure 7.37 with center G and radius will contain the points D, E , and F. This can be done by an indirect argument. If we suppose that these points do not lie on the circle, then we contradict the fact that an angle inscribed in a semicircle must be a right angle. Of course, , and were altitudes of , so inscribed angles at D, E, and F must measure 90°; in turn, these angles must lie inside semicircles. In Figure 7.35, intercepts an arc a semicircle determined by diameter . So D, E, and F are on the same circle that has center G and radius r. Thus, the circle described in the preceding paragraphs is the anticipated nine- point circle NZ ∠NFZ 䉭ABC CE BF , AD r = 1 2 NZ r = 1 2 NZ = NG. NZ Then must be parallel to With , it follows that must be perpendicular to as well. In turn, , and NMZY is a rectangle in Figure 7.35. It is possible to circumscribe a circle about any rectangle; in fact, the length of the radius of the circumscribed circle is one-half the length of a diagonal of the rectangle, so we choose This circle certainly contains the points N , M, Z, and Y. r = 1 2 NZ = NG. MZ ⬜ NM AD NM CB ⬜ AD MZ . NY To understand why the nine-point circle can be drawn, we show that the quadrilateral NMZY is both a parallelogram and a rectangle. Because joins the midpoints of and , we know that and Likewise, Y and Z are midpoints of the sides of , so and By Theorem 4.2.1, NMZY is a parallelogram. YZ = 1 2 CB. YZ 7 CB 䉭OBC NM = 1 2 CB. NM 7 CB AB AC NM A E F N C B P D M Z Y G O X A E F N C B P D M Z Y G O X A E F N C B P D M Z Y G O X Figure 7.35 Figure 7.36