Proof of Nine-Point Circle

by

Samuel Obara

Adapted from a proof on the website of the computer Science and Mathematics Department, Arkansas State University

Given triangle ABC. let L, M and N be the mid points of the sides ABC and PQR be the altitudes of the triangle ABC. Let H be the orthocenter, where as D,E and F be the midpoints of AH, BH and CH. O is the circumcenter with the circumcircle as shown below.

Still having L, M, N as the mid points of the sides, where as P,Q, R are the feet of the altitudes and H as the orthocenter. Let AH =20M, since D bisects AH, we have DH=OM as illustrated below.

Given DH =OM and DH parallel to OM because both are perpendicular to BC. Therefore the diagonals OH,

DM bisect each other at a point X as shown below.

Given OH and DM bisect each other at a point X, so XD =XM. Also XP =XD=XM because angle DPM is a right angle (AP is an altitude). Extend AP to meet the circumcircle at point K. Then we have HP=PK as indicated in the diagram below.

Given X is the midpoint of HO, and P is the midoint of HK, therefore, by similar triangle HXP and HOK, XP =OK/2=circumradius/2. Now if we do the same thing with Q,N and E that we just did with P, M and D, we get the same center X and the same radius circumradius/2. And similarlity with R, L and F as shown below

Therefore the nine points lie in the circle as shown below

Back writeup 5

04-20-2001