106 
Then, since three points determine a circle, Ma, Mp, Mc, Ha, Hp, He, Ha’’, 
H,’” and H.’ must all lie on the same circle. This circle, since it passes through 
nine definite points, is called the nine-point circle. 
A | to the midpoint of Ha Ma meets HM at its midpoint, say F. So Js to 
Hp Mp and He M¢ at their midpoints meet HM in F. 
.’. F, the midpoint of HM, is the centre of the nine-point circle. 
Since it is the circumcircle of rt. “, MaM»Me, which has just half the dimen- 
sions of /\ ABC, its radius will be just half the radius of the larger circle. 
9. The nine-point circle bisects any line drawn from H to the circumcircle 
of the fundamental triangle. 
Let MX and FY be any two | radii of the two circles. Now, since F is the 
mid-point of MH and FY—4MX, HYX is a straight line with Y as its mid- 
point. 
10. Simson’s line of P bisects PH. (Fig. 7.) 
Let us suppose that D is the midpoint of PH. Then D lies on the nine-point 
circle. Then we must prove it lies on P, P,. 
Since PP; and AH are | to BC, they are||. Also since D is midpoint of 
PH, /\ P, DHa is isosceles. Let E be midpoint of AH. Then DE=—3 AP. 
D, E, and Ha are on the nine-point circle. 
A, P, and C are on the circumcircle. 
Then since DE = 3 AP and radius of nine-point circle = 4R, Z EH, D, 
inscribed in nine-point circle, = 4 ACP, inscribed in circumcircle. 
2 Be D7" ACP = 2 DE. re, 
Let the intersection of P, D and AC be P,, 
Then: 7 P) D7 ACE == 7 ECP: 
