105 
chords | AB are concurrent with Simson’s line of C’. Thus there is a triple in- 
finity of sets of three points on the circumcircle, the points of concurrency of the 
Simson’s lines of which lie in the sides of the fundamental triangle. 
7. Since, in the cosine circle (Fig. 6), F’ EDD’, FEE’ D’, and FF’ E’” D 
are all rectangular, it follows at once that the Simson’s line of D’, with regard to 
rt. A, DEF, is FD, of E’, DE and of F’, EF. Also Simson’s line of D, with re- 
gard to rt. \ D’E’F’, is D’ B’, of E, KH’ F’ and of F, YD’. 
8. In Fig. 2, Ma, Mn, Mc are the midpoints of the sides of fundamental 
triangle opposite A, B, and C, respectively. Ha’’, Hp’, He” are the midpoints of 
AH, BH and CH, respectively, where H is the ortho-center. 
Now MpA = MpHa. 
* £ MpHaA= Z MpAHa. 
Likewise 7 McHaA = Z McAHa. 
- Z McHaMp= Z A. 
We also know Z McMaMp=Z A. 
*. Mc, Mp, Ma and Ha are concyelic. 
In the same way we can show 
Mc, Mp, Ma and Hy and Me, Mp, Ma, and H¢ to be concyclic. 
.*. Since three points determine a circle, these six points are all concyclic. 
Now Ha, Hp, He are the feet of the altitudes of (. AHB. 
But we have just shown that, in any triangle, the feet of its altitudes and the 
midpoints of its sides are all concyclic. 
. Ha, Hp and He and Ha”, Hp”, and H-” are concyclic. 
