116 
27. If Ha’’ Hi’” He” are the points on the nine-point circle opposite 
Ha Hp and He respectively, then the Simson’s lines of Mc, H-’”’, and He’ concur 
in Muc, for Ha Hp» is Simson’s line of H-’”. Likewise the Simson’s lines of Mb, 
Hy” and Hy” concur in Mp» and the Simson’s lines of Ma, Ha’’, and Ha’” con- 
cur in Mya. 
28. Now considering Ma Mp Mc as the reference /\ in the nine-point circle, 
let us prove that the Simson’s lines of these same points, 7. e., Mc, Hc’ and H-””, 
ete., concur. 
Chord Me H-’”” is|| He Hc’ and.*. | toMaMp. This is true because Z He 
M. H-”” is a right Z. 
The Simson’s line of M- will be this chord Mc H.’”’, the Simson’s line of H¢’’” 
will pass through E, the foot of this altitude, and the Simson’s line of H-% will 
be side Ma Mp» since it is a point opposite the vertex Me. 
So, also, the Simson’s lines of Mp, Hp’ and H»’” will concur in R, and the 
Simson’s lines of Ma, Ha’, and Ha’” concur in S. 
29. Now by noticing the lettering and arrangement of (Fig. 17) it will be 
seen that H.’”, Hp’ and H.’” correspond to Ha’, Hy’ and H_’ of that figure, 
and that Ha, Hp, and He correspond to A”, B’, and C” of that figure. There- 
fore we know at once that the Simson’s lines of Ha, Hp» and He and of Ha’, 
H,’” and H-<’” concur just as in that case by threes in four different points, the 
point of concurrency of HaH» and He being the ortho-center S’” of the triangle 
formed by the intersection of the Simson’s lines of the other three points. 
30. Also since /\ Ha” Hp’. and points Ha, Hp and H¢- bear the same re- 
lation to the nine point circle that /\ ABC as points Ha’, Hy’, and H-’ do to the 
circle in Fig. 5, it follows at once that what was true of the Simson’s lines of 
those points is also true of these. 
Depending upon this same comparison between Figs. 5 and 9 it follows that 
the Simson’s lines of points A, B and C in Fig. 5 concur at in-center of rt. A 
Ma’ Mp Me. 
31. Now let us prove that 8S” (Fig. 5), the point of concurrency of Simson’s 
lines of A’, B’’, C’, is also the in-center of (\ Ma MyMev. 
We have already proven that the Simson’s lines of Ha’, Hy’ and H.’ form a 
Simson triangle X YZ, of which the Simson’s lines of A’, B’ and O” are the alti- 
tudes, and that /\ XYZ, is equivalent to /(, A” B”’C” with their sides respect- 
ively |. Now, since Ha, Hp and He are the midpoints of the sides of /(\ XYZ, 
/\ HaHpHe will be equal in every respect and similarly placed to /\ MayMp-Mey. 
We have also proven H to be the in-center of this /(\ HaHpHe. Again, since the 
Simson’s lines of A”, B’ and C” bisect A” H, B’” H and C” H, respectively, it is 
