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H.”, being midpoint of AH, is the center and therefore chord Ha’”’Hp — chord 
H.’’He. Therefore HaHa” bisects 2 Ha, HpHv” bisects 7 Hp, and H-H-” 
bisects 7 He. 
H.”’, Hp” and H-” are points on the nine-point circle opposite Ma, Mp, Me, 
respectively, for, since MMa and HaHa” are | and the center of the nine-point 
circle is F, the midpoint of MH, a line from M, through F, will meet AHa on the 
circumference of the nine-point circle, necessarily at H,’’. In like manner H)/” 
can be shown to be opposite Mp, and H” opposite Me. 
This proves Ha’’H»” || to MaMp and equal to it, and therefore || to AB, 
Hy’H-’ equal to MpMc and parallel to both MpMe and BC, and H.” Ha” equal to 
McMa, and parallel to both MceMa and CA. 
Now, the Simson’s line of Ma will be | to BC and | to AHa, for HaC is isogo- 
nal conjugate to HaMa since AH, bisects 7 HpHaHe, thus making 7 CHaHy = 
Z MaHaHe. Therefore the Simson’s line of Ha”, since it is conjugate to Simson’s 
line of Ma, is | to MeMp, Hp’ He and BC. In like manner we can prove the Sim- 
son’s line of My | AC and | to BH» and, therefore, the Simson’s line of Hy” | to 
MaMce, He’ Ha” and AC; and Simson’s line of Mc | AB and | to CHe and, therefore, 
the Simson’s line of He” || to MaMp, Hp” Ha” and AB. 
Now (A MnaMupMuc is oppositely similar to (A) HaHpHe. Also, from what 
we have proven before, the Simson’s lines of Ma and Ha” must both pass through 
Mua. Then the Simson’s line of Ma being || to AHa, will bisect 7 MurMua Mue. 
In like manner we can show that the Simson’s lines of Mp and Me bisect /s MuHa 
Map Mae and MuaMuceMup. Now the Simson’s lines of Ha’’, Hp’ and He” are | 
to Simson’s lines of Ma, Mp and Mc, respectively, and therefore they are the exter- 
nal bisectorsof the /sof thert. /\ MHaMupMne. Since the internal and external 
bisectors of a /\ meet by threes in four points, we may conclude the Simson’s lines 
of Ma, Hp”, and H-” concur in X, Simson’s lines of Mp, He’, and Ha” concur in 
Y, Simson’s lines of Mc, Ha”, and Hy” concur in Z, and Simson’s lines of 
Ma My Mc concur in S’”. 8”, we see, is the orthocenter of (| XYZ and in-center 
of A. Mua MupMuce. 
A\ XYZ has its sides || to sides of (\ MaMpMc and oppositely || to /\s Ha’ Hy” 
H.” and ABC. 
26. Let us prove /\ XYZ equivalent to (js MaMpMe and Ha” Hp He” and, 
therefore, inscribable in the nine-point circle. 
A\ MuaM pM ue bears the same relation to /\ XYZ that (\ HaHpHe does to 
A ABC. Therefore, since (\ MaaMupMuc is } the size of (\ HaHpHe, /\ XYZ 
must be } the size of rt. (\ ABCand thus equivalent to (\ MaMpMe, and rt. /\ Ha” 
H)”H-” and hence inscribable in the nine-point circle of the fundamental ING 
