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B”’ BY is a median and so is BM). 
.. B’ BY passes through E. 
In the same way we can show that C’’CY also passes through FL. 
.*. Eis the center of perspective of A’ B’C” and AY BY CY. 
VI. All the lines in AY BY CY are just one-half the corresponding lines in 
A” BY C0”. 
This is an immediate consequence of the fact OM —}3 HQ. (Schwatt, 
Geomet. Curves, page 40.) 
_ VII. The sides of the triangle AY BY CY are oppositely parallel to the cor- 
responding sides of A’’ B’’C”’, i. ¢., AY BY is parallel to B’ A”, ete. 
OM is parallel to HQ. 
HA” is perpendicular to AA’. 
MAY is perpendicular to AA’. 
AC OR OMA S. 
In the same way 
Vogts O=—-, OMB™ 
ZO IO = A OME ; 
This shows that the points AY, BY, CY are located with respect to O, just as 
A’, BY’, C” are located with respect to Q. 
Z (OM, BY AY) is measured by 4 (are OBY + are AY CY 4 are CY M). 
Z (HQ, A’ B’’) is measured by 4 (are B’Q + are A”’C’ + are C’H). 
But arc OBY measures the same angle in the circle on OM as diameter, that the 
are B’’(@ measures in the circle on HQ as diameter. 
The same is also true of the arcs AY CY and A’’C”, and CY M and C’ H. 
ee COMME AY ) =" (HO) AB). 
But since OM is parallel to HQ, we have at once A’ B” parallel to BY AY. 
In the same way we may prove that B’C” is parallel to CY BY and C’ A” 
parallel to AY CY. 
.*. the sides of A¥ BY CY are oppositely parallel to the corresponding sides of 
A” BY’ OC”, 
VIII. The triangle AY BY CY is Nagel’s triangle for the triangle Ma Mp Me. 
It is known (Schwatt, page 41) that O is Nagel’s point in the triangle 
M, My M., and that M is the orthocenter. The circle on OW as diameter is Nagel’s 
circle for the triangle MaMpMc. We know that the sides of MaMpMc- are oppo- 
sitely parallel to the sides of ABC, and we have proven that AY’ BYC’, inscribed 
in the Nagel’s circle of MaMy Me, has its sides oppositely parallel to the sides of 
Nagel’s triangle for ABC. 
.. AY BYCY is Nagel’s triangle for Ma My Me. 
