90 
Hence MAY O is a right angle. 
Similarly MBY O, and MC¥ O are right angles. 
... a circle upon OM as diameter will pass through AY, BY, CY. 
ll. The triangle AY BY CY is similar to Nagel’s triangle A’’ B’C”. 
Z BY CY A* is the supplement 7 BYOAY. 
ZL BUOAY == AOD: 
—=7—($A+4B). 
=A+tB+C—4(A+ B). 
== @-+-4(4-_ 8): 
z— £ BY OAY—=x—[C+3(A-+ B)]. 
=A+ B+ C—[C+4(A+ B)]. 
=34(A+ B) 
aie De OAT FCA APB). 
MA* is perpendicular to AA’. 
MC’ is perpendicular to CC’. 
PCM At MO y= AAS OC). 
be Oxy Lo AY MCV A’ OC. 
=3(A+C). 
Bat: 7 4* CY. AY BY Ce. 
Sa AB OY = FCA G,) 
Similarly 7 BY AYC’Y=3(B+(C). 
The angles of the triangle A” B’C” (Nagel’s triangle) are 4 (B+ (C), 
3 (A+C), } (A+B), respectively. (Schwatt’s Geometric Treatment of Curves, 
page 39.) 
.'. AY BYC’ is similar to A” B’C”. It is also similar to A’B’C’, for A’ B’C’ * 
and A’ B’C” are similar. ; 
III. O is the centre of perspective of AY BY CY and A’ B/C’. 
IV. J, the centroid of ABC, is the internal center of similitude of the cir- 
cumscribing circles of d’”B’’C” and AY BY CY. 
OM is parallel to HQ. 
It is known that H, F, M, are collinear, as are also O, E, Q. 
.. HM and O@ intersect at EF. 
.. Fis the internal center of similitude. 
V. Lis also the center of perspective of AY BY CY and A” B’C”, 
For, consider the triangle AA’ A’, 
A” A¥ is a median and so is A Ma. 
.. A” AY passes through L. 
Now consider the triangle Bb” b’. 
