BD 
be also the nine-point circle of the /\ EKF, and therefore the second intersection 
of the nine-point circle with the side, EF, must be the foot of the altitude to EF, 
or KS’. 
Hence S” is the foot of the altitude KN. But any side and its altitude is a 
pair of conjugate Simson’s lines, and since EF is a Simson’s line of a point on the 
circumcircle of ABC, KN is the Simson’s line conjugate to EF. 
Any triangle like EKF formed by three Simson’s lines, the altitudes of which 
are Simson’s lines conjugate to the sides, and haying the nine-point circle in 
common with the triangle ABC, we shall call 2 Simson Triangle. 
Since the nine-point circle is common to both triangles ABC and EKF, the 
radius of the nine-point circle is one-half the radius of the circumcircle of either 
triangle; therefore the radius of the circumcircle of any Simson triangle is equal 
to the radius of the circumcircle of the original triangle. 
19. The common vertex S’” of the pair of limiting Simson’s lines belonging 
to TS is on the same straight line as K, N and 8’. For, since T’S’” is a diameter 
of the nine-point circle, 7 T’S’S’’” — 90°, or S’8’” | to EF. .°. S’” is on the 
altitude KS’. 
20. A’, BY’ and C” are points on the circumference opposite Ha’, Hp’ and 
H-’ respectively. (Fig. 5.) Prove that the Simson’s lines of A’’, B’’, and C” are 
|| respectively to AA’, BB’ and CC” and that they are | respectively to B/C”, 
A”’C’, A’ B’. Now the angle between the Simson’s lines uf A and A” will be 
equal to an angle measured by 3 arc AA”. But the angle between AA’ and 
AHa, the Simson’s line of A, is measured by an are equaltothis. Therefore Sim- 
son’s line of A” is|| to AA’. So the Simson’s line of B” is|| to BB’ and Simson’s 
line of C’’ || to CC’. 
Now are HH,’ CB” =are H-’ BC” = 180°. 
2, Ato JB GY (OOH Ss pire, 18a) B38 
ae oly? Hes || BC”, also Ep? His’ || A’ Band 
H,’ H-’ || A’C’’. Therefore (A, A” BY’ C” is equivalent to /\ Ha’ Hv’ He’, 
being inscribed in same circle and having sides equal and parallel. 
Z H’jCA = Z Hy’ BA. (From similarity of (Ais ABH» and ACH). 
-. are AH)’ —arc AH,’, .*. since AA’ is a diameter it must be | to chord 
H.’ Hp’ and therefore to B’ C’. So we may prove BB’ | A” C” and CC’ | 
A” B”. 
21. The Simson’s lines of Ha’ Hi’ He’ form a Simson’s triangle XYZ of 
which the Simson’s lines of A’, B’’, C’ are the altitudes, A”, C’, B” being 
points opposite Ha’, Hy’, He’ respectively. 
