A sequel to the first six books of the Elements of Euclid, containing an easy introduction to modern geometry, with numerous examples. By John Casey.

LETMOINE'S, TUCKER'S, AND TAYLOR'S CIRCLES. 187
7n1, cuts the circle T orthogonally, and similarly the
circles whose centrog', C, and radii w2,, wr3 cut T orthogonally. Hence the proposition is proved.
Prop. 8.-ITf o- be the semiperimeter of the triangle a/3y,
and p, pi, P2, p3, the, radii of its incircie and circumoircies;
then the squares of the radii of Taylor's circle are, re8)OectiVely, p2 + -12, pi2 ~ (o- - )2, p22+ (o- -/3)2, p3 I+(o- _y)2.
Dem.-Since the triangle A'N"C' is right-angled at
N", and A'C is bisected in /3, N"/3 is equal to A'/3;
that is [-Euc. I. xxxiv.], equal to ay. In like manner,
aK" is equal to fly. Hence N"K" = 2o-; and since the
circle T passes through the points iN", TK", and is concentric with the incirele of a/3-y, we have the square of
the radius of T = p2 ~ 1N"IKi"2 = p2 + ay2.
Again, if M1"C' be joined, the figure C'M"/B'K" is a
rectangle. Hence M"'K" = 2aB' = 2/37y = 2a, but N"IK"
= 2o-;. -. N"M" = 2 (o- - a), and, as bef ore, the square of
the radius of T, = p,2 + (o — a)2. Hence the proposition
is proved.
CoRn-The sum of the squares of the radii of Taylor's
circles is equal to the, square of the diameter of the circumcircle.
For it is easy to see that the squares of the radii of
the four circles are, respectively, equal to,
4R2(sin2A sin-B sin2C + cos2A cos2B cos2C),
41R(cos'A sin'B sin2C + sin-A cos1B cos2C),
4iR2(sin 2A cos2B sin2C + COS2A sin2B cos2C),
4112(sin2A Sin2B cos2C + cos2A cos2 B sin2C),
and the sum of these is 4R2.
Exercises.
1. The chords DE, EF, FD of Lemoine's hexagon meet the
chords F'D', D'E' E'F', respectively in three points forming a
triangle homothetic with ABC.
2. The triangle formed by the three alternate sides iDF', FE',
ED', produced, is homothetic with the orthocentric triangle, and
their ratio of similitude is 1: 4 cos A cos B cos C.