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.

THEORY OF HARMONIC POLYGONS. 211
If n be odd, the intercept made by one of the sides
on Ka is equal to Ka. Hence, in each case, the sum
of the reciprocals of the intercepts made by all the sides
on Ka is equal to n/Ka. Therefore a is a point on the
locus of P. Similarly, the pole of each symmedian
line is a point on the locus. Hence the locus passes
through the poles of all the symmedian lines; and since
it must be a right line [Ex. 62, page 155], it is the
right line through these points. Hence the proposition
is proved.
CoR. 1.-The point K and its polar, with respect to the
circumcircle, are harmonic pole and polar with respect to
the polygon. (See Salmon's Higher Curves, Third
Edition, p. 115.)
The harmonic pole and polar are called by French
geometers The Lemoine point and line of the polygon.
CoE. 2.-If a harmonic polygon be reciprocated with
respect to its Lemoine point, the pole of its Lemoine line
is the mean centre of the vertices of the reciprocal polygon.
This follows from Prop. 5 by reciprocation.
Prop. 6.-If the lengths of the sides of a harmonic polygon be a, b, c, sc., and the perpendiculars on them from
any point P in the Lemoine line be a, /3, y, sc., then the
sum a/a + P/b + y/c + oc., = 0.
Dem. —Let KP intersect the sides in the points R1,
R2, &c. Then we have
( l/KR1-l/KP)+(l/KR2-l/KP)+...(1/KiR,,- 1/KP)=0.
Hence
PRI/KR, + PR2/KR2 +.. PR,,/KR, = 0.
Now, if the perpendiculars from K on the sides
of the polygon be a', 3', y', &c., PR,/KR, = a/a',
PR2/KER = P/', &c. Hence a/a' + /313'+ y/y', &c.,
= 0; but at ', r/', &c., are proportional to a, b, c, &c.
Hence the proposition is proved.