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.

215

Dem.-Let KA be divided in A" in the ratio 1: m.
Join AO, OK, and draw A"O' parallel to AO, and let
A"U be parallel to the tangent AT. Then we have
lR
O'U2 = O'A"2 + A"U2; but O'A" =, and A"U
+ m'
KU -+- - Rtano n. Hence (l+m)2 O'U2 = R2
I + m l+ l Hm
(12 + m2 tan2w). Hence O'U is constant.
Again, if B"V be parallel to the tangent at B, the
triangle O'B"V is in every respect equal to O'A"U.
Hence the angle TJO' is equal to AOB. Therefore
the points U, V... are the angular points of a polygon
similar to that formed by the points A, B... Hence
they form a harmonic polygon. It is evident, by proceeding in the opposite direction from A, that we get
another harmonic polygon. Hence the proposition is
proved.
CoR. 1.-The intercept which the circle 0' makes on the
side AB is 2h (I sin A - m cos A tan o)/(t + m), where A
denotes the angle of intersection of the side AB with the
circumcircle.
For the perpendiculars from 0 and K on AB are, respectively,
R cos A, R sin A tan o,
and OK is divided in O' in the ratio m: I. Hence the
perpendicular from O' on AB is
R (I cos A + m sin A tan wo)/(l + im);
and subtracting the square of this from the square of
the radius of O' we get
R2 (I sin a - m cos A tan W)2/(l + m)2.
Hence intercept = 2R (I sin A - m cos A tan o)/(l + m).
COR. 2.-By giving special values to the ratio I: m,
we get some interesting results. Thus1~. If I = 0 we get intercepts proportional to cos A,
Q2