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.

28

A SEQUEL TO EUCLID.

6. In a quadrilateral the sum of the squares of two opposite
sides, together with the sum of the squares of the diagonals, is equal
to the sum of the squares of the two remaining sides, together with
four times the square of the line joining their middle points.
7. Divide a given line AB in C, so that the rectangle under BO
and a given line may be equal to the square of AC.
8. Being given the rectangle contained by two lines, and the
difference of their squares: construct them.
9. Produce a given line AB to C, so that AC. CB is equal to the
square of another given line.
10. If a line AB be divided in C, so that AB. BO = AC2, prove
AB2 + BC2 = 3AC2, and (AB + BC)2 = 5AC2.
11. In the fig. of Prop. xi. prove that(1). The lines GB, DF, AK, are parallel.
(2). The square of the diameter of the ( about the A FHK
= 6HK2.
(3). The square of the diameter of the 0 about the A FHD
= 6FD2.
(4). The square of the diameter of the ( about the A AHD
= 6AH2.
(5). If the lines EB, CH intersect in J, AJ is 1 to CH.
12. If ABC be an isosceles A, and DE be II to the base BC,
and BE joined, BE2 - CE2 = BC. DE.
13. If squares be described on the three sides of any A, and
the adjacent angular points of the squares joined, the sum of the
squares of the three joining lines is equal to three times the sum
of the squares of the sides of the triangle.
14. Given the base AB of a A, both in position and magnitude,
and mAC2 - nBC2: find the locus of C.
16. If from a fixed point P two lines PA, PB, at right angles
to each other, cut a given 0 in the points A, B, the locus of the
middle point of AB is a 0.
16. If CD be any line 1] to the diameter AB of a semicircle, and
if P be any point in AB, then
CP2 + PD2 = AP2 + PB2.
17. If 0 be the mean centre of a system of points A, B, C, D,
&c., for a system of multiples a, b, c, d, &c.; then, if L and M
be any two 11 lines,
2 (a. AL2) -: (a. AM2) = (a). (OL2 - OM2).