right angles (C 10), or altogether six, the exterior angles will contain the difference, or four.

As every rectilineal figure may be divided into triangles less by two than the number of sides, the like demonstration can be given, whatever the number of sides.

43. The interior angles are together equal to twice the number of right angles, less four, that the figure has sides. A triangle (Fig. 29).

The entire angles = six right angles (Sect. 41), and the interior angles two (C 10), or four less than double the number of sides.

With four sides (Fig. 30).

The entire angles = eight right angles (Sect. 41).

The exterior angles four right angles (Sect. 42).

=

The interior angles four right angles, in respect of the

=

two triangles into which it is divided, and being four less than double the number of sides.

With five sides (Fig. 31).

The entire angles ten right angles (Sect. 41).

=

The exterior angles = four right angles (Sect. 42).

=

The interior angles six right angles, in respect of the three triangles, and being four less than double the number of sides.

It is obvious that the like proof could be given with any number of sides.

44. To find the side of a square equal in area to the balance in area of several squares, some added and some deducted.

This is so interesting and so useful an application of

Sect. 27 and the deductions from it that the operation is now shown.

In Fig. 32 let it be required to show the side of a square which shall equal in area the balance of area on the addition to the square with side a of with sides b and squares then C, the deduction of a square with side d, and finally the addition

of a square with side f.

1. Draw the base line = side a.

2. Perpendicular to it draw line =b, and draw hypothenuse m.

This gives side of square = a2 + b2.

3. Perpendicular to m draw line = c, and draw hypothenuse n.

This gives a2 + b2 + c2.

4. To deduct square with side = d make line n hypothenuse, and with same as diameter draw semicircle, as shown in figure, and from point q set off side d, and from r draw rs (side o), and this (C 27) will give a2+ b2 + c2- d2.

5. From r draw side = f, and from end of same draw

side p.

Then p2 = a + b2 + c2 - d2 + f as required.

45. In any parallelogram the sum of the squares of the cross diagonals is equal to the sum of the squares of the sides. The demonstration is given in Div. D, Sect. 36, as it depends on a preliminary demonstration which has relation to Differentials.

DIVISION D.

DIFFERENTIALS.

1. INTRODUCTORY REMARKS.

The previous Division (C) having shown the equivalence (including equality) of lines and surfaces, the present Division will explain the differences.

Sectns. to No. 14 have relation to triangles; Sectns. 15 to 22 to the division of lines and the squares and rectangles of the parts; Sectns. 23 to 25 the differences in the squares of the different sides of triangles not being right-angled; Sect. 26 has relation to the different lengths of the sides of triangles, treated of in an entirely new manner; and Sectns. 27 to 35 apply to differences in relation to circles.

2. Any two sides of a triangle are together longer than the

third side.

In the triangle abc, Fig. D 1, let fall on the longest side the perpendicular bd, dividing the triangle into two rightangled triangles.

Then by Sect. 27 the square of the side ab hypothenuse is greater than that of the part ad of the longest side, and square of side bc is greater than the square of the remainder (dc) of the longer side, and the squares being greater, the two sides will be greater than the longest side (ac).

The demonstration in Euclid (1, 20) is more complex, the side ab being continued to f, making bf = bc, and consequently af equal to the two sides ab and bc, then joining of to form

the isosceles triangle cbf, in which the angles bef and bfe will be equal (Euclid 1, 5; Div. C ante, 15).

The angle acf being greater than bef, the opposed side will be greater than side ac (Euclid 1, 18 post, Sect. 8), and such. opposed side (af) being equal to the two sides ab and bc, they will be greater together than the side ac.

3. If any side of a triangle be extended, the exterior angle is greater than either of the interior and opposed angles. (Euclid 1, 16.)

This is proved in Div. C, Sect. 11, Fig. C 5.

The Section referred to proves that the external angle (angle 7 in Fig. 5, C) is equivalent with both the included and opposed angles (1 and 2), and must consequently be greater than either of them.

The external angle is supplemental to the adjacent angle (3), making with it two right angles, and angle 3 with the other angles included in the triangle (1 and 2) will also equal two right angles, as shown in Sect. 10 of Div. C.

4. Any two angles of a triangle are together less than two right angles. (Euclid 1, 17.)

This is a necessary deduction from the demonstration in C 10, that the three angles are together equal to two right angles, and any two must be less.

5. If one side of a triangle be greater than a second, the angle opposed to the first will be greater than the angle opposed to the second. (Euclid 1, 18.)

Let abc (Fig. D 2) be a triangle in which side ac is greater than side ab.

Then angle abc (that opposed to the longer side ac) will be greater than angle acb (that opposed to side ab).

On the longer side (ac) set off ad = ab and join bd, completing the isosceles triangle bad.

Then angle abd = angle adb (C 15). But adb as external angle of triangle bed is greater than angle at c, the angle opposed (C 11).

Angle abd (the equal of adb) will also be greater than the angle at c, and much more so will the angle abc (that opposed to side ac) be greater than the angle at c opposed to side ab.

56. The Converse. If one angle of a triangle be greater than a second, the side opposed will be greater. (Euclid 1, 19.) In Fig. D 2 let angle bac be greater than angle cab. Then side be will be greater than side ac.

If side bc be not greater than side ac it must be equal or less.

If equal, the angles cab and cba would be equal (C 15), which by hypothesis they are not.

If less, the angle bac would be less than angle cab (by Sect. 5), but it is greater.

Consequently side be not being equal or less than side ac it must be greater.

6. If from the ends of any side of a triangle lines be drawn to a point within the triangle, (1) the sides so drawn will be less than the outer sides, but (2) they will contain a greater angle. (Euclid 1, 21.)

In Fig. D 3 let abc be the triangle and bd and cd the lines to the point (d) within the triangle.

Then (1) the lines bd and cd will together be less than the