PROPOSITION XXV. THEOREM.

Two parallels are every where equally distant.

Two parallels AB, CD being с_H given, if through two points assumed at pleasure, the straight lines EG, FH be drawn perpendicular to AB, those straight lines will (Prop. 23.) at the same time be perpendicular to CD:

G D

and we are now to shew that they will be equal to each other. For, if GF be joined, the angles GFE, FGH, considered in reference to the parallels AB, CD, will be alternate interior angles, and therefore (Prop. 23. Schol.) equal to each other. Also, because the straight lines EG, FH are perpendicular to the same straight line AB, and consequently parallel, the angles EGF, GFH, considered in reference to the parallels EG, FH, will be alternate interior angles, and therefore equal. Hence the two triangles EFG, FGH have a common side, and two adjacent angles in each equal; hence these triangles (Prop. 7.) are equal; hence the side EG, which measures the distance of the parallels AB and CD at the point E, is equal to the side FH, which measures the distance of the same parallels at the point F.

PROPOSITION XXVI. THEOREM.

If two angles BAC, DEF have their sides parallel, each to each, and lying in the same direction, those angles will be equal.

Produce DE, if necessary, till it meets AC in G. The angle DEF is equal to DGC (Prop. 23.), since EF is parallel to GC; and the angle DGC is equal to BAC, since DG is parallel to AB: hence the angle DEF is equal to BAC.

H

A

B

G

C

Scholium. The restriction of this proposition to the case where the side EF lies in the same direction with AC, is necessary, because if FE were produced towards H, the angle DEH would have its sides parallel to those of the angle BAC, but would not be equal to it. In that case, DEH and BAC would be together equal to two right angles.

PROPOSITION XXVII. THEOREM.

In every triangle, the sum of the three angles is equal to two right angles.

Let ABC be any triangle. Produce the side CA towards D; and, at the point A, draw AE parallel to BC.

Since AE, CB are parallel, and CAD cuts them, the exterior angle DAE will be equal to C

its interior opposite one ACB; in like manner, since AE, CB are parallel, and AB cuts them, the alternate interior angles ABC, BAE will be equal: hence the three angles of the triangle ABC make up the same sum as the three angles, CAB, BAE, EAD; hence (Prop. 2. Cor. 3.) that sum is equal to two right angles.

Cor. 1. Two angles of a triangle being given, or merely their sum, the third will be found by subtracting that sum from two right angles.

Cor. 2. If two angles of one triangle are respectively equal to two angles of another, the third angles will also be equal, and the two triangles will be mutually equiangular.

Cor. 3. In any triangle there can be but one right angle; for if there were two, the third angle must be nothing. Still less can a triangle have more than one obtuse angle.

Cor. 4. In every right-angled triangle, the sum of the two acute angles is equal to one right angle.

Cor. 5. Since every equilateral triangle (Prop. 12.) is also equiangular, each of its angles will be equal to the third part of a right angle; so that if the right angle is expressed by unity, the angle of an equilateral triangle will be expressed by.

Cor. 6. In every triangle ABC, the exterior angle BAD is equal to the sum of the two interior opposite angles B and C. For, AE being parallel to BC, the part BAE is equal to the angle B, and the other part DAE is equal to the angle C.

PROPOSITION XXVIII. THEOREM.

The sum of all the interior angles of a polygon is equal to as many times two right angles, as there are units in the number of sides diminished by two.

Let ABCDEFG be the proposed polygon. If from the vertex of any one angle A, diagonals B AC, AD, AE, AF, be drawn to the vertices of all the opposite angles, it is plain that the A polygon will be divided into five triangles, if it has seven sides; into six triangles, if it has eight;

E

and, in general, into as many triangles, all but two, as the polygon has sides; for those triangles may be considered as having the point A for a common vertex, and for bases, the several sides of the polygon, excepting the two which form the angle A. It is evident, also, that the sum of all the angles in those triangles does not differ from the sum of all the angles in the polygon: hence the latter sum is equal to as many times two right angles as there are triangles in the figure; in other words, as there are units in the number of sides diminished by two.

Cor. 1. The sum of the angles in a quadrilateral is equal to two right angles multiplied by 4-2, which amounts to four right angles: hence if all the angles of a quadrilateral are equal, each of them will be a right angle; a conclusion which sanc tions our seventeenth Definition, where the four angles of a quadrilateral are asserted to be right, in the case of the rectangle and the square.

Cor. 2. The sum of the angles of a pentagon is equal to two right angles multiplied by 5-2, which amounts to six right angles: hence when a pentagon is equiangular, each angle is equal to the fifth part of six right angles, or to g of one right angle.

Cor. 3. The sum of the angles of a hexagon is equal to 2x(6-2), or eight right angles: hence in the equiangular hexagon, each angle is the sixth part of eight right angles, or

one.

Scholium. When this proposition is applied to polygons which have re-entrant angles, each re-entrant angle must be regarded as greater than two right angles. But to avoid all ambiguity, we shall henceforth limit our reasoning to polygons with salient

of

angles, which might otherwise be named convex polygons. Every convex polygon is such that a straight line, drawn at pleasure, cannot meet the contour of the polygon in more than two points.

PROPOSITION XXIX. THEOREM.

D

The opposite sides and angles of a parallelogram are equal. Draw the diagonal BD. The triangles ADB, DBC have a common side BD; since AD, BC are parallel, they have also the angle ADB=DBC (Prop. 23.); and since AB, CD A

are parallel, the angle, ABD BDC: hence they are equal (Prop. 7.); hence the side AB, opposite the angle ADB, is equal to the side DC, opposite the angle DBC; and in like

manner, AD the third side is equal to BC: hence the opposite sides of a parallelogram are equal.

Again, since the triangles are equal, it follows that the angle A is equal to the angle Č; and also that the angle ADC, composed of the two ADB, BDC, is equal to ABC, composed of the two DBC, ABD: hence the opposite angles of a parallelogram are equal.

Cor. Therefore two parallels AB, CD, included between two other parallels AD, BC, are equal.

PROPOSITION XXX. THEOREM.

If the opposite sides of a quadrilateral ABCD are equal, namely, AB=CD, and AD=BC, the equal sides will be parallel, and the figure will be a parallelogram.

For, having drawn the diagonal BD (see the preceding figure), the triangles ABD, BDC have all the sides of the one equal to the corresponding sides of the other; therefore they are equal; therefore the angle ADB, opposite the side AB, is equal to DBC, opposite CD; therefore (Prop. 23.) the side AD is parallel to BC. For a like reason, AB is parallel to CD: therefore the quadrilateral ABCD is a parallelogram.

PROPOSITION XXXI. THEOREM.

If two opposite sides AB, CD of a quadrilateral are equal and parallel, the remaining sides will also be equal and parallel, and the figure ABCD will be a parallelogram.

Draw the diagonal BD (see the last figure). Since AB is parallel to CD, the alternate angles ABD, BDC are equal (Prop. 23.); moreover, the side BD is common, and the side AB-DC; hence the triangle ABD is equal (Prop. 6.) to DBC; hence the side AD is equal to BC, the angle ADB to DBC, and consequently AD is parallel to BC; hence the figure ABCD is a parallelogram.

PROPOSITION XXXII. THEOREM.

The two diagonals AC, DB of a parallelogram divide each other into equal parts.

For, comparing the triangles ADO, COB, we find the side AD-CB, the angle ADO=CBO (Prop. 23.), and the angle DAO-OCB; hence (Prop. 7.) those triangles are equal; hence

B

AO, the side opposite the angle ADO, is equal to OC opposite OBC; hence also DO is equal to OB.

Scholium. In the case of the rhombus, the sides AB, BC being equal, the triangles AOB, OBC have all the sides of the one equal to the corresponding sides of the other, and are therefore equal; whence it follows that the angles AOB, BOC are equal, and therefore, that the two diagonals of a rhombus cut each other at right angles.

BOOK II.

THE CIRCLE, AND THE MEASUREMENT OF ITS ANGLES.

Definitions.

I. The circumference of a circle is a curve line, all the points of which are equally distant from a point within, called the centre.

The circle is the space terminated by A this curve line.

Note. In common language, the circle is sometimes confounded with its circumference but the correct expression may always be easily recurred to, if we bear in

mind that the circle is a surface which has length and breadth, while the circumference is but a line.

II. Every straight line, CA, CE, CD, drawn from the centre to the circumference, is called a radius or semidiameter; every line which, like AB, passes through the centre, and is terminated on both sides by the circumference, is called a diameter.

From the definition of a circle, it follows that all the radii are equal; that all the diameters are equal also, and each double of the radius.