26. To bisect a given straight line, or to find the middle F15 of it. Let A B (fig. 15) be the given line. By the method before described (24) fix two points C and D at equal distances from the extremities A and B. Then draw the straight line C D and it will be perpendicular to the middle of A B (23). Therefore the point E is in the middle of A B.

F16

27. To bisect a given arc or angle- Let A (fig. 16) be this angle. With the centre A and any radius, make an arc B C to measure this angle. Draw the chord B C. A is equally distant from B and C. Fix another point D equally distant from B and C, and draw A D. AD is perpendicular to the middle of B C (23). Therefore E, one of its points, is equally distant from B and C, and the chord E B=the chord E C. Then (13) the arc B E-the arc E C, that is, the arc B C is bisected at E. Also the angle A is bisected, for B A E=E A C, having the same measure.

28. —If a perpendicular be erected upon the middle of a chord, it will pass through the middle of the arc, and through the centre of the circle to which the arc belongs. The truth of this proposition is evident from the preceding conF16 struction. A D (fig. 16) is perpendicular to the middle of B C, it passes through the centre A, and it bisects the arc B C. It is moreover evident that no line can be perpendicular to the middle of B C which does not pass through A; for if there were such a line, it would differ from A F, and then the angle which it would make with F C, would be either greater or less than a right angle, which would contradict the supposition of its being perpendicular. Hence the proposition enunciated must be universally true.

29. The preceding proposition being admitted, we can solve the following problem-to find the centre of a given arc or circumference, or of one which shall pass through any three points not in the same straight line. Let the circumF17 ference A D C B (fig. 17), or any portion of it be given. In order to find the centre, take any three points A, D, C and join them by the chords A D, D C. Erect a perpendicular upon the middle of A D and it will pass through the centre (28). Erect another upon the middle of D C and it must also pass through the centre. But a point which is in two lines at once must be at their intersection

Therefore E is the centre sought. If we had any three points given, as A, D, C, not in the same straight line, the process for finding the centre of the circumference passing through them, would be the same; for the point E thus found, is equally distant from the three points in question (23), and therefore the circumference described with the centre E so as to pass through one, must pass through the other two. Moreover as the above construction is independent of any particular situation of the three points A, D, C, it is possible so long as these points can be joined by two straight lines, which can be made two chords of a circle, since all that is necessary is to bisect these two chords by perpendiculars. But if the three points were in the same straight line, there would be no longer two straight lines joining them, but only one, and the construction would be manifestly impossible.

30. A perpendicular measures the shortest distance of a point from a line. Let A (fig. 18) be the point in ques- F18 tion, and B E the line. We say that the perpendicular A D is shorter than any oblique line as A C. This, as well as the following proposition, is very nearly self evident, but it is usual to give a demonstration of them. With A as a centre and a radius A C, make the arc C F, and produce it till it cuts C E in E. Now it is evident from the definitions of a curve and a straight line (3, 10), that the chord C E can never coincide with the arc C F E, so long as CE is of any perceptible magnitude, that is so long as C is taken at any appreciable distance from D. But so long as these do not coincide, A D will be less than A F, and therefore less than its equal A C. Accordingly the perpendicular A D is shorter than any oblique line, A C, however near to D the point C be taken.

31. Two oblique lines drawn equally distant from the perpendicular are equal, and of two oblique lines drawn unequally distant, the more remote is the greater. First let A C and A E (fig. 18) be drawn at equal distances from A F 18 D. We say they are equal. By the supposition A D is perpendicular to the middle of CE. Therefore (23) the point A is equally distant from C and E; in other words A CA E which was first to be demonstrated. Again let A B be more remote from the perpendicular than A C. We say that A B is greater than A C. Let A B be supposed to turn about A as a centre till it coincides in direction with

A C. The point B will describe the arc B G, and this arc will always differ from the straight line B C so long as B C is of any perceptible magnitude. Accordingly AC will be less than A G, and therefore less than its equal A B. Or, as the proposition was enunciated, A B will be greater than A C, so long as it is more remote, by however small a quantity, from A D.

32.

-There can be only one perpendicular let fall from a point to a straight line, and there can be only one perpendicular erected at a point in a line. The first part of this proposition is true, because there can be but one shortest distance from a point to a line. The second part is true, because all right angles are equal, which would not be the case if all perpendiculars erected at the same point did not coincide and actually form but one perpendicular.

Parallel Lines.

F 19 33. When two straight lines as A B and C D (fig. 19), are so drawn as to be throughout at the same distance from each other, they are said to be parallel. But we have already shown (30) that the shortest distance of a point from a line is measured by a perpendicular. Thus the shortest distance of the point L from the line C D is the perpendicular L M, and the same is true with respect to any other point. Accordingly we shall assume the following definition as the basis of our reasoning upon parallel lines-two straight lines are parallel when all the perpendiculars let fall from points in one to the other, are equal. It follows as an immediate consequence from this definition, that—two parallel straight lines can never meet each other, however far produced-for if they have any distance at first, they must always have it. The last proposition is called a corollary, which may be defined to be a proposition which follows immediately as a consequence from a preceding proposition—.

F20

34. If two parallel straight lines, as A B and C D (fig. 20), are cut by a third straight line E F, the angles thus formed are known by particular names, which it is important to remember. The angles A H G and H G D, when named together, are called alternate-internal angles, because they are on opposite sides of the single line E F and within the parallels A B and C D. Again the angles F

H B and H G D are called internal-external angles, because one is within and the other without the parallels, and both are on the same side of the single line. For the same reason A H G and C G E are internal-external angles. These explanations being kept in view, we proceed to demonstrate the following proposition-two alternate-internal angles are always equal, and two internal-external angles are always equal We are first to prove that A H G is equal to H G D From H. let fall the perpendicular H L to C D. From G let fall the perpendicular G M to A B. By the definition (33) G M=H L. Produce H L making L K=H L. Produce G M making M. I=G M. Then H K=G I. Moreover C D is perpendicular to the middle of H K, and A B is perpendicular to the middle of GI. Therefore if with the centre H and radius H G an arc GOI be described, this arc will be bisected at O (28). Also if with the centre G and the same radius G H, the arc H N K be described, this arc will be bisected at N. Now as the chord H K= the chord G I, the arc HN K= the arc GO I (13); and since these are bisected at N and O, it follows that the arc H N= the arc G 0. But the arc H N measures the angle H G D, and the arc G O measures the angle A H G. Therefore A H G=HG D which was the first thing to be demonstrated. And it only remains to prove that F H B⇒H G D. Now FHBA H G because they are vertical (22). But we have just proved that A H G=H G D. Consequently F HB=H GD; for it is an axiom that—two things, each of which is equal to a third, are equal to each other—.

35. If two parallel straight lines A B and C D (fig. 21) F21 meet a third straight line G H, the two angles A I K and CK I are called interior on the same side, because they are within the parallels and on the same side of the single line. For the same reason L I K and I K M are called interior on the same side. We shall now demonstrate the following proposition-the sum of two interior angles on the same side is always equal to two right angles. We are to prove that A IK+IK C=2 right angles. Now AI GA I K=2 right angles (20). But A I G=I K C being internal-external angles (34). Therefore, substituting IKC in the place of its equal A I G, we have A IK+I K C=2 right angles, which was to be demonstrated.

36. —If a straight line is perpendicular to one of two

parallels it is also perpendicular to the other, and if two lines are perpendicular to a third they are parallel. First we F21 say that if E F is perpendicular to A B (fig. 21) it is also perpendicular to CD. We here take it for granted that ALM is a right angle, and we are to show that L MC is also a right angle. Now A L M+L M C=2 right angles, being interior on the same side (35). Then if from 2 right angles we take 1 right angle A L M, there must remain a right angle L M C. Hence E F is perpendicular to C D, which was the first thing to be demonstrated. The second part hardly needs demonstration, but it can be demonstrated as follows. If A B and C D are perpendicular to E F, we say they are parallel. For if A B is not parallel to C D, there can be a line drawn through L different from A B, which shall be parallel to C D. But then if it differ from A B, it cannot make an angle A L M, which added to L M C, shall make the two interior angles on the same side equal to two right angles (35). Therefore no line different from A B can be parallel to C D, and A B itself must be parallel to C D.

37. We are now prepared to solve the following problem-through a given point to draw a straight line parallel to a given straight line. Let A B be the given line and F22 C the given point (fig. 22). With C as a centre and any convenient radius as C D, make an indefinite are D F. With D as a centre and the same radius make the arc C G. Then make the arc D F C G (14). Through C and the point F thus determined, draw C F and it will be the parallel required. For if C D be drawn, the angle A D C the angle D C F, because their arcs are equal (18). Now we have proved (34) that if A B and C F were parallel, A D C would be equal to D C F being alternateinternal; and since no line different from C F could make DC F=A D C, we conclude that C F, which fulfils this condition, must be the parallel required.

38. Two parallels comprehended between two other F23 parallels are equal. Let A B and C D (fig. 23) be two parallels and E F and G H two parallels drawn between them. We wish to prove that E F G H. If they are perpendicular to the other two, this is evident from the definition (33). But suppose they are oblique. Still we say that E F G H. For since F I=H B by definition, FI may be placed upon H B. Then since Ě I FG B