E A I B C, the center of the circle, and from it draw CF, CG, perpendiculars to AB, DE. Join CA, CD; then, because the radius CF is perpendicular to the chord AB, it bisects it (Prop. VI.). Hence AF is the half of AB; and, for the same K reason, DG is the half of DE. But AB is equal to DE; therefore AF is equal to DG (Axiom 7, B. I.). Now, in the right-angled triangles ACF, DCG, the hypothenuse AC is equal to the hypothenuse DC, and the side AF is equal to the side DG; therefore the triangles are equal, and CF is equal to CG (Prop. XIX., B. I.); hence the two equal chords AB, DE are equally distant from the center. Secondly. Let the chord AH be greater than the chord DE; DE is further from the center than AH. For, because the chord AH is greater than the chord DE, the arc ABH is greater than the arc DE (Prop. V.). From the arc ABH cut off a part, AB, equal to DE; draw the chord AB, and let fall CF perpendicular to this chord, and CI perpendicular to AH. It is plain that CF is greater than CK, and CK than CI (Prop. XVII., B. I.); much more, then, is CF greater than CI. But CF is equal to CG, because the chords AB, DE are equal ; hence CG is greater than CI. Therefore equal chords, &c. Cor. Hence the diameter is the longest line that can be in scribed in a circle. A straight line perpendicular to a diameter at its extremity, is a tangent to the circumference. G Let ABG be a circle, the center of which is C, and the di. ameter AB; and let AD be drawn from A perpendicular to AB; AD will be a tangent to the circumference. In AD take any point E, and join CE; then, since CE is an oblique line, it is longer than the perpendicular CA (Prop. XVII., B. I.). Now CA is equal B to CK; therefore CE is greater than CK, and the point E must be without H. the circle. But E is any point whatever in the line AD; therefore AD has only the point A in common with the cucumference, hence it is a tangent (Def. 9). Therefore, a straight line, &c. Scholium. Through the same point A in the circumference, only one tangent can be drawn. For, if possible, let a second tangent, AF, be drawn; then, since CĀ can not be perpendicular to AF (Prop. XVI., Cor., B. I.), another line, CH, must be perpendicular to AF, and therefore CH must be less than CA (Prop. XVII., B. I.; hence the point H falls within the circle, and AH produced will cut the circumfer ence. PROPOSITION X. THEOREM. Two parallels intercept equal arcs on the circumference. The proposition admits of three cases: H Α. cants, as AB, DE. Draw the radius CH perpendicular to AB; it will also be perpendicular to DE (Prop. XXIII., Cor. D E IC 1, B. I.); therefore, the point H will be at the same time the middle of the arc AHB, and of the arc DHE (Prop. VI.). Hence the arc DH is equal to the arc HE, and the arc AH equal to HB, and therefore the arc AD is equal to the arc BE (Axiom 3, B. I.). Second. When one of the two par H allels is a secant, and the other a tan- D E gent. To the point of contact, H, draw the radius "CH; it will be per А, \B pendicular to the tangent DE (Prop. C IX.), and also to its parallel AB. But since CH is perpendicular to the chord AB, the point H is the middle of the arc AHB (Prop. VI.); therefore the F G K arcs AH, HB, included between the parallels AB, DE, are equal. Third. If the two parallels DE, FG are tangents, the one at H, the other at K, draw the parallel secant AB; then, according to the former case, the arc AH is equal to HB, and the arc AK is equal to KB; hence the whole arc HAK is equal to the whole arc HBK (Axiom 2, B. I.). It is also ev, ident that each of these arcs is a semicircumference. There fore, two parallels, &c. If two circumferences cut each other, the chord which joins the points of intersection, is bisected at right angles by the straight line joining their centers. Let two circumferences cut each other in the points A and B; then will the -D ine AB be a com C D mon chord to the two circles. Now, if B a perpendicular be erected from the middle of this chord, it will pass through C and D, the centers of the two circles (Prop. VI., Schol.). But only one straight line can be drawn through two given points ; therefore, the straight line which passes through the centers, will bisect the common chord at right angles. If two circumferences touch each other, either externally or internally, the distance of their centers must be equal to the sum or difference of their radır. It is plain that the centers of the circles and the point of contact are in the same straight line; for, if possible, :et the point of contact, A, be without the straight line CD. From A let fall upon CD, or CD produced, the perpendicular AE, and produce it to B, making BE equal to AE. Then, in the triangles ACE, BCE, the side AE is equal to EB, CE is common, and the angle AEC is equal to the angle BEC; therefore AC is equal to CB (Prop. VI., B. I.), and the point B is in the circumference ABF. In the same manner, it may be shown to be in the circumference ABG, and hence the point B is in both circumferences. Therefore the two circumfe . rences have two points, A and B, in common; that is, they cut each other, which is contrary to the hypothesis. Therefore, the point of contact can not be without the line joining the centers; and hence, when the circles touch each other externally, the distance of the centers CD is equal to the sum of the radii CA, DA; and when they touch internally, the distance CD is equal to the difference of the radii CA, DA. Therefore, if two circumferences, &c. Schol. If two circumferences touch each other, externally or internally, their centers and the point of contact are in the same straight line. If two circumferences cut each other, the distance between their centers is less than the sum of their radii, and greater than their difference. Let two circumferences cut each other in the point A. Draw the radii CA, DA; then, because any two sides of a triangle are together greater than the third side (Prop. VIII., B. I.), CD must be less than the sum of AD and AC. Also, DA must be less than the sum of CD and CA; or, subtracting CA from these unequals (Axiom 5, B. I.), CD must be greater than the difference between DA and CA. Therefore, if two circumfe rences, &c. In equal circles, angles at the center have the same ratio with the intercepted arcs. Case first. When G H the angles are in the ratio of two whole numbers. м Let ABG, DFH be equal circles, and let the angles ACB, A B DEF at their cen D ters be in the ratio of two whole numbers; then will the angle ACB : angle DEF :: arc AB: arc DF. M. С . B А DEI Suppose, for example, that the angles ACB, DEF are to each other as n to 4; or, which is the same thing, suppose that the angle M, which may serve as a common measure, is contained seven times in the angle ACB, and four times in the angle DEF. The seven partial angles into which ACB is divided, being each equal to any of the four partial angles into which DEF is divided, the partial arcs will also be equal to each other (Prop. IV.), and the entire arc AB will be to the entire arc DF as 7 to 4. Now the same reasoning would apply, if in place of 7 and 4 any whole numbers whatever were employed ; therefore, if the ratio of the angles ACB, DEF can be expressed in whole numbers, the arcs AB, DF will be to each other as the angles ACB, DEF. Case second. When the ratio of the angles can not be ex pressed by whole numbers. Let ACB, ACD be two an с gles having any ratio whatever. Suppose ACD to be the smaller angle, and let it be placed on the greater; then will the angle ACB : angle A D ACD :: arc AB : arc AD. For, if this proportion is not true, the first three terms remaining the same, the fourth must be greater or less than AD. Suppose it to be greater, and that we have Angle ACB : angle ACD :: arc AB : arc AI. Conceive the arc AB to be divided into equal parts, each less than DI; there will be at least one point of division betweeu D and I. Let H be that point, and join CH. The arcs AB, AH will be to each other in the ratio of two whole numbers, and, by the preceding case, we shall have Angle ACB : angle ACH : : arc AB : arc AH. Comparing these two proportions with each other, and observing that the antecedents are the same, we conclude that the consequents are proportional (Prop. IV., Cor., B. II.) ; therefore, Angle ACD : angle ACH :: arc AI: arc AH. But the arc AI is greater than the arc AH; therefore the angle ACD is greater than the angle ACH (Def. 2, B. II.), that is, a part is greater than the whole, which is absurd. Hence the angle ACB can not be to the angle ACD as the arc AB to an arc greater than AD. In the same manner, it may be proved that the fourth term of the proportion can not be less than AD; therefore, it must be AD, and we have the proportion Angle ACB : angle ACD :: arc AB : arc AD. Cor. 1. Since the angle at the center of a circle, and the |