APPENDIX TO THE FIRST SIX, AND THE ELEVENTH AND TWELFTH BOOKS OF EUCLID'S ELEMENTS. BOOK I.* DEFINITIONS. 1. Lines, angles, and spaces are said to be given in magnitude, when they are either exhibited, or when the method of finding them is known. 2. Points, lines, and spaces are said to be given in position, when their situation with respect to each other is always the same, and is known. 3. A circle is said to be given in magnitude, when its radius is given; and in position, when its centre is given. Magnitudes, instead of being said to be given in magnitude, or given in position, are often said simply to be given, when no ambiguity arises from the omission. 4. A ratio is said to be given, when it is the same as that of two given magnitudes. 5. A rectilineal figure is said to be given in species, when its several angles, and the ratios of its sides are given. 6. When a series of unequal magnitudes, unlimited in number, agree in certain relations, the greatest of them is called a maximum ; the least a minimum. Thus, of chords in a given circle, the diameter is the maximum; and of straight lines drawn to a given straight line, from a given point without it, the minimum line is the perpendicular. This book contains a number of miscellaneous propositions, several of which are curious and important. The first five demonstrate the converses of the 34th proposition in the first book of Euclid, and the 11th, 12th, 21st, 22d, and 35th propositions, and the corollary to the 36th, in the third book. They are often used in demonstrating other propositions. PROP. I. THEOR. Ir the opposite sides, or the opposite angles of a quadrilateral be equal, it is a parallelogram. 1. Let AD be a quadrilateral; and first, let AB be equal to CD, and AC to BD: AD is a parallelogram. A B Join BC. Then in the triangles ABC, DCB, AB, AC are equal to DC, DB, each to each, and BC is common: therefore (I. 8.) the angle ABC is equal to DCB, and ACB to DBC; and (I. 27.) AB is parallel to CD, and AC to BD. Hence (I. def. 24.) AD is a parallelogram. C D 2. Again, if the angle BAC be equal to BDC, and ABD to ACD, AD is a parallelogram. : For (I. ax. 2.) the angles CAB, ABD are equal to BDC, DCA and therefore, all four being equal (I. 32. cor. 1.) to four right angles, CAB, ABD are equal to two right angles, and (I. 28.) AC, BD are parallel. In the same manner it would be proved that AB, CD are parallel: and therefore (I. def. 24.) AD is a parallelogram. PROP. II. THEOR. IF two circles cut one another, the straight line passing through their centres bisects their common chord. Let the circles ABC, ADC cut one another in A and C ; join AC, and through the centres E, F, draw EFG: AC is bisected in G. B D G A F E Join EA, EC, FA, FC. Then the triangles AEF, CEF have the sides EA, AF equal to EC, CF, each to each, and EF common : therefore (I. 8.) the angles AEF, CEF are equal. Again, the triangles AEG, CEG have EA, EC equal, EG common, and the contained angles equal: therefore AG, GC are equal. Cor. Hence, when the circles cut one another, the straight line passing through their centres does not pass through either point of intersection and therefore, if the line passing through the centres pass also through a point in which the circumferences meet, the circles must touch one another in that point. PROP. III. THEOR. IF two triangles on the same base, and on the same side of it, have equal vertical angles, the vertex of each is in the circumference of the circle described about the other. Let ABC be a triangle inscribed in the circle CAB: any other triangle on the base AB, and having its vertical angle equal to C, will have its vertex on the arc ACB. Let If possible, let ADB be a triangle on the base AB, and having the angle ADB equal to C, but the vertex D not on the arc ACB. AD, produced if necessary, meet the circumference in E, and join BE. Then (III. 21.) D E B the angle AEB is equal to C: wherefore ADB is equal to AEB, which (I. 16.) is impossible. Therefore, if the angle ADB be equal to C, D must fall on the arc ACB. PROP. IV. THEOR. IF two opposite angles of a quadrilateral be together equal to two right angles, a circle may be described about it. Let ADBF (see the figure for the preceding proposition) be a quadrilateral having the angles ADB, AFB together equal to two right angles: a circle described through A, F, B will also pass through D. For, if possible, let D be within the circle passing through A, F, B: produce AD to E, and join BE. Then (III. 22.) the angles E, F are equal to two right angles, and therefore to ADB and F. Take away F, and there remains ADB equal to E, which (I. 16.) is impossible. In the same manner, if D were supposed to be without the circle, it would be shown that the angle at D would be equal to AEB, which (I. 16.) is also impossible. Therefore D must be on the circumference. PROP. V. THEOR. IF two straight lines intersect each other, so that the rectangle under the segments of one of them is equal to the rectangle under the segments of the others, their extremities lie in the circumference of a circle. Let the straight lines BC, AD cut one another in G, and let the rectangles AG.GD, BG.GC be equal: a circle described through A, B, C will also pass through D. For, if possible, let D not fall on the circumference; and let AD, produced, if necessary, meet the circumference in E. Then (III. 35.) the rectangle AG. GE is equal to BG.GC, and therefore to AG.GD; which (VI. 1.) is impossible, unless D and E coincide: and therefore D must be on the circumference. E D B Schol. By means of the corollary to the 36th proposition of the third book, it would be shown in a similar manner, that if two straight lines meet in a point, and if they be so divided, that the rectangle under one of them and its segment next the common point, is equal to the rectangle under the other, and its corresponding segment, the points of section and the extremities remote from the common point lie in the circumference of the same circle. PROP. VI. THEOR. FROM AB, the greater side of the triangle ABC, cut off AD equal to AC, and join DC: draw AE bisecting the vertical angle BAC, and join DE: draw also AF perpendicular to BC, and DG parallel to AE. Then (1.) the angle DEB is equal to the difference of the angles at the base, ACB, ABC; or of BAF, CAF, or of AEB, AEC and DE is equal to EC: (2.) the angles BCD, EAF are each equal to half the same difference: (3.) ADC or ACD is equal to half the sum of the angles at the base, or to the complement of half the vertical angle: (4.) BG is equal to the difference of the segments BE, EC, made by the line bisecting the vertical angle. D H 1. In the triangles AED, AEC, AD, AC are equal, AE common, and the contained angles equal: therefore (I. 4.) DE is equal to EC, the angle ADE to ACE, and AED to AEC. But (I. 32.) because BD is produced, the angle ADE is equal to B and BED: therefore BED is the difference of B and ADE, or of B and ACB. Also BED is the difference of B AEB, AED, or of AEB, AEC. Again, ABF, BAF are equal to ACF, CAF, each pair being (I. 32. cor. 3.) equal to a right angle. Take away ABF: then, because the difference of ACF, ABF is BED, there remains BAF equal BED, CAF; that is, BED is the difference of BAF, CAF. G EF |