A B an arc which is greater, and the other line be drawn cutting the circle, the by an arc which is less than a quadrant. angles which it makes with the tangent Cor. 3. If upon the base of a tri- shall be equal to the angles which are angle there be described a segment of a contained in the alternate segments of circle, the vertex of the triangle shall the circle. fall without, or within, or upon the are Let the straight line A B touch the of the segment, according as the verti- circle C D E in the point C, and from cal angle of the triangle is less than, or C let there be greater than, or equal to the angle in drawn the straight the segment. line CD cutting For it may easily be shown, (I. 8. the circle: the anCor. 1.) that if the vertex fall within gle DCA shall the arc of the segment, the vertical be equal to the angle must be greater than the angle of angle in the seg. the segment, and if without it, less. ment DFC, and the angle D C B to the angle in the segPROP. 16. ment DEC. From C draw C E at right angles to the If any chord be drawn in a circle, the angles contained in the two opposite tangent A B, and therefore (2. Cor. 2.) segments shall be together equal to two passing through the centre of the circle: let CE meet the circumference in E: right angles. take any point F in the arc of the oppoLet ADB be à cir site segment, and join CE, ED, DF, cle, and let it be di FC. Then, because CDE is a semivided by the chord AB circle, the angle C D E is å right angle into the segments AD (15. Cor. 1.): therefore the remaining B, A EB: the angles angles of the triangle CDE (I. 19.), ADB, AEB contained that is, the angles DEC and DCE, are in these segments shall together equal to a right angle. But the be together equal to angles D C B and DCE are likewise two right angles. For the angle ADB is measured by fore the two latter angles are equal to together equal to a right angle: therehalf the arc A E B upon which it stands the two former, and the angle D C B is (14. Cor. 1.), and in like manner the equal to D E C, that is, to the angle in angle A E B is measured by half the arc the alternate segment. AD B. Therefore the angles ADB, And because (16.) the angles in the AEB together are measured by half the two segments are together equal to circumference, that is, by two quadrants, two right angles, that is (I. 2.), to the and are consequently equal to two right angles D CB, DC A, the angle DCA angles. is equal to the angle in the other segTherefore, &c. ment DFC. Cor. 1. (Euc. iii. 22.) If a quadri Therefore, &c. lateral figure be inscribed in a circle, Cor. The converse is also true : that either pair of its opposite angles shall is, if from the extremity of a chord there be equal to two right angles. be drawn a straight line, such that the Cor. 2. And conve versely, if the oppo. angles which it makes with the chord site angles of a quadrilateral be together are equal to the angles in the alternate equal to two right angles, a circle may segments of the circle, that straight line be described about it. For, if the cir- must touch the circle. cle described through the three points A, D, B (5. Cor.) were to cut the side Scholium. BE in any other point than E, suppose The theorem which has been just F, the angles A F'B, A D B being equal demonstrated, states no more than is to two right angles, would be equal to the contained in Prop. 15., if the tangent be angles A E B, A D B, and therefore the considered as a chord in which the angle A F B to the angle A E B; where points of section are coincident. For, as one of them, being exterior, must if the point F be supposed to move (I. 8. Cor. 1.) be greater than the other, up to the point C, the chord C F will tend more and more to coincide in Prop. 17. (Euc. iii. 32.) position with the tangent C B. But If a straight line touch a circle, and if E F be joined, then, by Prop. 15, if from the point of contact a straight the angle DCF is always equal to the BE E А. B c D angle DEF. Therefore, when F coin- arcs, according as the point in which cides with C, that is, when the chord they meet is within or without the C F becomes a tangent at C, the angle circle. DC B is equal to the angle D EC. As Let ABC be a cirthis, however, was a case not contem- cle, and let the plated in the demonstration given of chords A B, CD A that proposition, the inference could meet one another not have been directly drawn from it. in the point E: the The proposition (2.) that the tangent angle A EC shall is at right angles to the radius is an be measured by instance of the same kind. Others may half the sum or by be seen in the corollaries of the two half the difference following propositions, and in certain of the arcs, AC, properties of tangents which will be BD, according as found in the next section. the point E is with in or without the PROP. 18. circle. Parallel chords intercept equal arcs ; Through B draw B F parallel to DC, and conversely. and let it meet the circumference in F: Let A B C be a then (18.) the arc F C is equal to BD, circle, and let the and therefore the arc A F is equal to the chords A B, C D be sum or to the difference of A C, BD, according as the point E is within or parallel to one ano without the circle. But, because B F is ther: the arc AC shall be equal to the parallel to DC, the angle AEC is equal to ABF (I. 15.); and ABF is arc BD. measured by half the arc AF,(14.Cor.1.): Join BC. Then, because AB is therefore the angle A E C is measured parallel to CD, the angle A B C (I. 15.) by half the sum or by half the differis equal to the angle B C D: therefore ence of the arcs A C, B D, according as (14. Cor. 2.) the arc A C is equal to the the point E is within or without the arc B D. circle. And conversely, if the arc A C be Therefore, &c. equal to the arc BD, the angle When the point E is in the circumA B C will (14. Cor. 2.) be equal to the ference, the result of this proposition angle B C D, and therefore (Î. 15.) A B coincides with that of 14. Cor. 1. will be parallel to C D. Cor. By a similar demonstration Therefore, &c. (18. Cor.) if a chord meet a tangent in Cor. If one of the chords, as A B, be a point which is not the point of contact, supposed to move parallel to itself until the angle contained by them will be the points A and B in which it cuts the measured by half the difference of the circle coincide, as at E, the same and intercepted arcs. its converse will be true: that is, if a The case of a chord meeting a tanchord and tangent be parallel, they gent in the point of contact, has been shall intercept equal arcs; and con- already contemplated in Prop. 17. It versely. may be considered, however, as included For, because E F is parallel to CD, under the above rule, the measuring arç the angle FEC is equal (I. 15.) to the in this case being the same by this coangle É C D, which stands upon the arc rollary as by Prop. 17. ED: but, because EF is a tangent, (17.) the same FE C is equal to EDC which stands upon the arc E C. There SECTION 3.- Rectangles under the fore the arc EC is equal to ED, Segments of Chords. (14. Cor. 2.) And the proof of the converse is similarly varied. PROP. 20. (Euc. iii. 35.) PROP. 19. If two chords of a circle cut one another, the rectangles under their seg. If two chords of a circle meet one ments terminating in the points of another, the angle contained by them section shall be equal, whether they shall be measured by half the sum, or. cut one another within or without the by half the difference of the intercepted circle. B Let `ABC be a EC is equal to the rectangle under circle, and let the A E, EB. chords AB, CDA Therefore, &c. cut, or be produced The same remark may be made here to cut, one another as at the end of the preceding proposiin the point E: the tion: viz., that an easy demonstration rectangle under AE, is likewise afforded by I. 39. and I. 36. EB shall be equal Cor. 1. to the rectangle un Cor. And hence, conversely, if two der CE, ED. straight lines A B, C E cut one another Join AD, BC. in a point E, and if the points A, B and Then, because the C, be so taken, that the square of EC angle EAD is equal be_equal to the rectangle under A E, to the angle ECB E B, the straight line EC shall touch in the same segment (15.), and that the circle which passes through the the angles at E, which are vertical (I. 3.) points A, B, C. as in the upper figure, or coincide as in Prop. 22. (Euc. vi. B.) the lower, are equal to one another, the triangles AED, CEB are equiangular. If the vertical or exterior-vertical Therefore (II. 31.) AE : ED :: EC: angle of a triangle be bisected by a E B, and (II. 38.) the rectangle under straight line, which cuts the base, or AE, E B is equal to the rectangle under the base produced, the square of that CE, ED. straight line shall be equal to the Therefore, &c. difference of the rectangles under the We may remark that an easy demon- two sides, and under the segments of stration of this proposition is likewise the base, or of the base produced.* afforded by I. 39.; for the rectangles in Let A B C be a triangle, and let the question are each of them equal to the vertical or exterior-vertical angle be difference of the squares of the radius, bisected by the and of the distance of the point E from straight line the centre of the circle. A D, Cor. And hence, conversely, if two meets the base straight lines A B, C D cut one another the base in a point E, and if the points A, B and produced in D: C, D be so taken, that the rectangle the square of under A E, E B be equal to the rect- AD shall be angle under CE, ED; the points equal to the A, B, C, D shall lie in the circumference difference of of the same circle. the rectangles BA, A C, and Let A E C be the circle which (5.Cor.) to cut a tangent to the same circle, the passes through the points A, B, C, and square of the tangent shall be equal to let AD he produced to meet the cirthe rectangle under the segments of the cumference in E, and join E C. chord. Then, because the angles B AD, Let ABC be a circle, and let the E A C are halves, or supplementary to chord A B be pro the halves of the bisected angle, they duced to meet the are equal to one another : also the anglė tangent C E in E: A B D is equal to the angle A EC in the square of CE the same segment (15.): therefore, the shall be equal to triangles B A D, E A C being equianguthe rectangle under lar, (11.31.) BA: AD :: EA: A C, and A E, E B. (II. 38.) the rectangle under B A, AC is Join CA, C B. Then, because the equal to the rectangle under E A, A D. angle ECB is equal to the angle EAC Again, because the chords B C, E A in the alternate segment of the circle, cut one another in D (20.), the rect(17.) and that the angle at E is com * This, as is evident from the enunciation, is a mon to the two triangles ECB, EAC, property' not of the circle, but of a triangle, and these two triangles are equiangular. belongs as such to I. 5 6. The required demonstra tion has, however, in this and one or two other in: Therefore (II. 31.) AE:EC::EC: stances rendered an infringement of the classification EB, and (II. 38, Cor. 1.) the square of unavoidable. which of D or angle under BD, DC is equal to the EAC, BAD being equiangular, BA : rectangle under ED, DA: therefore, AD:: EA: AC (II. 31.), and (II. 38.) the difference of the rectangles under the rectangle under BA, AC is equal BA, A C and B D, DC is equal to the to the rectangle under E Á, A D. difference of the rectangles under EA, Therefore, &c. AD, and ED, DA, that is, to the square Cor. If two triangles be inscribed in of AD (I. 31.). the same, or in equal circles, the rectTherefore, &c. angle under the two sides of the one, It should be observed in the case of shall be to the rectangle under the two exterior bisection (see the lower figure), sides of the other, as the perpendicular, that the bisecting line AD must, if pro- which is drawn from the vertex to the duced, cut the circumference in a second base of the one, to the perpendicular point E, in all cases in which it cuts the which is drawn from the vertex to the base B C produced in a point D; that base of the other (II. 35.). is, in all cases in which the sides A B, PROP. 24. AC are unequal. For when AB is equal to A C, the angles A B C, ACB are If a quadrilateral be inscribed in a likewise equal"(I. 6.), and therefore circle, its diagonals shall be to one an. (I. 19. and I. ax. 5.) equal to the halves other as the sums of the rectangles under of the exterior angle: therefore, the the sides adjacent to their extremities. angle CAD beingequal to A CB, AD Let ACBD be a is parallel to BC (I. 15.), and the same quadrilateral figure, C AD being equal to the angle ABC inscribed in the circle in the alternate segment, AD touches A B C and A B, C D the circle in A (17. Cor.). But, when its diagonals : A B as one of the sides, as A B, is greater than shall be to CD, the other, the angle ACB is also greater the sum of CAXA D than ABC (I. 9.); therefore the angle and C BxB D to the CAD, which (I. 19.) is equal to half the sum of AC X CB, sum of the two ABC, ACB, is less than and AD X D B. ACB, and greater than ABC; and Let A B, and C D cut one another in because the angle CAD is not equal to the point E: and, first, let A B cut CD ACB, AD is not parallel to BC (I.15.); at right angles. Then, because A CD, and because the same CAD is not equal BCP, CA B, and D A B are triangles to ABC, that is to the angle in the alter- inscribed in the same circle, the pernate segment, A D does not touch the pendiculars A E, BE, C E and D E, are circle in A, but cuts it and meets the to one another as the rectangles A CX circumference in a second point E, as A D, B C XBD, CAXC B, and D AX was observed. DB: : therefore, (II. 25. Cor. 3.) the sum of AE and B E, that is A B, is to the PROP. 23. (Euc. vi. C.) sum of C E and D Е, that is CD, as the If a triangle be inscribed in a circle, sum of AC XAD, and B CRB D to and if a perpendicular be drawn from In the next place, let A B cut CD, but the sum of CAXC B and D AXD B. the vertex to the base; the rectangle under the two sides shall be equal to the not at right angles: and let the perrectangle under the perpendicular and pendiculars A a, B6, Cc, and D d be the diameter of the circle. drawn. Then, as before, it may be Let A B C be a triangle inscribed in shown that A a+B6 is to Cc+D d, the circle ABC; from A draw A D as ACX AD+BC x B D to CAX perpendicular to BC, and CB+D AD B. But, because the triAE through the centre of angles A Ea, B Eb, C EC, D Ed are the circle to meet the equiangular, A a, B 6, C c, and D d are circumference in E: the to one another as A E, BE,C E, and rectangle under BA, AC DE (II. 31.). Therefore, A a + B b is shall be equal to the rect to Cc+D d, as A E B E to CE+DE, angle under EA, AD. that is, as AB to CD. Therefore, Join C E. Then, because ACE is a (II. 12.) A B:CD :: A CAD+BC semicircle, (15. Cor. 1.) the angle ACE XBD:CAXCB+DARD B. is a right angle: but ADB is likewise Therefore, &c. a right angle, and the angle A EC is PROP. 25. (Euc. vi. D.) equal to the angle ABD in the same segment (15.); therefore, the triangles circle, the rectangle under its diagonals If a quadrilateral be inscribed in a : a shall be equal to the sum of the rect- SECTION 4.- Regular polygons, and angles under its opposite sides. approximation to the area of the Let ACBD be circle. a quadrilateral inscribed in the circle Def. 11. A regular polygon is that ABC; and let AB, which has all its sides equal, and likewise CD be its diagonals: all its angles equal. the rectangle under A figure of five sides is called a pentaAB, CD shall be gon; a figure of six sides a hexagon; of equal to the sum of ten sides a decagon; and of fifteen sides the rectangles under a pente-decagon. There is so seldom AD, BC, and AC, any occasion, however, to specify the BD. number of sides of an irregular figure, At the point A make the angle DAF as distinct from a multilateral figure in equal to B AC, and let AF meet CD general, that it has become common to in F. appropriate these names with others of Then, because the angle ABC is similar derivation (as by way of preequal to AD F in the same segment eminence) to the regular figures (15.), and that B A C was made equal hexagon,” for instance, is understood to to DAF, the triangles ABC, ADF mean a regular figure of six sides, and are equiangular: therefore, (II.31.) AB: so of the rest. BC:: AD:DF, and (II. 38.) the rect It is evident, that regular polygons, angle under A B, D F is equal to the which have the same number of sides, rectangle under AD, BC. are similar figures; for their angles are Again, because the angles BAC; equal, each to each, because they are DAF are equal to one another, let the contained the same number of times in angle B AF be added to each ; therefore the same number of right angles (I. 20.); the whole angle FAC is equal to the whole and their sides about the equal angles angle D AB; and the angle FC A is are to another in the same ratio, viz. the equal to the angle D BA in the same ratio of equality. segment (15.); therefore, the triangles 12. The centre of a regular polygon AFC, AD B are equiangular. There is the same with the common centre of fore (II. 31.) A B:BD::AC:CF, the inscribed and circumscribed circles and (II. 38.) the rectangle under AB, (see Prop. 26.): and the perpendicular CF is equal to the rectangle A C, B D. which is drawn from the centre to any Therefore, the sum of the rectangles one of the sides is called the apothem. under AB, D F and AB, C F, that is, 13. Similar arcs of circles are those (I. 30. Cor.) the rectangle under AB, which subtend equal angles at the centre. C D, is equal to the sum of the rectangles Similar sectors and segments are those under AD, BC, and A C, B D. which are bounded by similar arcs. Therefore, &c. Cor. Hence, a quadrilateral may be PROP. 26. (Euc. iv. 13 and 14,). constructed, which shall have its sides If any two adjoining angles of a equal to four given straight lines, in a regular polygon be bisected, the intergiven order, each to each, and its angular section of the bisecting lines shall be the points lying in the circumference of a common centre of two circles, the one circle. For, by the 24th proposition, the circumscribing, the other inscribed in, ratio of the diagonals, and by that which the polygon. has been just demonstrated, their rect- Let A B C DEF angle is given: therefore, (II. 63.) the be any regular podiagonals may be found, and (1.50.) the lygon, and let the quadrilateral constructed. angles at A and B It is only essential to the possibility be bisected by the of the construction that of the four given straight lines AO, straight lines, every three be greater 0; which meet than the fourth (I. 10. Cor. 2.). It is in some point 0, remarkable that, although the diagonals (I. 15. Cor. 4.) because each of the anwill be different in different orders of the gles FAB, CB A is less than two right given sides, the circumscribing circle angles, and therefore each of their halves has the same magnitude whatsoever be OAB, OBA less than a right angle, and their order. (See Sect. 5. Prop. 41. the two together less than two right anScholium.) gles The point 0 shall be the centre 1) B |