an arc which is greater, and the other by an arc which is less than a quadrant. Cor. 3. If upon the base of a triangle there be described a segment of a circle, the vertex of the triangle shall fall without, or within, or upon the are of the segment, according as the vertical angle of the triangle is less than, or greater than, or equal to the angle in the segment. For it may easily be shown, (I. 8. Cor. 1.) that if the vertex fall within the arc of the segment, the vertical angle must be greater than the angle of the segment, and if without it, less. Therefore, &c. Cor. 1. (Euc. iii. 22.) If a quadrilateral figure be inscribed in a circle, either pair of its opposite angles shall be equal to two right angles. Cor. 2. And conversely, if the opposite angles of a quadrilateral be together equal to two right angles, a circle may be described about it. For, if the circle described through the three points A, D, B (5. Cor.) were to cut the side BE in any other point than E, suppose F, the angles A FB, ADB being equal to two right angles, would be equal to the angles AE B, A D B, and therefore the angle A FB to the angle A EB; whereas one of them, being exterior, must (I. 8. Cor. 1.) be greater than the other, PROP. 17. (EUc. iii. 32.) If a straight line touch a circle, and if from the point of contact a straight From C draw CE at right angles to the tangent A B, and therefore (2. Cor. 2.) passing through the centre of the circle: let CE meet the circumference in E: take any point F in the arc of the opposite segment, and join CE, ED, DF, FC. Then, because C D E is a semicircle, the angle CD E is a right angle (15. Cor. 1.): therefore the remaining angles of the triangle CDE (I. 19.), that is, the angles DEC and D C E, are together equal to a right angle. But the angles D C B and DCE are likewise fore the two latter angles are equal to together equal to a right angle: therethe two former, and the angle D C B is equal to DEC, that is, to the angle in the alternate segment. And because (16.) the angles in the two segments are together equal to two right angles, that is (I. 2.), to the angles DCB, DC A, the angle D CA is equal to the angle in the other seg ment D F C. Therefore, &c. Cor. The converse is also true that is, if from the extremity of a chord there be drawn a straight line, such that the angles which it makes with the chord are equal to the angles in the alternate segments of the circle, that straight line must touch the circle. Scholium. demonstrated, states no more than is The theorem which has been just contained in Prop. 15., if the tangent be considered as a chord in which the points of section are coincident. For, if the point F be supposed to move up to the point C, the chord C F will tend more and more to coincide in position with the tangent CB. But if E F be joined, then, by Prop. 15, the angle DCF is always equal to the angle DEF. Therefore, when F coincides with C, that is, when the chord CF becomes a tangent at C, the angle DCB is equal to the angle D E C. As this, however, was a case not contemplated in the demonstration given of that proposition, the inference could not have been directly drawn from it. The proposition (2.) that the tangent is at right angles to the radius is an instance of the same kind. Others may be seen in the corollaries of the two following propositions, and in certain properties of tangents which will be found in the next section. Cor. If one of the chords, as AB, be supposed to move parallel to itself until the points A and B in which it cuts the circle coincide, as at E, the same and its converse will be true: that is, if a chord and tangent be parallel, they shall intercept equal arcs; and conversely. For, because EF is parallel to CD, the angle FEC is equal (I. 15.) to the angle E C D, which stands upon the arc ED: but, because EF is a tangent, (17.) the same FEC is equal to EDC which stands upon the arc E C. Therefore the arc EC is equal to ED, (14. Cor. 2.) And the proof of the converse is similarly varied. PROP. 19. If two chords of a circle meet one another, the angle contained by them shall be measured by half the sum, or by half the difference of the intercepted arcs, according as the point in which they meet is within or without the circle. Through B draw B F parallel to DC, and let it meet the circumference in F: then (18.) the arc FC is equal to BD, and therefore the arc A F is equal to the sum or to the difference of A C, B D, according as the point E is within or without the circle. But, because B F is parallel to DC, the angle AEC is equal to ABF (I. 15.); and ABF is measured by half the arc AF, (14.Cor.1.): therefore the angle A E C is measured by half the sum or by half the difference of the arcs A C, BD, according as the point E is within or without the circle. Therefore, &c. When the point E is in the circumference, the result of this proposition coincides with that of 14. Cor. 1. Cor. By a similar demonstration (18. Cor.) if a chord meet a tangent in a point which is not the point of contact, the angle contained by them will be measured by half the difference of the intercepted arcs. The case of a chord meeting a tangent in the point of contact, has been already contemplated in Prop. 17. It may be considered, however, as included under the above rule, the measuring arc in this case being the same by this corollary as by Prop. 17. SECTION 3.-Rectangles under the Segments of Chords. PROP. 20. (EUc. iii. 35.) If two chords of a circle cut one another, the rectangles under their segments terminating in the points of section shall be equal, whether they 'cut one another within or without the circle. CE, ED. Therefore, &c. We may remark that an easy demonstration of this proposition is likewise afforded by I. 39.; for the rectangles in question are each of them equal to the difference of the squares of the radius, and of the distance of the point E from the centre of the circle. Cor. And hence, conversely, if two straight lines A B, C D cut one another in a point E, and if the points A, B and C, D be so taken, that the rectangle under A E, EB be equal to the rectangle under CE, ED; the points A, B, C, D shall lie in the circumference of the same circle. Let ABC be a circle, and let the chord A B be produced to meet the tangent CE in E: the square of CE shall be equal to the rectangle under AE, EB. Join CA, CB. Then, because the angle E C B is equal to the angle EAC in the alternate segment of the circle, (17.) and that the angle at E is common to the two triangles E C B, EA C, these two triangles are equiangular. Therefore (II. 31.) AE: EC::EC: E B, and (II. 38, Cor. 1.) the square of EC is equal to the rectangle under AE, EB. Therefore, &c. The same remark may be made here as at the end of the preceding proposition: viz., that an easy demonstration is likewise afforded by I. 39. and I. 36. Cor. 1. Cor. And hence, conversely, if two straight lines A B, CE cut one another in a point E, and if the points A, B and C, be so taken, that the square of E C be equal to the rectangle under A E, EB, the straight line EC shall touch the circle which passes through the points A, B, C. PROP. 22. (Euc. vi. B.) If the vertical or exterior-vertical angle of a triangle be bisected by a straight line, which cuts the base, or the base produced, the square of that straight line shall be equal to the difference of the rectangles under the two sides, and under the segments of the base, or of the base produced.* Let A B C be a triangle, and let the vertical or exterior-vertical angle be bisected by the straight line A D, which B meets the base or the base produced in D: the square of A D shall be equal to the difference of the rectangles BA, A C, and BD, D C. B Let A E C be the circle which (5. Cor.) passes through the points A, B, C, and let AD be produced to meet the circumference in E, and join E C. Then, because the angles BAD, EAC are halves, or supplementary to the halves of the bisected angle, they are equal to one another: also the angle ABD is equal to the angle AEC in the same segment (15.): therefore, the triangles BAD, EA C being equiangular, (II. 31.) BA: AD :: EA: A C, and (II. 38.) the rectangle under B A, AC is equal to the rectangle under EA, A D. Again, because the chords B C, E A cut one another in D (20.), the rect This, as is evident from the enunciation, is a property not of the circle, but of a triangle, and belongs as such to I. § 6. The required demonstra tion has, however, in this and one or two other instances rendered an infringement of the classification unavoidable. angle under BD, DC is equal to the rectangle under ED, DA: therefore, the difference of the rectangles under BA, AC and BD, DC is equal to the difference of the rectangles under E A, A D, and ED, DA, that is, to the square of AD (I. 31.). Therefore, &c. It should be observed in the case of exterior bisection (see the lower figure), that the bisecting line AD must, if produced, cut the circumference in a second point E, in all cases in which it cuts the base B C produced in a point D; that is, in all cases in which the sides A B, AC are unequal. For when AB is equal to A C, the angles ABC, ACB are likewise equal (I. 6.), and therefore (I. 19. and I. ax. 5.) equal to the halves of the exterior angle: therefore, the angle CAD beingequal to AC B, AD is parallel to BC (I. 15.), and the same CAD being equal to the angle ABC in the alternate segment, AD touches the circle in A (17. Cor.). But, when one of the sides, as A B, is greater than the other, the angle ACB is also greater than ABC (I. 9.); therefore the angle CAD, which (I. 19.) is equal to half the sum of the two ABC, ACB, is less than ACB, and greater than ABC; and because the angle CAD is not equal to ACB, AD is not parallel to BC (I. 15.); and because the same CAD is not equal to ABC, that is to the angle in the alternate segment, AD does not touch the circle in A, but cuts it and meets the circumference in a second point E, as was observed. PROP. 23. (Euc. vi. C.) If a triangle be inscribed in a circle, and if a perpendicular be drawn from the vertex to the base; the rectangle under the two sides shall be equal to the rectangle under the perpendicular and the diameter of the circle. Let A B C be a triangle inscribed in the circle ABC; from A draw A D perpendicular to BC, and AE through the centre of the circle to meet the circumference in E: the rectangle under BA, AC shall be equal to the rectangle under EA, AD. B D Join C E. Then, because ACE is a semicircle, (15. Cor. 1.) the angle A CE is a right angle: but ADB is likewise a right angle, and the angle AEC is equal to the angle AB D in the same segment (15.); therefore, the triangles EAC, BAD being equiangular, BÁ : AD::EA: AC (II. 31.), and (II. 38.) the rectangle under B A, A C is equal to the rectangle under E A, A D. Therefore, &c. Cor. If two triangles be inscribed in the same, or in equal circles, the rectangle under the two sides of the one, shall be to the rectangle under the two sides of the other, as the perpendicular, which is drawn from the vertex to the base of the one, to the perpendicular which is drawn from the vertex to the base of the other (II. 35.). PROP. 24. If a quadrilateral be inscribed in a circle, its diagonals shall be to one an other as the sums of the rectangles under the sides adjacent to their extremities. Let ACBD be a quadrilateral figure, inscribed in the circle ABC and A B, C D its diagonals: A B shall be to CD, as the sum of CAXADC and CBX BD to the sum of AC × CB, and A D x D B. B but Let A B, and CD cut one another in the point E: and, first, let A B cut CD BCD, CA B, and D A B are triangles at right angles. Then, because A CD, inscribed in the same circle, the perpendiculars A E, B E, CE and D E, are to one another as the rectangles A Cx AD, BCX BD, CAx C B, and D Ax DB: therefore, (II. 25. Cor. 3.) the sum of AE and B É, that is A B, is to the sum of C E and D E, that is CD, as the the sum of CAXC B and D AXD B. sum of AC XAD, and BCX BD to In the next place, let A B cut CD, not at right angles: and let the perpendiculars A a, Bb, Cc, and D d be drawn. Then, as before, it may be shown that Aa+Bb is to Cc+Dd, as ACX AD+BCX BD to CAX CB+DAXD B. But, because the triangles AE a, BEb, CEC, DEd are equiangular, A a, B b, C c, and D d are to one another as AE, BE,CE, and DE (II. 31.). Therefore, Aa + Bb is to Cc+Dd, as A E+BE to CE+DE, that is, as AB to CD. Therefore, (II. 12.) ABCD::AC×AD+BC xBD: CAXCB+DA×D B. Therefore, &c. PROP. 25. (EUc. vi. D.) circle, the rectangle under its diagonals If a quadrilateral be inscribed in a shall be equal to the sum of the rect- SECTION 4.-Regular polygons, and angles under its opposite sides. Let ACBD be a quadrilateral inscribed in the circle ABC; and let A B, CD be its diagonals: the rectangle under AB, CD shall be equal to the sum of the rectangles under AD, BC, and A C, BD. At the point A make the angle DAF equal to BA C, and let A F meet CD in F. Then, because the angle ABC is equal to A D F in the same segment (15.), and that B A C was made equal to DAF, the triangles A B C, ADF are equiangular: therefore, (II. 31.) AB: BC::AD: D F, and (II. 38.) the rectangle under AB, DF is equal to the rectangle under A D, B C. Again, because the angles BA C, DAF are equal to one another, let the angle B A F be added to each; therefore the whole angle FAC is equal to the whole angle D AB; and the angle FCA is equal to the angle DBA in the same segment (15.); therefore, the triangles AFC, ADB are equiangular. There fore (II. 31.) AB: BD::AC: CF, and (II. 38.) the rectangle under AB, CF is equal to the rectangle A C, B D. Therefore, the sum of the rectangles under A B, DF and AB, C F, that is, (I. 30. Cor.) the rectangle under AB, CD, is equal to the sum of the rectangles under A D, B C, and A C, B D. Therefore, &c. Cor. Hence, a quadrilateral may be constructed, which shall have its sides equal to four given straight lines, in a given order, each to each, and its angular points lying in the circumference of a circle. For, by the 24th proposition, the ratio of the diagonals, and by that which has been just demonstrated, their rectangle is given therefore, (II. 63.) the diagonals may be found, and (I. 50.) the quadrilateral constructed. It is only essential to the possibility of the construction that of the four given straight lines, every three be greater than the fourth (I. 10. Cor. 2.). It is remarkable that, although the diagonals will be different in different orders of the given sides, the circumscribing circle has the same magnitude whatsoever be their order. (See Sect. 5. Prop. 41. Scholium.) approximation to the area of the circle. Def. 11. A regular polygon is that which has all its sides equal, and likewise all its angles equal. A figure of five sides is called a pentagon; a figure of six sides a hexagon; of ten sides a decagon; and of fifteen sides a pente-decagon. There is so seldom any occasion, however, to specify the number of sides of an irregular figure, as distinct from a multilateral figure in general, that it has become common to appropriate these names with others of similar derivation (as by way of preeminence) to the regular figures" a hexagon," for instance, is understood to mean a regular figure of six sides, and so of the rest. It is evident, that regular polygons, which have the same number of sides, are similar figures; for their angles are equal, each to each, because they are contained the same number of times in the same number of right angles (I. 20.); and their sides about the equal angles áre to another in the same ratio, viz. the ratio of equality. 12. The centre of a regular polygon is the same with the common centre of the inscribed and circumscribed circles (see Prop. 26.): and the perpendicular which is drawn from the centre to any one of the sides is called the apothem. 13. Similar arcs of circles are those which subtend equal angles at the centre. Similar sectors and segments are those which are bounded by similar arcs. PROP. 26. (Euc. iv. 13 and 14,). If any two adjoining angles of a regular polygon be bisected, the intersection of the bisecting lines shall be the common centre of two circles, the one circumscribing, the other inscribed in, the polygon. Let ABCDEF be any regular polygon, and let the angles at A and B be bisected by the straight lines AO, BO; which meet in some point O, (I. 15. Cor. 4.) because each of the angles FA B, C BA is less than two right angles, and therefore each of their halves OAB, OBA less than a right angle, and the two together less than two right angles The point O shall be the centre B |