will be greater than a Right Angle; and is in the Segment ADC, which is less than a Semicircle. I fay, moreover, the Angle of the greater Segment contained under the Circumference A BC and the Right Line A C, is greater than a Right Angle; and the Angle of the leffer Segment, contained under the Circumference ADC, and the Right Line AC, is lefs than a Right Angle. This manifeftly appears; for, because the Angle contained under the Right Lines B A, A C, is a Right Angle; the Angle contained under the Circumference ABC, and the Right Line AC, will be greater than a Right Angle. Again, because the Angle contained under the Right Lines CA, A F, is a Right Angle, therefore the Angle which is contained under the Right Line A C, and the Circumference A D C, is lefs than a Right Angle. Therefore, in a Circle, the Angle that is in the Semicircle is a Right Angle; but the Angle in a greater Segment is lefs than a Right Angle; and the Angle in a leffer Segment, greater than a Right Angle: Moreover, the Angle of a greater Segment is greater than a Right Angle; and the Angle of a leffer Segment is less than a Right Angle; which was to be demonftrated. PROPOSITION XXXII. If any Right Line touches a Circle, and a Right Line be drawn from the Point of Contact cutting the Circle; the Angles it makes with the Tangent Line, will be equal to those which are made in the alternate Segments of the Circle. LE ET any Right Line EF touch the Circle ABCD in the Point B, and let the Right Line BD be any how drawn from the Point B, cutting the Circle. I fay, the Angles which B D makes with the Tangent Line E F, are equal to thofe in the alternate Segments of the Circle; that is, the Angle F BD is equal to an Angle made in the Segment D A B, viz. to the Angle DAB; and the Angle D B E equal to the Angle DCB, made in the Segment DCB. For, Draw * 11. 1. 32. I. Draw * B A, from the Point B, at Right Angles to E F; and take any Point C, in the Circumference B D; and join A D, D C, C B. Then, because the Right Line EF touches the Circle ABCD in the Point B: and the Right Line B A is drawn from the Point of Contact B, at Right Angles to the Tangent Line; the Centre of the Circle † 19 of this. A BCD will be in the Right Line BA; and fo BA is a Diameter of the Circle, and the Angle † 31 of this, A DB in a Semicircle is a Right Angle. Therefore the other Angles BAD, ABD, are equal to one Right Angle. But the Angle A B F is alfo a Right Angle: Therefore the Angle A B F is equal to the Angles B A D, ABD; and if A B D, which is common, be taken away, then the Angle DBF, remaining, will be equal to that which is in the alternate Segment of the Circle, viz. equal to the Angle B AD. And because ABCD is a quadrilateral Figure in a Circle, the oppofite Angles thereof are † equal to two Right Angles; therefore the Angles D B F, DBE, will be equal to the Angles BAD, BCD. But BAD has been proved to be equal to D B F; therefore the Angle DBE is equal to the Angle made in DCB, the alternate Segment of the Circle, viz. equal to the Angle DCB. Therefore, if any Right Line touches a Circle, and a Right Line be drawn from the Point of Contact cutting the Circle; the Angles it makes with the Tangent Line, will be equal to thofe which are made in the alternate Segments of the Circle; which was to be demonstrated. † 22 of this. PROPOSITION XXXIII. PROBLEM. To defcribe upon a given Right Line, a Segment equal ET the given Right Line be A B, and C the given Right-lined Angle. It is required to defcribe the Segment of a Circle upon the given Right Line AB, containing an Angle, equal to the Angle C. At At the Point A, with the Right Line A B, make the Angle B A D equal to the Angle C, and draw 23. 1. *AE from the Point A, at Right Angles to AD. * 11. 1. Likewife bifect + A B in F, and let F G be drawn † c. 1. from the Point F, at Right Angles to A B; and join GB. Then, because A F is equal to F B, and F G is common, the two Sides A F, F G, are equal to the two Sides B F, F G ; and the Angle AFG is equal to the Angle B F G. Therefore the Bafe A G is ‡ ‡ 4. 1. equal to the Bafe G B. And fo, if a Circle be described about the Centre G, with the Diftanee AG, this fhall pass through the Point B. Defcribe the Circle, which let be AB E, and join E B. Now, becaufe A D is drawn from the Point A, the Extremity of the Diameter A E, at Right Angles to A E, the faid AD will touch the Circle. And fince the * Cor. 16. of Right Line A D touches the Circle A B E, and the this. Right Line A B is drawn in the Circle from the Point of Contact A, the Angle D A B is + equal to the An- † 32 of this. gle made in the alternate Segment, viz. equal to the Angle A E B. But the Angle D A B is equal to the Angle C. Therefore the Angle C will be equal to the Angle A E B. Wherefore, the Segment of a Circle AEB is defcribed upon the given Right Line, A B, containing an Angle ♬ È B, equal to a given Angle C; which was to be done. PROPOSITION XXXIV. PROBLEM. To cut off a Segment from a given Circle, that fball contain an Angle, equal to a given Rightlined Angle. LET the given Circle be A B C, and the Right lined Angle given D. It is required to cut off a Segment from the Circle A B C, containing an Angle equal to the Angle D. * Draw the Right Line E F, touching the Circle in 17 of this. the Paint B, and make the Angle FBC, at the Point B, equal to the Angle D. Then, because the Right Line E F touches the Circle ABC in the Point B, and B C is drawn from the 23. 1. *32 of this, the Point of Contact B; the Angle FBC will be * equal to that in the alternate Segment of the Circle; but the Angle F B C is equal to the Angle D. Therefore the Angle in the Segment BAC will be equal to the Angle D. Therefore, the Segment BAC is cut off from the given Circle ABC, containing an Angle equal to the given Right-lined Angle D; which was to be done. PROPOSITION XXXV. THEOREM. If two Right Lines in a Circle mutually cut each other, the Rectangle contained under the Segments of the one is equal to the Rectangle under the Segments of the other. I N the Circle A B C D, let two Right Lines mutually cut each other in the Point E. I fay, the Rectangle contained under AE and EC is equal to the Rectangle contained under D E and E B. If AC and D B pafs through the Centre, so that E be the Centre of the Circle ABCD; it is manifeft, fince A E, EC, DE, E B, are equal, that the Rectangle under A E and E C is equal to the Rectangle under D E and E B. But if A C, D B, do not pass through the Centre, affume the Centre of the Circle F; from which draw FG, FH, perpendicular to the Right Lines A C, DB; and join F B, FC, F E. Then, because the Right Line G F, drawn through the Centre, cuts the Right Line A C, not drawn thro' * 3 of this. the Centre, at Right Angles, it will alfo bisect * the + 5.2. fame. Wherefore A G is equal to GC: And because the Right Line A C is cut into two equal Parts in the Point G, and into two unequal Parts in E, the Rectangle under A E and E C, together with the Square of E G, is † equal to the Square of G C. And if the common Square of GF be added, then the Rectangle under A E and E C, together with the Squares of E G and GF, is equal to the Squares of CG and $47. 1. GF. But the Square of FE is equal to the Squares of E G and G F, and the Square of F C equal ‡ to the I. Squares Squares of CG and G F. Therefore the Rectangle. under A E and E C, together with the Square of FE,. is equal to the Square of F C; but C F is equal to FB. Therefore the Rectangle under A E and E C, together with the Square of E F, is equal to the Square of F B. For the fame Reafon the Rectangle under DE and EB, together with the Square of F E, is equal to the Square of F B. But it has been proved, that the Rectangle under A E and E C, together with the Square of F E, is alfo equal to the Square of F B. Therefore the Rectangle under A E and EC, together with the Square of F E, is equal to the Rectangle under D E and E B, together with the Square of F E. And if the common Square of FE be taken away, then there will remain the Rectangle under A E and EC, equal to the Rectangle under DE and EB. Wherefore, if two Right Lines in a Circle mutually cut each other, the Rectangle contained under the Segments of the one, is equal to the Rectangle under the Segments of the other; which was to be demonftrated. PROPOSITION XXXVI. THEOREM. If fome Point be taken without a Circle, and from that Point two Right Lines fall to the Circle, one of which cuts the Circle, and the other touches it; the Rectangle contained under the whole Secant Line, and its Part between the Convexity of the Circle and the affumed Point, will be equal to the Square of the Tangent Line. L ET any Point D be affumed without the Circle A B C, and let two Right Lines D C A, D B, fall from the faid Point to the Circle whereof DCA cuts the Circle, and D B touches it. I fay, the R-&angle under A D and D C is equal to the Square of DB. Now DCA either paffes thro' the Centre, or not. In the firft Place, let it país through the Centre of the Circle A BC, which let be E, and join E B. Then the Angle EBD is a Right Angle. And fo, fince 18 of thisa the Right Line A C is bifected in E, and C D is added H thereto, |