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. A E from the Point A, at Right Angles to A D. * 11. 1. Likewife bifect + AB in F, and let FG be drawn + 10. 1. from the Point F, at Right Angles to A B; and join G B. Then, because AF is equal to FB, and FG is common, the two Sides A F, F G, are equal to the two Sides B F, FG; and the Angle AFG is equal to the Angle BFG. Therefore the Bafe AG is ‡‡ 4. . equal to the Bafe G B. And fo, if a Circle be described about the Centre G, with the Diftance AG, this fhall pass thro' the Point B. Defcribe the Cir cle, which let be A B E, and join E B. Now, because 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 EB. But the Angle D A B is equal to the Angle C. Therefore the Angle C will be equal to the Angle A EB. Wherefore, the Segment of a Circle AEB is defcribed upon the given Right Line AB, containing an Angle AEB, 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. LE ET the given Circle be A BC, and the Rightlined 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. I the Point B, and make the Angle F B C, at the Point * 23. 1. B, equal to the Angle D. Then, because the Right Line EF touches the Circle A B C in the Point B, and BC is drawn from the 32 of this, the Point of Contact B; the Angle F BC will be * equal to that in the alternate Segment of the Circle; but the Angle FBC is equal to the Angle D. Therefore the Angle in the Segment BAC will be equal to the Angle D. Therefore, the Segment B A C 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. If two Right Lines in a Circle mutually cut each IN the Circle'A BCD, let two Right Lines mutually cut each other in the Point EI say, the Rectangle contained under A E and EC is equal to the Rectangle contained under D E and E B. If A C and D B pafs through the Centre, fo 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 AE and EC is equal to the Rectangle under DE 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 AC, DB; and join F B, F C, 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 bifect * the † 5.2. $47. I. fame. Wherefore AG is equal to GC: And becaufe the Right Line AC is cut into two equal Parts in the Point G, and into two unequal Parts in E, the Rectangle under AE and EC, together with the Square of E G, is + equal to the Square of G C. And if the common Square of G F be added, then the Rectangle under A E and EC, together with the Squares of E G and GF, is equal to the Squares of C G and GF. But the Square of F E is equal to the Squares of E G and G F, and the Square of FC equal to the Squares Squares of CG and GF. Therefore the Rectangle under A E and E C, together with the Square of FE, is equal to the Square of F C; but CF 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 Reason, the Rectangle under DE and E B, 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 FE. 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 D E and E B. 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. LET any Point D be affumed without the Circle A B C, and let two Right Lines DCA, DB, fall from the faid Point to the Circle; whereof DCA cuts the Circle, and DB touches it. I fay, the Rectangle under AD and DC is equal to the Square of DB. Now D C A either paffes thro' the Centre, or not. In the firft Place, let it pafs thro' the Centre of the Circle A B C, which let be E, and join E B. * Then the Angle E BD is a Right Angle. And fo, fince * 18 of this. the Right Line AC is bifected in E, and CD is added #6.2. thereto, the Rectangle under AD and DC, together with the Square of E C, fhall be equal to the Square of E D. But E C is equal to E B; wherefore the Rectangle under A D and DC, together with the Square of E B, is equal to the Square of ED. But the Square t47. 1. of ED is + equal to the Squares of EB and B D, for the Angle EBD is a Right Angle: Therefore the Rectangle under AD and DC, together with the Square of E B, is equal to the Squares of EB and BD; and if the common Square of EB be taken away, the Rectangle under AD and DC remaining, will be equal to the Square of the Tangent Line B D. # 6.2. † 47. I. Now, let DC A not pass thro' the Centre of the 1 of bis. Circle ABC; and find t the Centre E thereof, and draw EF perpendicular to A C, and join E B, EC, ED. Therefore EFD is a Right Angle. And becaufe a Right Line EF, drawn thro' the Centre, cuts a Right Line A C, not drawn thro' the Centre, at Right * 3 of this. Angles, it will* bifect the fame; and fo A F is equal to FC. Again, fince the Right Line A C is bifected in F, and CD is added thereto, the Rectangle under AD and DC, together with the Square of F C, will be* equal to the Square of FD. And if the common Square of EF be added, then the Rectangle under A D and DC, together with the Squares of F C and F E, is equal to the Squares of D F and F E. But the Square of DE is equal to the Squares of DF and FE, for the Angle EFD is a Right one; and the Square of CE is † equal to the Squares of CF and FE. Therefore the Rectangle under A D and DC, together with the Square of CE, is equal to the Square of ED; but CE is equal to E B. Wherefore the Rectangle under AD and DC, together with the Square of E B, is equal to the Square of ED. But the Squares of E B and B D are † equal to the Square of ED; fince the Angle EBD is a Right one. Wherefore the Rectangle under AD and DC, together with the Square of E B, is equal to the Squares of E B and B D. And if the common Square of E B be taken away, the Rectangle under AD and DC, remaining, will be equal to the Square of DB. Therefore, if any 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 under the whole Secant Line, and its Part between the Convexity of the Circle and the affumed Part, will be equal to the Square of the Tangent Line; which was to be demonftrated. PROPOSITION XXXVII. THEOREM. If fome Point be taken without a Circle, and two Right Lines be drawn from it to the Circle, fo that one cuts it, and the other falls upon it; and if the Rectangle under the whole Secant Line, and the Part thereof, without the Circle, be equal to the Square of the Line falling upon the Circle; then this laft Line will touch the Circle. L ET fome Point D be affumed without the Circle ABC, and from it draw two Right Lines DCA, DB, to the Circle, in fuch manner, that D C A cuts the Circle, and DB falls upon it: And let the Rectangle under A D and D C be equal to the Square of DB. I fay, the Right Line D'B touches the Circle. For, let the Right Line DE be drawn touching* 17 of this. the Circle A B C, and find F the Centre ‡ of the Cir-tx of this. cle; and join EF, F B, F D. Then the Angle FED is + a Right Angle. And + 18 of this. because D E touches the Circle ABC, and DCA cuts it, the Rectangle under AD and DC will be equal to the Square of D E. But the Rectangle under AD and DC is equal to the Square of DB. ‡ By Hyp. Wherefore the Square of DE fhall be equal to the Square of D B. And fo the Line DE will be equal to the Line DB. But E F is equal to FB: Therefore the two Sides DE, EF, are equal to the two Sides DB, B F; and the Bafe F D is common. Wherefore the Angle D E F is equal to the Angle DBF: 8. But DEF is a Right Angle; wherefore DB F is allo a Right Angle, and FB produced is a Diameter. But a Right Line drawn at Right Angles, on the End of the Diameter of a Circle, touches the Circle; therefore BD neceffarily touches the Circle. We prove this in the fame manner, if the Centre of the Circle be |