PROPOSITION XXXV. THEOREM XXIX. IF F in a circle (DAEB) two chords (AB, DE,) cut one another; the rectangle contained by the fegments (AF, FB,) of one of them, is equal to the rectangle contained by the segments (DF, FE,) of the other. Hypothefis. Thesis. 1. AB, DE, are two chords of the same O DAEB. The Rgle AF. FB is = to II. And these chords cut one another in a point F. the Rgle DF , FE. CASE I. If the two chords pass thro' the center F of the O. Fig. s. DEMONSTRATION. HEN, the straight lines AF, FB, DF, FE, are = to one another. D. 15. B. 2. Conlequently, the Rgle AF.FB is = to the Rgle DF.FE. Ax. 2 B.2 CASE II. If one of the chords AE, passes thro' the center & cuts the other DE which does not pass thro' the Preparation. Pof. i. ECAUSE the straight line AB is cut equally in C & unequally in F. 1. The Pgle AF. Fb + the of CF is = to the of CB, or is SP. 5. B. 2. to the of Ce. Ax. IB. I. But the o of FE + the of CF is also = to the o of ce (P. 47. B. 1.). 2. From whence it follows, that the Rgle AF.FB + the D of CF = to the of FE + the O of CF. Ax. 1.B. 1. 3. Contiquently, the Rgle AF. FB is = to the [ of Fe. And since DF is = to FE (P. 3. B. 3.), or DF. FE = to the o of FE (Ax. 2. B. 2.). 4. The Rgle AF. FB is also = to the Rgle DF.FE. is = Ax. 3. B.1. Ax,1.B. 1. Because CASE III. If one of the chords AB, passes thro' the cen ter & cuts the other DE which does not pass Preparation. P. 12. B. I. 2. And draw the ray CD. Pos. 1. DEMONSTRATION. ECAUSE DH is to HE (Prep. 1. & P. 3. B. 3.). 1. The Rgle DF . FE + the of FH is = to the D of DH. P. 5. B. 2. 2. Wherefore, the Rgle DF.FE+ of FH + of CH is = to the O of DH + > of CH. Ax. 2. B. 1. But the o of FH+ of CH is = to the of CF, & the of DH + the of CH is = to the of CD (P. 47. B. 1.). 3. Therefore, the Rgle DF. FE+of CF is = to the of CD or to the o of CB. Ax. 1. B.1. Moreover, the Rgle AF. FB + of CF being = to the fame o of CB (P. 5. B. 2.). 4. The Rgle DF . FE + O of CF is also = to the Rgle AF . FB + O of CF. Ax. 1.B. I. 5. Or taking away the common o of CF, the Rgle DF. FE is = to the Rgle AF FB. Ax. 3. B. I. CASE IV. If neither of the chords AB, DE, passes thro' the center (Fig. 4.). Preparation. Pol. 1. Which was to be demonstrated, BECAUS PROPOSITION XXXVI. THEOREM XXX. F from any point (E) without a circle (ABD) 'two straight lines be drawn, one of which (DE) touches the circle, & the other (EA) cuts it; the rectangle contained by the whole secant (AE), & the part of it (EB) without the circle, shall be equal to the square of the tangent (ED). Hypothesis. Thesis. 1. The point É is taken without the O ABD. The Rgle AE.EB is = to the II. From this point E, a tangent ED & a fe O of ED. Preparation. Pos. 1. P. 18. B. 3And because the straight line AB is bisečted in C, & produced to the point E. 2. The Rgle AE. EB + the of CB is = to the of CE. P. 6. B.2 Moreover, the of CE being also= to the of De+the of CD (P.47.B.1.), or to the of DE+the of CB (P. 46. Cor. 3. B. 1.). 3. The Rgle AE. EB + the of CB is = to the of DE + the of CB. Ax. 1. B. I. The of CB being taken away from both sides. 4. The Rgle AE. EB will be = to the of DE. Ax. 3.B.1. Preparction. Pol, I. P. 12. B.L BECAUSE . ECAUSE the straight line AB is bisected in F (Prep. 1. & P. 3. B. 3.) and produced to the point E. 1. The Rgle AE.EB + of FB is = to the of FE. P. 6. B. 2. 2. Consequently, the 'Rgle AE. EB + of FB + of FC is = SAx. 2.B.1. to the o of Fe + 0 of FC, or is = to the o of CE. But since the of DE + of CD is = to the D of CE, and the of FB + of FC is = to the o of CB (P. 47. B. 1.), or is = to the of CD (D 15. & P. 46. Cor. 3. B. 1.) 3. The Rgle AE.EB + of CD is = to the [] of De + of CD. 4. Consequently, the Rgle AE.EB is = to the of DE. Ax. 3. B. 1. Which was to be demonstrated. {: COROLLA R r 1. LE F (fig. 3.) from any point (E) without a circle (ADBF), there be drawn several Firaighi lines (AE, EG, &c) culting it in (B & F, &c): the retlangles contained by the whole secants (AE, GE), and the parts of them (EB, EF) without the circle, are equal to one another; for drawing from the point (E) the tangent (ED), these redangles will be equal to the square of the same tangent (ED). COROLLA R r II. F from any point (E), without a circle (ADBF), there be drawn to this circle two tangents (ED, EC), they will be equal io one another, because the square of each is equal to the same rectangle (AE.EB). S PROPOSITION XXXVII. THEOREM XXXI. I from a point (C), without a circle (ADH), there be drawn two ftraight lines, one of which (AE) cuts the circle, and the other (ED) meets it; if the rectangle contained by the whole fecant (AE) and the part of it without the circle (EB), be equal to the square of the line (ED) which meets it: the line which meets shall touch the circle in D. Hypothesis. Thelis. Preparation. 1. From the point E to the O ADH draw the tangent EH. P. 17. B. 3 Pof. 1. DEMONSTRATION, Because ECAUSE the Rgle of AE.EB is = to the of ED (Hyp. 3.) and the Rgle AE. EB is also = to the of EH (Prep. 1 & P. 36. B. 3) 1. The of ED is = to the of EH (Ax.'1. B. 1.) or EĎ = EH52.46.B. 1. Cor. 3. And moreover, since in the A CDE, CHE, the fide CD is = to the side CH (D. 15. B. 1), and CE is common to the two A. 2. The V CDE is = to the V CHE. P. 8. B. I. 3. Wherefore, V CHE being a L (Prep. 1. & P. 18. B. 3), V CDE is also a L Ax 1. B.1. P. 16. B. 3. 4. And the straight line ED touches the O ADH in the point D. Cor. 3. |