IN PROPOSITION XXXI. THEOREM XXVII. N a circle, the angle (ADB) in a semicircle (ADEB), is a right angle; but the angle (DAB) in a segment (DAB) greater than a semicircle, is less than a right angle, & the angle (DEB) in a segment (DEB) less than a semicircle, is greater than a right angle: also the mixtilineal angle (BDA) of the greater segment, is greater than a right angle, & that (BDE) of the lesser segment, is less than a right angle. CASE I. Thesis This Y ADB is a L. Pof. 2. DEMONSTRATION, 1 Pos. 1. BE CAUS ECAUSE in the A ADC the side CA is = to the side CD (D. 15. B. 1.). 1. The V CDA is = to the V CAD. P. 5. B. 1. Again, in the ACDB; the fide CD being = to the side CB, D.15. B. 1. 2. The V CDB is = to the V CBD. P. 5. B. 1. 3. Consequently, V ADB is = to V CAD + CBD. Ax. 2. B. i. But VNDB is also = to V CAD + CBD (P. 32. B. 1.). 4. Wherefore, this V NDB is = to V ADB. Ax. 1.B.1. $. From whence it follows, that V ADB is a L. D. 10. B. I. Thesis. This DAB is <aL DEMONSTRATION. BECAUSE E CAUSE in the A ADB, the V ADB is a L (Cafe I.). 1. The Y DAB will be <a La R P. 17. 2. 1. HE the opposite V DAB + DEB of the quadrilateral figure ADEB are = to 2 L. P. 22. B. 3 2. Wherefore, V DAB being <a L (Case II.), DEB will be neceffarily > al CASE IV. Thesis. The V BDA is > L,Elbe formed by the straight line BD E V BDE is <a L. the arcbes DA, DE, DEMONSTRATION. ECAUSE the rectilineal V ADB, NDB, are L (Case I.). 1. The mixulineal V EDA will be necessarily > a L, & the inixtilineal V BDE <aL Which was to be demonstrated Ax. 8.B.I. Because PROPOSITION XXXII. THEOREM XXVIII. F a straight line (AB) touches a circle (ECF), & from the point of conta& (F) a chord (FD) be drawn; the angles (DFB, DFA,) made by this chord & the tangent, shall be equal to the angles (FED, FCD,) which are in the alternate segments (FED, FCD,) of the circle. Hypothefis. Thesis. 1. BA is a tangent of the O ECF. 1. The FED is = to V DFB. II. And FD is a chord of this O II. The V FCD is = to y DFA. drawn from the point of contact, Preparation. P. 11. B.I. 2. Take any point C in the arch DF, & draw ED, DC, CF. Pol. 1. DEMONSTRATION. ECAUSE the straight line AB touches the O ECF (Hyp. 1.), and FEis a Lerected at the point of contact F in the line AB (Prep. 1.). 1. The straight line FE is a diameter of the O ECF. P. 19. B. 3. 2. Consequently, V FDE is a La. P. 31. B. 3. 3, Wherefore, the V DEF + DIE are = to a La. P. 32. B. i. But EFB or V DFE + V DFB being also = to a L (Prep. 1.). 4. The V DEF + DFE are = to the V DFB + DFE. Ax. 1. B. i. Ś. Wherefore, the V DEF is = to V DFB, or the V in the segment s 4x. 3.B.1. DEF is = to the made by the tangent BF & the chord DF. P.21. B. 3. Which was to be demonstrated. I. The V FED + FCD being = to 2 L (P. 22. B. 3.), & the adjacent V DFB + DFA being also = to 2 L (P. 13. B. 1.). Ax.1. B 1. 6. The V FED + FCD are = to the V DFB + DFA. 7. Wherefore, V FED being = to the DFB ( Arg. 5.), the V FCD is also = to the V DFA; or they in the segineni FCD is = to s Ax.3.B. 1. the contained by the tangent AF & the chord DF. P.21. B. 3. Which was to be demonstrated. II. I Pol. 3. PROPOSITION XXXIII. PROBLEM V. PON a given straight line (AB), to describe a segment of a circle (ADB) containing an argle equal to a given re&ilineal angle (N). Given Sought The straight line AB together with V N. The segment ADB described uper AB, containing an V = 10 V N. CASEI. If the given V is a L. (Fig. 1.). T fuffices to describe upon AB a semni O ADB. 1. This semi O will contain an V = to the given right V N. P. 31. B. 3. CASE II. If the given Vis acute (Fig. 2.) or obtufe (Fig. 3.) Refolution. 1. At the point A in AB, make the VBAE = to the given VN. P. 23. B. 1, 2. At the point A in AE, erect the I AG. P. 1. B. 1, 3. Divide AB into two equal parts in the point F. P. 10. B. I. 4. At the point F in AB, erect the IFC, which will cut AG in C. P.U.B. 1. 5. From the center C at the distance CA, describe the O ADG. Pof. 3. Preparation. Pos. 1. DEMONSTRATION. ECAUSE in the A ACF, BCF, the side AF is = to the side BF (ROf. 3.), FC common to the two A, & the V AFC = the V BFC (Ax. 10. B. 1. & Res. 4.). 1. The base CA is = to the base CB. P. 4. B. r. 2. Consequently, the O described from the center C at the distance CA, will pass thro' the point B, & ADB is a segment described D. 15.B.1. AB. upon But AE touching the O ADB in A (Res. 2. & P. 16. Cor. B. 3.), B. 1. 19. and AB being a chord drawn from this point of contact A ( Arg. 2.). 3. The V contained in the alternate segment ADB is = the V BAE. P. 32. B. 3. Wherefore, V BAE being = to the given VN (Ref. 1.), the contained in the segment ADB described upor AB, is also = to the given V N. Ax. 1, B. 1. Which was to be done. BECAUSE PROPOSITION XXXIV. PROBLEM VI. To O cut off a segment (BED) from a given circle (BDE), which shall contain an angle (DEB) equal to a given re&tilineal angle (N). Given Sought The OBDE, & she rettilineal V N. The segment BED cut off from this , containing an V DEB=to ibegiven VN. Resolution. P. 17. B. 3. 1. From any point A to the O BDE, draw the tangent ABC. = to the given V N. P. 23. B. I. DEMONSTRATION. Because the given V N is = to the V DBA (Ref. 2.), & V DEB = to the V DBA (P. 32. B. 3.). 1. The V DEB & N are = to one another. Ax. 1. B.. 2. Wherefore, the segment BED is cut off from the O BDE, containing an V DEB = io the given y N. P. 21. B. 3: Which was to be done. |