14 FRO PROPOSITION II. PROBLEM II. ROM a given point (A), to draw a straight line (AL), equal to a given traight line (BC). Given 1. The point A. Sought 2. The ftraight line BC. Refolution. 1. From the point A to the point B, draw the ftraight line AB. BECA DEMONSTRATION. USE the lines BC and BG, are drawn from the center B to Pof. 1. the 1. Those two lines are rays of the fame CGM, D. 16. B. 1. 2. Confequently, BCBG. D. 15. B. 1. And because the lines DG and DL, are drawn from the center D to the O GLN (Ref. 5.). 3. Thofe lines, are alfo rays of the fame GLN. D. 16. B. 1. 4. Confequently, DGDL. D. 15. B. 1. But the lines DA & DB, being the fides of an equilateral 5. The line DA, is to the line DB. D. 24. B. 1. Cutting off therefore from the equal lines DG, DL, (Arg. 4.); their equal parts DB, DA, (Arg. 5.). Since therefore the line AL is to the line BG (Arg. 6.), and the line BC is alfo to the fame line BG (Arg. 2.). But it is manifeft that this line AL, is a line drawn from the given point A (Ref. 3.). 8. Wherefore from the given point A, a ftraight line AL, equal to the given ftraight line BC, has been drawn, Ax. 1. Which was to be done. Two WO unequal straight lines (A & CD) being given; to cut off from the greater (CD) a part (CB) equal to the lefs A. Given the line CD > line A. Sought from CD to cut off CB = A. Refolution. 1. From the point C draw the ftraight line CE to the given one A. P. 2. B. 1. 2. From the center C and at the diftance CE, defcribe O CEB; Pof. 1. which cuts the greater CD in B. DEMONSTRATION. THE ftraight lines CB, CE, being drawn from the center C to the O BEF (Ref. 2.). 1. They are rays of the fame O BEF. 2. Confequently, CB = CE. But the ftraight line A being to the ftraight line CE (Ref. 1.); and the straight line CB being likewife to CE (Arg. 2.). 3. The ftraight line A is to the straight line CB. And fince CB is a part of CD. 4. From CD the greater of two straight lines, a part CB has been cut off to A the lefs. D. 16. B. 1. Ax. 1. Which was to be done. IF PROPOSITION IV. THEOREM I F two triangles (BAC, EDF,), have two fides of the one, equal to two fides of the other, (i. e. AB = DE, & AC DF), & have likewife the angle contained (a) equal to the angle contained (d): they will alfo have the bafe (BC), equal to the base (EF); & the two other angles (b & c) equal to the two other angles (e & f) each to each, viz. thofe to which the equal fides are oppofite; and the whole triangle (BAC) will be equal to the whole triangle (EDF). Hypothefis. 1. ABDE. II. ACDF. III. Va= Vd. Preparation. Suppose the ABAC to be laid upon the ▲ EDF, in fuch a man SINCE DEMONSTRATION. INCE the line AB is to the line DE (Hyp. 1.), & the point A falls upon the point D (Prep. 1.), & the line AB upon the line DE (Prep. 2.). 1. The point B will fall neceffarily upon the point E. Because the Va≈ Vd (Hyp. 3.), & the point A falls upon the point D (Prep. 1.), & the fide AB upon the fide DE (Prep. 2.). 2. The fide AC will fall neceffarily upon the fide DF. Moreover, fince this fide AC is to the fide DF. 3. The point C muft fall also upon the point F. 4. Wherefore, the extremities B and C of the base BC, coincide with the extremities E and F of the base EF. 5. And confequently, the whole bafe BC coincides with the whole base EF; for if the base BC did not coincide with the base EF, though the extremities B and C of the bafe BC, coincide with the extremities E and F of the bafe EF; two ftraight lines would inclose a space (EXF or EYF); which is impoffible. Since therefore, the bafe BC coincides with the base EF (Arg. 5.). 6. This bafe BC will be to the bafe EF. Which was to be demonftrated. I Again, the bafe BC coinciding with the bafe EF (Arg. 5.), & the two other fides AB, AC, of A BAC, coinciding with the the two other fides DE, DF, of A EDF (Prep. 2, Arg. 2.). 7. Those two A BAC, EDF, are neceffarily equal to one another. Which was to be demonftrated. III. In fine, fince the V&Ve to which the equal fides AC, DF are oppofite (Hyp. 2.); likewife, the Vc & f to which the equal fides AB, DE, are oppofite (Hyp. 1.), coincide both as to their vertices and their fides (Arg. 1, 2, 3, 5.) 3. It follows, that the Vb & Ve, as alfo the Vc & Vf, to which the equal fides are oppofite, are equal to one another. Which was to be demonftrated. II. Ax. 9. Ax. 9. Ax 9. PROPOSITION V. THEOREM II. IN every ifofceles triangle (BAC): the angles (a & b) at the bafe (BC) are equal, & if the equal fides (AB, AC,) be produced: the angles (c + e & d+f) under the base (EC) will be alfo equal. Hypothefis. 1. The ABAC is an ifofceles A. II. AB & AC are produced indefinitely. Preparation. Thefis. 1. Va & bare equal. II. Vc +e & \d+fare alfo equal. 1. In the fide AB produced take any point D. 2. Make AE = AD. P. 3. B. 1. 3. Through the points B & E, as alfo C & D, draw BE, CD. Pof. 1. BECAU DEMONSTRATION. USE in the ADAC the two fides AD, AC, are equal to the two fides AE, AB of ▲ EAB, each to each (Prep. 2. Hyp. 1.); and the VA contained by thofe equal fides is common to the two A. 1. The base DC is to the bafe BE; & the two remaining Vm & b+d of ▲ DAC, are equal to the two remaining Vn & a + c of ▲ EAB, each to each of thofe to which the equal fides are oppofite. And because the whole line AD is to the whole line AE (Prep. 2.), and the part AB = to the part AC (Hyp. 1.); cutting off &c. 2. The remainder BD will be to the remainder CE. Again, fince in the DBC the fides DB, DC, are equal to the fides CE, EB, of A ECB, each to each (Arg. 2. and 1.), & likewife contained m is equal to V contained n (Arg. 1.). 3. The two remaining of the one, are to the two remaining of the other, each to each, viz. Vc + e=\d+f & \d = √c. The whole Va+c & b+d being therefore to one another, as alfo their parts Vc & Vd (Arg. 1. & 3.); cutting off &c. 4. The remaining Va & b are likewife to one another. But thofe are the two V at the bafe BC. 5. Therefore Va & Vb at the bafe BC are to one another. Which was to be demonftrated. I. Moreover, fince Ve+c=\d+f(Arg. 3.) are the V under the base. 6. It is evident that the Ve+c & \d+ funder the bafe, are alfo = to one another. Which was to be demonftrated. II. P. 4. B. 1. Ax. 3. P. 4. B. 1. Ax. 3. |