PROPOSITION VI. THEOREM III. F a triangle (ACB) has two angles (a & b + c) equal to one another; the fides which are oppofite to thofe equal angles, will be alfo equal to one another. 1. The fides CA, BA, will be neceffarily unequal, C. N 2. Confequently one of them, as BA, will be> the other CA. C. N. Preparation. 1. Cut off therefore from the > fide BA, a part to the <fide CA. IN the ▲ ACB, DBC, the fide BD = to the fide CA (Prep. 1.), the fide BC is common to the two A, & contained a contained bc (Hyp. 1.). 1. Confequently, the two ▲ ACB, DBC, have two fides of the one equal to two fides of the other, each to each, & Vontained a = contained b+c. 2. Wherefore the AACB is to A DBC. But the ACB being the whole, & the A DBC its part. 3. It follows, that the whole would be to its part. 4. Which is impoffible. Therefore as the fides CA, BA, which are opposite to the equal Ya & bc, cannot be unequal. 5. Thofe fides are equal to one another, or CA = BA. C. N. Which was to be demonftrated. PROPOSITION VII. THEOREM IV. B FROM the extremities (A & B) of a ftraight line (AB), from which have been drawn to the fame point (C), two straight lines (AC, BC,): there cannot be drawn to any other point (D) fituated on the fame fide of this line, two other straight lines (AD, BD,), equal to the two firft each to each. There is on the fame fide of the line AB a point D fo fitu- point will be placed, CASE 1. Either in the fide AC, or BC. Fig. 1. CASE 2. Or within the ▲ ACB. Fig. 2. CASE 3. Or laftly without the AACB. Fig. 3. Confequently this CASE I. If the point D be fuppofed to be in one of the fides, as in AC. Fig. 1. BECAUSE ECAUSE the point D is fuppofed to be a point in AC different from the point C. 1. The line AD is either > or << the line AC. 2. Confequently it is impoffible that AD AC. Which was to be demonftrated. C. N. CASE II. If the point D be fupposed to be fituated within the ▲ ACB. Fig. 2. Preparation. 1. From the point D to the point C, draw the ftraight line DC. Pof. I. 2. Produce at will BD to E & BC to F. Pof. 2. BECAUSE ECAUSE AC is fuppofed AD. 1. The ACAD will be an ifofceles A. D. 25. B. 1. 2. Confequently the V at the base a + b &c will be equal to one another. P. 5. B. 1. And because BC is fuppofed BD. 3. The ACBD will be likewife an ifofceles A. D. 25. B. 1. 4. Hence the V under the bafe b & c+d, will be alfo equal to one another. P. 5. B. 1. Wherefore, if from Ve+d be taken its part Vd. 5. Vb will be > Vc. And if to the fame Vb be afterwards added Va. 6. Much more then will the whole Va+b be > Vc. 7. Confequently Va+b & Vc are not equal. But it has been demonftrated that in confequence of the fuppofition of this cafe, Va+b & Ve fhould be equal (Arg. 2.). 8. From whence it follows that this fuppofition cannot fubfift, unless those angles at the fame time be equal and unequal, 9. Which is impoffible. 10. Therefore the fuppofition which makes AC AD & BC BD, is in itself impoffible. Which was to be demonftrated. C. N. C. N. C. N. C. N. CASE III. If the point D be fuppofed to be without the AACB. Fig. 3. Preparation. From the point D to the point C let there be drawn the straight line DC. BECAUSE ECAUSE AC is fuppofed AD. 1. The A CAD will be an ifofceles A. 2. Confequently Vb & d+c at the bafe are equal to one another. Again, because BC is likewife fuppofed = BD; 3. The ACBD will be an ifofceles A. 4. Hence Ve & vba at the bafe will be equal to one another. If therefore we take from Vba its part Va. 5. The Vc will be > the remaining Vb. And if to this fame Vc be added Vd. 6. Much more then will the whole Vc + d be > Vb. 7. Wherefore V + d & Vb are not equal to one another. But it has been proved that in confequence of the fuppofition of this cafe, Vdc & b are equal to one another, (Arg. 2.). 8. From whence it follows that this fuppofition cannot fubfift, unless those angles be at the same time equal and unequal. 9. Which is impofiible. 10. Therefore, the fuppofition which makes AC AD & BC BD is impoffible. Which was to be demonstrated. F two triangles (FHG, ACB,), have the three fides (FH, HG, GF,) of the one equal to the three fides (AC, CB, BA,) of the other, each to each, they are equal to one another, & the angles contained by the equal fides are likewife equal, each to each. DEMONSTRATION. BECAUSE the point F coincides with the point A (Prep. 1.), & the line FG with the line AB (Prep. 2.), & thofe lines are equal (Hyp.3.). 1. The point G muft coincide with the point B. The extreme points F & G of the fide FG, coinciding therefore with the extreme points A & B of the fide AB (Prep. 1. Arg. 1.); & the ftraight lines FH, GH, being equal to the ftraight line AC, BC, each to each. 2. The ftraight lines FH, GH, will neceffarily coincide with the ftraight lines AC, BC, each with each. If not; then from the extremities A & B of a line AB, there may be drawn to two different points C & D, on the fame fide of AB, two ftraight lines AC, BC, equal to two other ftraight lines AD, BD, each to each. Which is impoffible. 3. Those fides therefore coincide. 4. But the bafe FG coinciding with the bafe AB (Prep. 2.), the fide FH 5. It follows, that the ▲ ACB, FGH, are equal to one another; as like- Ax. 9. P. 7. B. 1. Ax. 9. To PROPOSITION IX. PROBLEM. IV. O divide a given rectilineal angle (ECF), into two equal angles (ECD, FCD,). Given A rectilineal V ECF. Sought V ECDV FCD. Refolution. 1. Take CA of any length. 2. Make CB = CA. 3. From the point A to the point B, draw the ftraight line AB. BECAUSE = DEMONSTRATION. ECAUSE AC BC (Ref. 2.), DA DB (Ref. 4.), and the fide DC common to the two A CAD, CBD. 1. Those two A have the three fides of the one equal to the three fides of the other, each to each. 2. Confequently the VFCD, ECD, contained by the equal fides CA, CD; & CB, CD, are equal to one another. Which was to be done. |