• Post. 1. + 1 of this, + Poft. 2. * Poft. 3. + Def. Draw the Right Line AC from the Point A to C *, upon it describe the Equilateral Triangle DAC+; produce DA and DC directly forwards to E and C+; about the Center C, with the Distance BC, describe the Circle BGH *; and about the Center D, with the Distance DG, describe the Circle GKL. Now because the Point C is the Center of the Circle BGH, BC will be equal to CG+; and because D is the Center of the Circle GKL, the whole DL. will be equal to the whole DG, the Parts whereof DA and D Care equal; therefore the Remainders AL, #Axiom 3. GC are also equal ‡. But it has been demonstrated that BC is equal to CG; wherefore both ALand BC are each of them equal to CG. But Things that are equal to one and the same Thing, are equal to one another; and therefore likewise A L is equal to BC. * 2 of this. Whence the Right Line AL is put at the given Point A, equal to the given right Line BC, which was to be done. PROPOSITION III. PROBLEM. Two unequal right Lines being given, to cut off a Part from the greater Equal to the leffer. LET AB and C be the two unequal Right Lines given, the greater whereof is AB; it is required to cut off a Line from the greater AB equal to the leffer C. Put * a right Line AD at the Point A, equal to the Line C, and about the Center A, with the Distance + Post. 3. AD, describe a Circle DEF†. Then because A is the Center of the Circle DEF, AE is equal to AD; and so both AE and C are each Axiom 1. equal to AD; wherefore A E is likewise equal to C‡. And so there is cut off from AB the greater of two given Right Lines AB and C, a Line AE equal to the lesser Line C, which was to be done. A PROP PROPOSITION IV. THEOREM. If there are two Triangles that have two Sides of the one equal to two Sides of the other, each to each, and the Angle contained by those equal Sides in one Triangle equal to the Angle contained by the correspondent Sides in the other Triangle, then the Base of one of the Triangles shall be equal to the Base of the other, the whole Triangle equal to the whole Triangle, and the remaining Angles of one Equal to the remaining Angles of the other, each to each, which fubtend the equal Sides. LET the two Triangles be ABC, DEF, which have two Sides AB, AC, equal to two Sides DE, DF, each to each, that is, the Side AB equal to the Side DE, and the Side AC to DF; and the Angle BAC equal to the Angle EDF. I say, that the Base BC is equal to the Base EF, the Triangle ABC equal to the Triangle DEF, and the remaining Angles of the one equal to the remaining Angles of the other, each to its Correfpondent, fubtending the equal Sides, viz. the Angle ABC equal to the Angle DEF, and the Angle ACB equal to the Angle DFE. For the Triangle ABC being applied to DEF, fo as the Point A may co-incide with D, and the Right Line AB with DE, then the Point B will co-incide with the Point E, because AB is equal to DE. And fince AB co-incides with DE, the Right Line AC likewife will co-incide with the Right Line DF, because the Angle BAC is equal to the Angle EDF. Wherefore also C will co-incide with F, because the Right Line A C is equal to the Right Line DF. But the Point B co-incides with E, and therefore the Base BC co-incides with the Base EF. For if the Point B co-inciding with E, and C with F, the Base BC does not co-incide with the Base EF; then two Right Lines will contain a Space, which is impoffible*. Therefore * Ax. 10. the Base BC co-incides with the Base EF, and is equal thereto; and consequently the whole Triangle ABC will co-incide with the whole Triangle DEF, B 4 and 1 and will be equal thereto; and the remaining Angles +4x. 8. will co-incide with the remaining Angles †, and will be equal to them, viz. the Angle ABC equal to the Angle DEF, and the Angle ACB equal to the Angle DFE. Which was to be demonstrated. PROPOSITION V. THEOREM. The Angles at the Bafe of an Isosceles Triangle are equal between themselves: And if the equal Sides be produced, the Angles under the Base hall be equal between themselves. LET ABC be an Isofceles Triangle, having the Side AB equal to the Side AC; and let the equal Sides AB, AC, be produced directly forwards to D and E. I say the Angle ABC is equal to the Angle ACB, and the Angle CBD equal to the Angle BCE.. For affume any Point Fin the Line BD, and from *3 of this. AE cut off the Line AG equal * to A F, and join FC, GB. Then because AF is equal to AG, and AB to AC, the two Right Lines FA, AC, are equal to the two Lines GA, AB, each to each, and contain the com + 4 of this. mon Angle FAG; therefore the Base FC is equal † to the Base GB, and the Triangle AFC equal to the Triangle AGB, and the remaining Angles of the one equal to the remaining Angles of the other, each to each, fubtending the equal Sides, viz. the Angle ACF equal to the Angle ABG; and the Angle AFC equal to the Angle AGB. And because the whole AF is equal to the whole AG, and the Part AB equal to the Part AC, the Remainder BF is equal to 4x. 3. the Remainder CG. But FC has been proved to be equal to GB; therefore the two Sides BF, FC, are equal to the two Sides CG, GB, each to each, and the Angle BFC equal to the Angle CGB; but they have a common Base BC. Therefore also the Triangle BFC will be equal to the Triangle CGB, and the remaining Angles of the one equal to the remain, ing Angles of the other, each to each, which fubtend the the equal Sides. And so the Angle FBC is equal to the Angle GCB; and the Angle BCF equal to the Angle CBG. Therefore because the whole Angle ABG has been proved equal to the whole Angle ACF, and the Part CBG equal to BCF, the remaining Angle ABC will be* equal to the remaining Ax. 3. Angle ACB; but these are the Angles at the Base of the Triangle ABC. It hath likewise been proved that the Angles FBC, GCB, under the Base, are equal; therefore the Angles at the Base of Isosceles Triangles are equal between themselves; and if the equal Right Lines be produced, the Angles under the Base will be also equal between themselves. Coroll. Hence every Equilateral Triangle is also Equiangular. PROPOSITION VI. THEOREM. If two Angles of a Triangle be equal, then the Sides fubtending the equal Angles will be equal between themfelves. LET ABC be a Triangle, having the Angle ABC equal to the Angle ACB. I say the Side AB is likewise equal to the Side A C. • For if A B be not equal to AC, let one of them, as AB, be the greater, from which cut off BD equal to AC*, and join DC. Then because DB is equal to 3 of this. AC, and BC is common, DB, BC, will be equal to AC, CB, each to each, and the Angle D B C equal to the Angle ACB, from the Hypothesis; therefore the Base DC is equal † to the Base AB, and + + of this. the Triangle DBC equal to the Triangle ACB, a part to the whole, which is absurd; therefore AB is not unequal to AC, and consequently is equal to it. Therefore if two Angles of a Triangle be equal between themselves, the Sides subtending the equal Angles are likewise equal between themselves. Which was to be demonstrated. Coroll. Coroll. Hence every Equiangular Triangle is also PROPOSITION VII. THEOREM. On the Same Right Line cannot be constituted two Right FOR if it be poffible, let two Right Lines A D, DB, equal to two others AC, CB, each to each, be conftituted at different Points C and D, towards the fame Parts CD, and having the same Ends A and B which the first Right Lines have, so that CA be equal to A D, having the fame End A which CA hath; and CB equal to DB, having the same End B; and let CD be joined. Then because AC is equal to A D, the Angle * 5 of this. ACD will be equal to the Angle ADC, and consequently the Angle ADC is greater greater than the Angle BCD; wherefore the Angle BDC will be much greater than the Angle BCD. Again, because C B is equal to DB, the Angle BDC will be equal to the Angle BCD; but it has been proved to be much greater, which is impoffible. Therefore on the same Right Line cannot be constituted two Right Lines equal to two other right Lines, each to each, at different Points, on the fame Side, and having the same Ends which the first right Lines have; which was to demonstrated. PRO : |