angle ABC will co-incide with the whole Triangle DEF, and will be equal thereto; and the remaining + Ax. 8. Angles 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 demonftrated. PROPOSITION V. THEOREM. The Angles at the Base of an Isosceles Triangle LET ABC be an Ifofceles Triangle, having the A B equal to the Side A.C; 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 A C B, and the Angle CBD equal to the Angle BCE. For afsume any Point F in the Line BD, and from 3 of this. AE cut off the Line AG equal to AF, and join FC, GB. Then, because A F is equal to A G, and AB to AC, the two Right Lines FA, AC, are equal to the two Lines GA, A B, each to each, and contain the com † 4 of this mon Angle FAG; therefore the Base F C is equal + to the Bafe 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, subtending 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 Ax. 3. A B equal to the Part A C, the Remainder BF ‡ is equal to 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 B F C equal to the Angle CGB; but they have a common Base BC. Therefore also the Triangle BFC will be equal to the * 4 of this. Triangle CGB*, and the remaining Angles of the one equal 1 9 equal to the remaining Angles of the other, each to each, which fubtend 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 A B C will be * equal to * Ax. 3. the remaining Angle ACB, but these are the Angles at the Base of the Triangle ACB. It hath likewise been proved, that the Angles FBC, GCB, under the Base, are equal; therefore, the Angles at the Bafe of 1fofceles Triangles are equal between themselves; and if the equal Right Lines be produced, the Angles under the Bafe will be alfo equal between themselves; which was to be demonftrated. Coroll. Hence every Equilateral Triangle is also Equiangular. : PROPOSITION VI. THEOREM. If two Angles of a Triangle be equal, then the LET ABC be a Triangle, having the Angle For if A B be not equal to A C, let one of them, as A B, be the greater, from which cut off B D equal to AC+, and join DC. Then, because BD 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 DBC equal to the Angle ACB, from the Hypothefis; therefore the Base DC is equal ‡ to the Bafe A B, and 1 4 of this. the Triangle DBC equal to the Triangle ACB, a Part to the Whole, which is abfurd; therefore A B is not unequal to A C, and confequently is equal to it. Therefore, if two Angles of a Triangle be equal between themselves, the Sides fubtending the equal Angles are likewise equal between themselves; which was to be demonftrated. Coroll. Coroll. Hence every Equiangular Triangle is also PROPOSITION VII. THEOREM. On the fame Right Line cannot be constituted two FOR, if it be poffible, let two Right Lines AD, D B, equal to two others A C, CB, each to each, be constituted at different Points C and D, towards the fame Part CD, and having the same Ends A and B, which the first Right Lines have, so that C A be equal to AD, having the fame End A, which CA hath; and CB equal to DB, having the same End B. Cafe 1. The Point D cannot fall in the Line AC; for instance at F: For then (AD that is) AF would not be equal to A C. Cafe 2. If it be said that D falls within the Triangle ABC; draw CD, and produce BD, BC, to F, and E. Now, fince A D is affirmed to be equal to A C, the Angle ADC is equal to the Angle ACD*; and consequently the Angle ACD is greater than FDC: Moreover ECD is greater than ACD, therefore ECD is much greater than FDC. But it is also said,. that BD is equal to BC, and so the Angle ECD under * 5 of this. the Base of the Isosceles Triangle is equal to FDC*; whereas it hath been proved to be much greater, which is absurd: Therefore D doth not fall within the Triangle. Cafe 3. Suppose D fell without the Triangle ABC; join CD. Then, because AC is equal to AD, the Angle +5 oftbis. ACD will be equal † to the Angle ADC, and consequently the Angle ADC is greater than the Angle BCD; wherefore the Angle BDC will be much greater than the Angle BCD. Again, because CB is equal to D B, the Angle BDC will be equal to the Angle BCD; but it has been proved to be much greater, which is impossible. Therefore, on the fame Right Line cannot be conftituted two Right Lines equal to two other Right Lines, each to each, at different Points, on the fame Side, and having the fame Ends which the first Right Lines have; which was to be demonftrated. PROPOSITION VIII. THEOREM.. If two Triangles have two Sides of the one equal to two Sides of the other, each to each, and the Bases equal, then the Angles contained under the equal Sides will be equal. LET the two Triangles be ABC, DEF, having two Sides, AB, AC, equal to two Sides DE, DF, each to each, viz. A B equal to DE, and AC to DF; and let the Base BC be equal to the Base EF. I fay, the Angle BAC is equal to the Angle EDF. For, if the Triangle ABC be applied to the Triangle DEF, so that the Point B may co-incide with E, and the Right Line BC with EF, then the Point C will co-incide with F, because BC is equal to EF. And fo, fince BC co-incides with EF, BA and AC will likewise co-incide with ED and DF. For if the Base B C should co-incide with EE, and at the same Time the Sides BA, AC, should not co-incide with the Sides ED, DF, but change their Position, as EG, GF, then there would be constituted on the fame Right Line two Right Lines, equal to two other Right Lines, each to each, at several Points, on the same Side, having the same Ends. But this is proved to be otherwise †; therefore it is impossible for thet 7 of this. Sides BA, AC, not to co-incide with the Sides ED, DF, if the Base BC co-incides with the Base EF; wherefore they will co-incide, and consequently the Angle BAC will co-incide with the Angle EDF+, and will be equal to it. Therefore, if two Triangles have two Sides of the one equal to two Sides of the other, each to each, and the Bases equal, then the Angles contained under the equal Sides will be equal; which was to be demonftrated, PRO PROPOSITION IX. PROBLEM. To cut a given Right-lin'd Angle into two equal LET BAC be a given Right-lin'd Angle, which † 3 of this, cut off A E from the Line AC equal to A D +; join $1 of this, DE, and thereon make the equilateral Triangle DEF, and join AF. I say, the Angle BAC is cut into two equal Parts by the Line A F. K For, because A D is equal to A E, and AF is common, the two Sides DA, AF, are equal to the two Sides A E, AF, and the Base D F is equal to the #8 of this. Base EF; therefore the Angle DAF is equal to the Angle EAF. Wherefore, a given Right-lin'd Angle is cut into two equal Parts; which was to be done. * 1 of this. : PROPOSITION Χ. PROBLEM. To cut a given finite Right Line into two equal Parts. LET AB be a given finite Right Line, required to be cut into two equal Parts. Upon it make * an Equilateral Triangle A BC, and + 9 of this, bisect + the Angle ACB by the Right Line CD. I fay, the Right Line A B is bisected in the Point D. For, because A C is equal to CB, and CD is common, the Right Lines AC, CD, are equal to the two Right Lines BC, CD, and the Angle ACD I 4 of this. equal to the Angle BCD; therefore the Bafe A D is equal to the Base D B. And fo, the Right Line A B is bisected in the Point D; which was to be done. PRO |