PROPOSITION IV. If there THEOREM. 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 thofe equal Sides in one Triangle, equal to the Angle contained by the correfpondent Sides in the other Triangle; then the Bafe of one of the Triangles fhall be equal to the Bafe 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, A B, A C, equal to two Sides DE, DF, each to each, that is, the Side A B equal to the Side D E, and the Side AC to DF; and the Angle B A C equal to the Angle EDF. I fay, that the Bafe B C is equal to the Bafe E F, the Triangle ABC equal to the Triangle D E F, 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 A B C equal to the Angle D E F, and the Angle AC B equal to the Angle DFE. For the Triangle A B C being applied to D EF, fo as the Point A may co-incide with D, and the Right Line A B with DE, then the Point B. will co-incide with the Point E, because A B is equal to D E. And fince A B co-incides with D E, the Right Line A C likewife will co-incide with the Right Line D F, becaufe the Angle B A C is equal to the Angle ED F. Wherefore alfo C will co-incide with F, because the Right Line A C is equal to the Right Line D F. But the Point B co incides with E, and therefore the Bafe BC co-incides with the Bafe E F. For, if the Point B co-inciding with E, and C with F; the Bafe BC does not co-incide with the Bafe E F; then two Right Lines will contain a Space, which is impoffible *. * Ax. 10. Therefore, the Bafe B C co-incides with the Pafe EF, and is equal thereto; and confequently the whole Tri B 4 angle 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 A B C equal to the Angle DE F, and the Angle ACB equal to the Angle D F E; which was to be demonftrated. PROPOSITION V. The Angles at the Bafe of an Ifoceles Triangle ET ABC be an Ifoceles Triangle, having the Side A B equal to the Side AC; and let the equal Sides A B, A C be produced directly forwards to D and E. I fay, the Angle ABC is equal to the Angle A C B, and the Angle CBD equal to the Angle BCE. For affume any Point F in the Line B D, and from 3 of this. A E cut off the Line A G equal * to A F, and join F C, G B. Then, because AF is equal to AG, and AB to AC, the two Right Lines F A, A C, are equal to the two Lines GA, A B, each to each, and contain the com* 4 of this. mon Angle F AG; therefore the Base F C is equal + to the Bafe G B, and the Triangle A F C equal to the Triangle AG B, 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 AC F equal to the Angle A B G; and the Angle A FC, equal to the Angle A G B. And because the Whole AF is equal to the whole A G, and the Part A B equal to the Part A C, the Remainder BF is equal to the Remainder C G. But FC has been proved to be equal to G B; therefore the two Sides BF, F C, are equal to the two Sides C G, GB, each to each, and the Angle B F C equal to the Angle CGB; but they have a common Bafe BC. Therefore alfo the Triangle B F C will be equal to the †4 of this. Triangle CG B*, and the remaining Angles of the one equal equal to the remaining Angles of the other, each to each, which fubtend the equal fides. And fo the Angle FBC is equal to the Angle G CB; and the Angle BCF equal to the Angle CB G. Therefore, because the whole Angle A BG hath been proved equal to the whole Angle A C F, and the Part C B G equal_to * Ax. 3. BC F, the remaining Angle A C B * will be equal to the remaining Angle A B C; but these are the Angles at the Bafe of the Triangle A CB. It hath likewife been proved, that the Angles FBC, GCB, under the Bafe, are equal; therefore, the Angles at the Bafe of Ifoceles Triangles are equal between themfelves; 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 alfo Equiangular. PROPO 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 A Ct, and join DC. Then becaufe B D is equal to † 3 of this. A C, and B C is common, D B, BC, will be equal - to A C, CB, each to each, and the Angle D B C equal to the Angle A C B, from the Hypothefis; therefore the Bale DC is equal to the Base A B, and 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 tuo Angles of a Triangle be equal between themselves, the Sides fubtending the equal Angles are likewife equal between themselves; which was to be demonftrated. Coroll. 4 of this. Coroll. Hence every Equiangular Triangule is alfo PROPOSITION VII. THEOREM. On the fame Right Lines cannot be conftituted two FOR, if it be poffible, let two Right Lines A D, D B, equal to two others A C, C B, each to each, be conftituted at different Points C and D, towards the fame Part C D, and having the fame Ends A and B, which the firft Right Lines have, fo that C A be equal to A D, having the fame End A, which CA hath; and C B equal to D B, having the fame End B. Cafe 1. The Point D cannot fall in the Line A C; for inftance at F: For then (A D that is) A F would not be equal to A C. Cafe 2. If it be faid that D falls within the Triangle ABC; draw CD, and produce B D, 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 A CD*; and confequently the Angle ACD is greater than FDC: Moreover ECD is greater than A CD, therefore ECD is much greater than F DC. But it is alfo faid, that B D is equal to BC, and fo the Angle ECD under the Bafe of the Ifoceles Triangle is equal to * 5 of this. FDC*; whereas it hath been proved to be much greater, which is abfurd: Therefore D doth not fall within the Triangle. Cafe 3. Suppole D fell without the Triangle ABC; join C D. Then because A C is equal to AD, the Angle 15 of this. ACD will be equai + to the Angle ADC, and confequently the Angle A D C is greater than the Angle BCD; wherefore the Angle BDC will be much greater than the Angle B CD. Again, because C B is equal to DB, the Angle B D C will be equal to the Angle BCD; but it has been proved to be much greater, which is impoffible. 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 firft Right Lines have; which was to be demonftrated. PROPOSITION VIII. If two Triangles have two Sides of the one equal L ET the two Triangles be ABC, DEF, having two Sides, A B, A C, equal to two Sides DE, DF, each to each, viz. A B equal to D E, and A C to DF; and let the Bafe BC be equal to the Base EF. I fay, the Angle B A C is equal to the Angle EDF. For, if the Triangle ABC be applied to the Triangle D E F, fo 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, becaufe B C is equal to E F. And fo, fince BC co-incides with E F, BA and A C will likewife co incide with E D and D F. For if the Bafe B C fhould co-incide with E E, and at the fame Time the Sides B A, A C, fhould not co-incide with the Sides E D, DF, but change their Pofition, as EG, GF, then there would be conftituted on the fame Right Line two Right Lines, equal to two other Right Lines, each to each, at feveral Points, on the fame Side, having the fame Ends. But this is proved to be otherwise +; therefore it is impoffible for the 7 of this. Sides B A, A C, not to co-incide with the Sides E D, DF, if the Bafe B C co-incides with the Base EF; wherefore they will co-incide, and confequently the Angle B A C 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 Bafes equal, then the Angles contained under the equal Sides will be equal; which was to be demonstrated. PRO |