« ForrigeFortsett »
ibe one equal to two sides of ike other, each to
otber,each to each, which subtend the equal Sides. LET the two Triangles b: ABC, DE F, which
have two Sides, AB, AC, equal to two Sides DE, DF, each to each, that is, the Side A B equal to the Side D E, and the Side A C to DF; and the Angle B A C equal to the Angle EDF. say, that the Base B C is equal to the Base 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 correspondent, fubtending the equal Sides; viz. the Angle A B C equal to the Angle D E F, and the Angle ACB equal to the Angle D FE.
For the Triangle A B C being applied to D EF, so as the Point A may co-incide with D, and the Right Line A B with D E, 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 AC likewise will co-incide with the Right Line D F, because the Angle BAC is equal to the Angle E D F. Wherefore allo 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 E F. For, if the Point B co-inciding with E, and C with F; the Base BC does not co-incide with the Base E F; then two Right Lines will contain a Space, which is impoffible *. * Ax. 10. Therefore, the Baje B C co-incides with the Prije EF, and is equal thereto; and consequently the whole Tri
angle A B C will co-incide with the whole Triangle
D EF, and will be equal thereto ; and the remaining, t Ax. 8. Angles will co-incide with the remaining Angles t, and
will be equal to them, viz. the Angle A B C equal to the Angle D E F, and the Angle A C B equal to the Angle D FE; which was to be demonstrated.
are equal between themselves : And, if the equal
Side A B' equal to the Side A C; and let the equal Sides A B, A C be produced directly forwards to D and E. I say, the Angle A B C is equal to the Angle A C B, and the Angle CBD equal to the Angle BCE.
For assume any Point F in the Line B D, and from 3 of ibis. A E cut off the Line A G equal * to A F, and join
Then, because AF is equal to AG, and A B 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 common Angle F AG; therefore the Base F C is equal + to the Base G B, and the Triangle A F C 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 A C F equal to the Angle A B G; and the Angle A FC, equal to the Angle A GB. 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 B F# 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 Bale BC. There
fore also the Triangle B FC will be equal to the * 4 of this. Triangle CGB*, and the remaining Angles of the one
4 of Ibis.
equal to the remaining Angles of the other, each to each, which subtend the equal fides. And so the Angle FBC is equal to the Angle G CB; and the Angle BCF equal to the Angle CBG. Therefore, because the whole Angle A B G hath been proved equal to the whole Angle A CF, and the Part CBG equal to * Ax. 3. BCF, the remaining Angle ACB * will be equal to the remaining Angle A B C; 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 Base of Isoceles Triangles are equal between themselves ; and if the equal Right Lines be produced, the Angles under the Base will be also equal between themselves ; which was to be demonstrated.
Coroll. Hence every Equilateral Triangle is also Equiangular. PROPOSITION VI.
Sides subtending the equal Angles will be equal
between themselves. LET A B C be a Triangle, having the Angle
A B C equal to the Angle ACB. I say, the Side A B is likewise equal to the Side A C.
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 t, and join DC. Then hecause B D is equal to + 3 of tbis. A C, and B C is common, DB, BC, will be equal - to A C, CB, each to each, and the Angle D BC equal to the Angle ACB, from the Hypothesis; therefore the Bale DC is equal f in the Base A B, and the Triangle DB C equal to the Triangle ACB, a Part to the Whole, which is absurd: therefore A B is not unequal to A C, and consequently is equal to it.
Therefore, if tuo Angles of a Triangle be equal between themselves, the Sides subtending the equal Angles are likewise equal between themselves; which was to be demonstrated,
1 4 of ibis,
Coroll. Hence every Equiangular Triangule is also
Right Lines equal to two other Right Lines,
Right Lines have.
DB, equal to two others AC, CB, each to each,
Case 1. The Point D cannot fall in the Line AC;
Cafe 2. If it be said that D falls within the Triangle
under the base of the lloceles Triangle is equal 10 * 5 of tbis. FDC*; whereas it hath been proved to be much
greater, which is absurd: Therefore D doth not fall
Case 3. Suppole D fell without the Triangle A BC;
Then because A C is equal to AD, the Angle 15 of ibis. ACD will be equal t to the Angle A DC, and con
sequently the Angle ADC is greater than the Angle
Right Line cannot be constituted two Right Lines equal to two other Right Lines, each to each, at different Points, on the same Side, and having the same Ends which the first Right Lines have; which was to be demonstrated.
to two sides of the other, each to each, and the
equal Sides will be equal.
two Sides, A B, AC, equal to two Sides D E, DF, each to each, viz. A B equal to D Е, and AC to DF; and let the Base B C be equal to the Base EF. I say, the Angle B A C is equal to the Angle EDF
For, if the Triangle A B C be applied to the Triangle DEF, so that the Point B may coincide with E, and the Right Line B C with EF, then the Point C will co-incide with F, because B C is equal to E F. And so, since B C co-incides with EF, B A and AC will likewise coincide with E D and DF. Foi if the Base B C should co-incide with E E, and at the fame Time the Sides B A, A C, should not co-incide with the Sides ED, DF, but change their Pofition, as E G, G F, then there would be constituted on the Same 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 t; therelore it is impossible for the ty of ibis. Sides B A, A C, not to co-incide with the Sides ED, DF, if the Base B C co-incides with the Base EF ; wherefore they will co-incide, and consequently 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 Bases equal, then the Angles contained under the equal Sides will be equal; which was to be demonstrated.