Poft. 1. Draw the Right Line AC from the Point A to C*, +1 of this, upon it defcribe the Equilateral Triangle DAC + Poft. 2. produce DA and DC directly forwards to E and C+ about the Center C, with the Distance B C, defcribe the Circle BGH*; and about the Center D, with the Distance DG, defcribe the Circle G KL.

Poft. 3.

+ Def.

*

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 DC are equal; therefore the Remainders AL, Axiom 3. GC are alfo equal t. But it has been demonstrated that BC is equal to CG; wherefore both A Land BC are each of them equal to CG. But Things that are equal to one and the fame Thing, are equal to one another; and therefore likewife AL is equal to BC.

2 of this.

+ Poft. 3.

Whence the Right Line AL is put at the given Point A, equal to the given right Line B, 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.

LE

ET 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 AD, defcribe a Circle DEFt.

Then becaufe A is the Center of the Circle DEF. AE is equal to AD; and fo both AE and C are each Axiom 1. equal to AD; wherefore AE is likewife equal to C+.

And fo there is cut off from AB the greater of two given Right Lines AB and C, a Line AE equal to the leffer Line C, which was to be done.

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 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

ET 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 A B equal to the Side DE, and the Side AC to DF; and the Angle BAC equal to the Angle EDF. I fay, that the Bafe BC is equal to the Bafe 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, becaufe AB is equal to DE. And fince AB co-incides with D E, the Right Line A C likewife will co-incide with the Right Line DF, because the Angle BAC is equal to the Angle EDF. Wherefore alfo 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 Bafe EF. For if the Point B co-inciding with E, and C with F, the Bafe BC does not co-incide with the Bafe EF; then two Right Lines will contain a Space, which is impoffible*. Therefore ⋆ Ax. 10. the Bafe BC co-incides with the Bafe EF, and is equal thereto; and confequently the whole Triangle ABC will co-incide with the whole Triangle DEF,

B 4

and

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 Ifofceles Triangle are equal between themselves: And if the equal Sides be produced, the Angles under the Base shall be equal between themselves.

LE

ET ABC be an Ifofceles Triangle, having the Side A B equal to the Side A C; and let the equal Sides AB, AC, be produced directly forwards to D and E. I fay the Angle ABC is equal to the Angle ACB, and the Angle CBD equal to the Angle BCE..

For affume any Point F in the Line BD, and from *3 of this. A E cut off the Line AG equal to A F, and join FC, GB.

Then because AF is equal to A G, and AB to AC, the two Right Lines FA, AC, are equal to the two Lines GA, AB, each to each, and contain the com4 of this mon Angle FAG; therefore the Bafe FC is equal t 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 cach, fubtending the equal Sides, viz. the Angle ACF 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 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 Bafe BC. Therefore alfo 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 fo 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 C B G equal to BCF, the remaining Angle ABC will be equal to the remaining • Ax. 3. Angle ACB; but these are the Angles at the Bafe of the Triangle ABC. It hath likewife been proved that the Angles FBC, GC B, under the Bafe, are equal; therefore the Angles at the Bafe of Ifofceles Triangles are equal between themselves; and if the equal Right Lines be produced, the Angles under the Bafe will be alfo equal between themselves.

Coroll. Hence every Equilateral Triangle is alfo 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 fay the Side AB is likewife equal to the Side A C.

For if AB 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 becaufe DB is equal to 3 of this. AC, and BC is common, DB, BC, will be equal to A C, C B, each to each, and the Angle DBC equal to the Angle A CB, from the Hypothefis; therefore the Bafe DC is equal to the Bafe AB, and † 4 of this. the Triangle DB C'equal to the Triangle ACB, a part to the whole, which is abfurd; therefore AB is not unequal to AC, and confequently is equal to it.

Therefore if two Angles of a Triangle be equal between themselves, the Sides fubtending the equal Angles are likewife equal between themfelves. Which was to be demonstrated.

Coroll.

Coroll. Hence every Equiangular Triangle is alfo
Equilateral.

PROPOSITION VII.

THEOREM.

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.

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 fame Ends A and B which the first Right Lines have, fo that CA be equal to A D, having the fame End A which CA hath; and CB equal to DB, having the fame End B; and let CD be joined.

Then becaufe A C is equal to A D, the Angle *5 of this. ACD will be equal* to the Angle ADC, and confequently the Angle ADC 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 BDC will be equal to the Angle B CD; 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 first right Lines have; which was to demonftrated.

PRO