Book I. the distance BA (by post. 3.) let the circle ACE be described: and from the point C in which the circles cut one another (by post. 1.) let the straight lines CA, CB, be drawn to the points A and B.

Since therefore the point
A is the center of the circle
BCD (by Def. 15.) AC is
equal to AB again, because D
the point B is the center of
the circle ACE (by Def. 15.)
BC is equal to BA. But it
has been alfo demonftrated
that CA is equal to AB; each

C

A

B

E

therefore of the ftraight lines CA, CB is equal to AB; but (by com. not. 1.) magnitudes which are equal to the fame magnitude are equal to one another, and the ftraight line CA is therefore equal to the ftraight line CB; wherefore the three straight lines CA, AB, BC are equal to one another.

The triangle ABC is therefore (by Def. 24.) equilateral, and is described upon the given finite ftraight line AB. Which was to be done.

PROP. II.

At a given point, to place a straight line, equal to a given ftraight line.

Let the given point be the point A, and the given straight line the ftraight line BC: it is required at the point A to place a straight line equal to the ftraight line BC.

For let the straight line AB be drawn (by poft. 1.) from the point A to the point B: and upon it let the equilateral triangle DAB be described (by prop. 1.) and let the straight lines AE, BF be produced (by post. 2.) in ftraight lines with DA and DB; and then with the point B as a center and at the distance BC, let (by post. 3.) the circle CGH be described and again, with the point D as a center, and at the distance DG let (by post. 3.) the circle GKL be described.

Book I.

Since therefore the point B is the center of the circle CGH the Straight line BC is (by def. 15.) equal to BG: and again, because the point D is the center of the circle GKL the ftraight line DL is equal to the straight line DG (by Def. 15.) the parts of which viz. the ftraight line DA is equal to the ftraigt line DB, therefore the ftraight line AL the remainder is equal (by

com. not. 3.) to the ftraight line BG the remainder, but the Straight line BC has been fhewn to be equal to the ftraight line BG: each therefore of the straight lines AL, BC is equal to the ftraight line BG; but magnitudes equal to the fame, are alfo equal to one another: and therefore the straight line AL is equal to the Straight line BC.

Wherefore at the given point A, equal to the given straight line BC.

the straight line AL is placed
Which was to be done.

Two unequal straight lines being given, to cut off a part from the greater equal to the lefs.

Let the ftraight lines AB and C be the two unequal straight lines given it is required from the greater the straight line AB to cut off a straight line equal to the straight line C the less.

Place (by prop. 2.) at the point A a straight line AD equal to the ftraight line C; and then with the point A for a center, but at the distance AD, let the circle DEF be defcribed. /.

And

Book I.

And fince the point A is the center of the circle DEF, the ftraight line AE is equal to the ftraight line AD (by def. 15.) but the straight line C is alfo equal to the straight line AD: each therefore of the straight lines AE and C is equal to the ftraight line AD: wherefore also the straight line AE is equal (by com. not. 1.) to the Straight line C.

Wherefore two unequal ftraight lines AB and C being given, from the straight line AB the greater, the straight line AE has been cut off equal to the straight line C the lefs. Which was to be done.

If two triangles have the two fides equal to the two fides, each to each, and have the angle equal to the angle, the angle contained by the equal straight lines: they fhall also have the base equal to the base, and the triangle shall be equal to the triangle; and the other angles under which the two equal fides are extended, shall be equal to the other angles, each to each.

Let the triangles ABC, DEF be two triangles having the two fides the ftraight lines AB, AC equal to the two fides the ftraight lines DE, DF each to each, the one, the straight line AB to the ftraight line DE; the other the ftraight line AC to the ftraight line DF; and an angle, that contained by BA, AC or BAC equal to that contained by ED, DF or EDF: I say, that alfo the ftraight line BC a base is equal to the ftraight line EF a base: and the triangle ABC shall be equal to the triangle DEF, and the other angles, under which the two equal fides are extended, shall be equal to the other angles each to each; the one, the angle contained by AB, BC or ABC to that contained by DE, EF or DEF; the other, that contained by AC, CB or ACB to that contained by DF, FE or DFE.

For

For the triangle ABC being applied to the triangle DEF; and the point A in particular being put upon the point D, but the straight line

AB upon the upon the straight line DE, also the point B will apply itself to the point E,

because the straight line AB is equal (by fuppofition) to the Straight line DE: but the ftraight line AB having applied itself to the ftraight line DE, the ftraight line AC will also apply itself to the ftraight line DF, because the angle contained by BA, AC or BAC is equal to that contained by ED, DF or EDF (by supp.): wherefore also, because again (by fupp.) the straight line AC is equal to the straight line DF, the point C will apply itself to the point F: certainly also the point B has applied itself to the point E, wherefore the straight line BC a base, will apply itself to the straight line EF a base: for if, the one, the point B having applied itself to the point E, and the other, the point C to the point F, the Straight line BC a base will not apply itself to the firaight line EF, two ftraight lines will inclofe a fpace, which (by com. not. 12.) is impoffible: therefore the ftraight line BC a bafe will apply itself to the straight line EF, and (by com. not. 8.) will be equal to it: wherefore also the whole, the triangle ABC will apply itself to the whole, the triangle DEF, and (by com. not. 8.) will be equal to it: and the other angles will apply themselves to the other angles, and (by com. not. 8.) will be equal to them; the one contained by AB, BC or ABC to that contained by DE, EF or DEF; the other contained by AC, CB or ACB to that contained by DF, FE or DFE.

If therefore two triangles have the two fides equal to the two fides, each to each, and have the angle equal to the angle, the angle contained by the equal ftraight lines: they fhall alfo have the bafe equal to the bafe, and the triangle fhall be equal, to the triangle; and the other angles, under which the two equal fides are extended, shall be equal to the other angles, each to each. Which was to be demonstrated.

PROP.

Book I.

Book I.

PROP. V.

The angles at the base of isofceles triangles are equal to one another; and if the equal straight lines be produced, the angles under the base shall be equal to one another.

Let the triangle ABC be an isofceles triangle, having the fide AB equal to the fide AC, and let the ftraight lines BD, CE be produced in straight lines to AB, AC; I say that the one, the angle contained by AB BC or ABC is equal to the angle contained by AC, CB or ACB; the other, contained by CB, BD or CBD to the angle contained by BC, CE or BCE.

Let any point which you may accidentally happen upon, as the point F be taken in the line BD, and let the ftraight line AG be cut off (by prop. 3.) from the greater the ftraight line AE, equal to the less the straight line AF; and let the straight lines FC, GB be drawn.

A

B

C

Since therefore the one ftraight line AF is equal to the straight line AG and the other AB to the ftraight line AC, certainly the two FA, AC, are equal to the two, GA, AB, each to each, and they contain a common angle, the angle contained by FA, AG or FAG: therefore (by the 4th prop.) the ftraight line CF a base is equal to the ftraight line GB a base, and the triangle AFC will be F equal to the triangle AGB, and the other angles will be equal to the other angles, each to each, under which the equal fides are extended, the one contained by AC, CF or ACF to the angle contained by AB, BG or ABG, and the other contained by AF, FC or AFC to the angle contained by AG, GB or AGB.

D

G

E

And fince a whole the ftraight line AF is equal to a whole the Straight line AG, parts of which the ftraight line AB is equal to the ftraight line AC, wherefore (by com. not. 3.) the straight line BF a remainder is equal to the straight line CG a remainder: but the ftraight line FC has been proved equal to the ftraight line GB; cer