5. If equals be taken from unequals the remainders will be unequal.

6. Things which are double of the fame thing are equal to each other.

7. Things which are halves of the fame thing are equal to each other.

8. The whole is equal to all its parts taken together.

9. Magnitudes which coincide, or fill the fame space, are equal to each other.

REMARK S.

A PROPOSITION, is fomething which is either propofed to be done, or to be demonftrated.

A PROBLEM, is fomething which is proposed to be done.

A THEOREM, is fomething which is propofed to be demonftrated.

A LEMMA, is fomething which is previoufly demonftrated, in order to render what follows more eafy,

A COROLLARY, is a confequent truth, gained from fome preceding truth, or demonftration.

A SCHOLIUM, is a remark or obfervation made upon fomething going before it.

PROPOSITION I. PROBLEM.

UPON a given finite right line to describe an equilateral triangle.

Let AB be the given right line; it is required to defcribe an equilateral triangle upon it.

From the point A, at the distance AB, defcribe the circle BCD (Pof. 3.)

And from the point B, at the distance BA, defcribe the circle ACE (Pof. 3.)

Then, because the two circles pafs through each other's centres, they will cut each other.

And, if the right lines CA, CB be drawn from the point of interfection C, ABC will be the equilateral triangle required.

For, fince A is the centre of the circle BCD, AC is equal to AB (Def. 13.)

And, because в is the centre of the circle ACE, BC is also equal to AB (Def. 13.)

But things which are equal to the fame thing are equal to each other (Ax. 1); therefore AC as equal to CB.

And, fince AC, CB are equal to each other, as well as to AB, the triangle, ABC is equilateral; and it is described upon the right line AB, as was to be done.

PROP. II. PROBLEM.

From a given point to draw a right line equal to a given finite right line.

E

F

Let A be the given point, and BC the given right line; it is required to draw a right line from the point A, that fhall be equal to BC.

Join the points A, B, (Pof. 1.); and upon BA defcribe the equilateral triangle BAD (Prop. 1.)

.

From the point B, at the distance BC, defcribe the cir

cle CEF (Pof. 3.) cutting DE produced in F.

And from the point D, at the distance DF, defcribe the circle FHG (Pof. 3.); then, if DA be produced to Ģ, AG will be equal to BC, as was required,

For, fince B is the centre of the circle CEF, BC is equal to BF (Def. 13.)

And, because p is the centre of the circle FHG, DG is equal to DF (Def. 13.)

But the part DA is alfo equal to the part DB (Def. 16.), whence the remainder AG will be equal to the remainder BF (Ax. 3.)

And fince AG, BC have been each proved to be equal to BF, AG will also be equal to BC (Ax. 1.).

A right line AG, has, therefore, been drawn from the point A, equal to the right line BC, as was to be done.

SCHOLIUM. When the point A is at one of the extremities B, of the given line BC, any right line, drawn from that point to the circumference of the circle CEF, will be the one required,

From the greater of two given right lines, to cut off a part equal to the lefs.

R

Let AB and c be the two given right lines; it is required to cut off a part from AB, the greater, equal to c the lefs.

From the point A draw the right line AD equal të c (Prop. 2.); and from the centre A, at the diftance AD, defcribe the circle DEF (Pof. 3.) cutting AB in E, and AE will be equal to c as was required.

For, fince A is the centre of the circle EDF, AE will be equal to AD (Def. 15.)

But c is equal to AD, by conftruction; therefore AE will also be equal to c (Ax. 1.)

Whence, from AB, the greater of the two given lines, there has been taken a part equal to c the lefs, which was to be done.

SCHOLIUM. When the two given lines are so fituate, that one of the extremities of falls in the point A, the former part of the contruction becomes unneceffary.

PROP.

PROP. IV. THEOREM.

If two fides and the included angle of one triangle, be equal to two fides and the included angle of another, each to each, the triangles will be equal in all respects.

Let ABC, DEF be two triangles, having CA equal to FD, CB to FE, and the angle c to the angle F; then will the two triangles be equal in all respects.

For conceive the triangle ABC to be applied to the triangle DEF, fo that the point c may coincide with the point F, and the fide CA with the fide FD.

L

Then, because CA coincides with FD, and the angle c is equal to the angle F (by Hyp.), the fide CB will also coincide with the fide FE.

And, fince CA is equal to FD, and CB to FE (by Hyp.), the point A will fall upon the point D, and the point B upon the point E.

But right lines, which have the fame extremities, muft coincide, or otherwise their parts would not lie in the fame direction, which is abfurd (Def. 5.); therefore ab falls upon, and is equal to DE.

And, because the triangle ABC coincides with the triangle DEF, the angle a will be equal to the angle D,· the angle B to the angle E, and the two triangles will be equal in all refpects (Ax. 9.) Q. E. D.