7. Things, which are halves of the same, are equal to each other.

8. Things, which, being applied to each other, do coincide, are equal.

Corollary. The whole is equal to all its parts.

9. The whole is greater than its part.

10. Two right lines cannot enclose a space. 11. Two right lines (AB, CB) have not a common segment (BD).

See notes on 10th, 11th, and 12th, ax. and on Prop. 4. of this book.

12. If a right line (EF), intersecting two others (AB, CD),

make the two interior angles (AGH, GHC) on one side of the intersecting right line, taken together, not

A

C

H

A

B

-D

C

E

B

-D

equal to the two interior angles (BGH, GHD) on the other side of the intersecting right line, taken together; the right lines (AB, CD), so met by the third, may be so produced towards the part (B, D), on which the interior angles taken together are least, as to meet.

See note to Prop. 29th of this book.

PROPOSITION I. PROBLEM.

Upon a given right finite line (AB), to make an equilateral triangle.

From the centre A, at the distance AB, describe the circle BEC (post. 3). From the centre B, at the distance BA, describe the circle AED (post. 3). Produce AB both ways (post. 2), so as to meet these circles in the points C and D. Then, since the

C

point C is without the circle AED, and the point B within the same circle, and the circles BEC, AED, are continued lines on each side of the right line CD [def. 10], these circles intersect each other on each side of that right line, as in E and F. From one of these intersections E, draw the right lines EA, EB to the extremes of the given right line [Post. 1]. The triangle ABE, which is constituted on the given right line AB, is equilateral.

For AE is equal to AB, being both radiuses of the same circle BEC [Def. 10]; and BE is equal to AB, being both radiuses of the same circle AED [Def. 10]; whence AE and BE, being each equal to AB, are equal to each other [Ax. 1]. Therefore AB, AE and BE are equal to each other, and the triangle ABE is equilateral [Def. 28].

Scholium. AF and BF being drawn, the triangle ABF may in like manner be proved to be equilateral. See note on this proposition.

PROP. II. PROB.

At a given point (A), to put a right line, equal to a given right line (BC).

From the given point A, to either extreme C of the given right line, draw the G right line AC [Post. 1]. On AC make the equilateral triangle ADC [1.1]. From the centre C, at the distance CB, describe the circle EBF [Post. 3], and produce DC to meet its circumference in E [Post. 2]. From the centre D, at the distance DE,

F

A

D

H

B

C

describe the circle EGH [Post. 3]. Produce DA to meet its circumference in H [Post. 2]. Aй is equal to the given right

line BC.

For DH and DE are equal, being radiuses of the same circle EGH [Def. 10]; taking from them the parts DA, DC, which are equal, being sides of the equilateral triangle ADC, the residues AH, CE are equal [Ax. 3]. But BC and CE are equal, being radiuses of the same circle EBF [Def. 10]: therefore AH and BC, being each equal to CE, are equal to each other [Ax. 1]. There is therefore put at the given point A, a right line AH, equal to the given right line BC.

Scholium. The position of the right line AH is varied, according to the extreme of the given right line, to which the right line is drawn from the given point; and also, according to the part of the right line so drawn, to which the triangle is constituted.

PROP. III. PROB.

Two unequal right lines (AB, CD) being given, to cut off from the greater, a part equal to the less.

At either extreme A, of the greater of the given right lines, put AE equal to the less CD [2. 1]. From the centre A, at the distance AE, describe the circle EFG [Post. 31, meeting AB in F. The part AF, cut off from AB, is equal to CD.

G

F

D

For AF is equal to AE, being radiuses of the same circle EFG [Def. 10]; and AE is equal to CD [By constr.]; therefore AF and CD are each equal to AE, and therefore to each other [Ax. 1]; and so there is cut off from AB, a part AF, equal to CD.

PROP. IV. THEOREM.

If two triangles (ABC, DEF), have two sides (CA CB), and the angle (ACB) included by them, of one triangle; (ABC), severally equal to two sides (FD, FE), and the angle (DFE) included by them, of the other; the bases or third sides (AB, DE) are equal; as are also, the angles at the bases, opposite to the equal sides (CAB to FDE, and CBA to FED); and the triangles themselves.

For the triangle ABC, being applied to the triangle DEF, so that the point C may coincide in the point F, the right line CA with the right line FD, and the right lines CB, FE be to

the same part; the point A would coincide with the point D, for part of CA cannot coincide with FD, and part be without it, as in the direction GH, 'for then two right lines GH, GD would have a common segment FG, contrary to the 11th axiom; and if the point A went beyond or fell short of the point D, the right lines CA, FD would be unequal [Ax. 9], contrary to the supposition; and because the angles C, F are equal [Hyp.], the right line CB would coincide with FE; and, because, CB, FE are equal [Hyp.], the point B would coincide with the point E.

But the point A coinciding with D, and B with E, the right lines AB, DE would coincide; for, if AB did not coincide, in every part of it, with DE, two right lines would contain a space, contrary to the 10th axiom. Therefore the bases AB, DE are equal [Ax. 8].

And the legs of the angle CAB, coinciding with those of the angle FDE, and those of the angle CBA, with those of the angle FED; the angles CAB, FDE, as also the angles CBA, FED would coincide, and are therefore equal [Ax. 8].

And the right lines, which contain the triangle ABC, coinciding with the right lines which contain the triangle DEF, the triangles themselves would coincide, and are therefore equal [Ax. 8].

See note on this proposition.

PROP. V. THEOR.

The angles (CAB, CBA), at the base (AB) of an isosceles triangle (ABC), are equal: and, if the equal sides (CA, CB) be produced below the base, the angles under the base (BAD: ABE) are equal.

In either leg, as CA, produced, take any point D; on CB produced, take CE equal to CD [3. 1], and draw DB, AE.

D

B

In the triangles DCB, ECA, the sides CD, CE are equal [Constr.], also the sides CB, CA [Hyp.], and the angle C is common to both these triangles; therefore the angle CBD is equal to CAE, the angle CDB to CEA, and BD to AE [4. 1]. Whence, in the triangles ADB, BEA, the angle ADB is equal to BEA, the side DB to AE, and, taking the equals CA, CB, from the equals CD, CE, the side AD to BE [Ax. 3], therefore the angles DAB, ABE, which are the angles under the base, are equal [4. 1].

And, in the same triangles, the angles ABD, BAE are equal; which being taken from the equal angles CBD, CAE, the residues CBA, CAB, which are the angles at the base AB, of the triangle ABC, are equal [Ax. s].

Corollary. Hence every equilateral triangle is equiangular. For, whichever side be considered as base, the angles adjacent to it are equal, being opposite equal sides.

PROP. VI. THEOR.

If two angles (CAB, CBA) of a triangle (ABC) be equal, the sides (CA, CB), opposite to them, are equal.

For, if CA and CB be not equal, let one of them, if possible, as CA, be the greater, and take from it AD equal to BC [3. 1], and draw BD.

Because, in the triangles DAB, CBA, the sides DA, AB are severally equal to the sides CB, BA, and the angles DAB, CBA included A

by the equal sides, also equal [Hyp.]; the triangles DAB, CBA are themselves equal [4. 1], a part to the whole, which is absurd [Ax. 9], therefore the sides CA, CB are not unequal, they are therefore equal.

Cor.-Hence every equiangular triangle is equilateral. For, whichever side be considered as base, the angles adjacent to it are equal, and therefore the sides opposite to them.