37. An oblong, one, which has all its angles right, but not its sides equal.

38. A rhombus, one, which is equilateral, but not right angled.

39. A rhomboid, one, whose opposite sides and angles are equal, but which is neither equilateral nor right angled.

40. All other quadrilateral figures, besides these, are called trapeziums.

POSTULATES.

1. It is required to be granted, that a right line, may be drawn from any point to any other point.

2. That a terminated right line, may be produced at pleasure in a right line.

3. That from any point as a centre, at any distance from that centre, a circle may be described.

AXIOMS.

1. Things, which are equal to the same thing, are equal:> each other.

2. If to equals, equals be added, the wholes are equal.

3. If from equals, equals be taken away, the remainders are equal.

4. If to unequals, equals be added, the wholes are unequal, that, which arises from the addition to the greater, being the greater.

5. If from unequals, equals be taken away, the remainders are unequal, that, which arises from the subtraction from the greater, being the greater. And if from equals, unequals be taken away, the remainders are unequal, that, which remains from the subtraction of the greater, being the less.

6. Things, which are double of the same, are equal to each other.

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,

B

F

A

H

E

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. 3], meeting AB in F. The part AF, cut off from AB, is equal to CD.

F

-B

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.

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.