PROP. XXXIII. THEOR.

The straight lines which join the extremities of two equal and parallel straight lines towards the same parts, are also themselves equal and parallel.

Let AB, CD be equal and parallel straight lines, and joined towards the same parts by the straight lines AC, BD: AC, BD shall be equal and parallel.

Join BC; and because AB is parallel

B

to CD, and BC meets them, the alternate angles* #29. 1. ABC, BCD are equal: and because AB is equal to CD, and BC common to the two triangles ABC, DCB, the two sides AB, BC are equal to the two DC, CB, each to each and the angle ABC was proved to be equal to the angle BCD; therefore the base AC is equal to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles,* each to each, to which the equal sides are opposite: therefore the angle ACB is equal to the angle CBD: and because the straight line BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal to one another, AC is parallel to BD: and it was shown to be equal to it. Therefore, straight lines, &c. Q. E. d.

#4. 1.

#27. 1.

PROP. XXXIV. THEOR.

The opposite sides and angles of a parallelogram are equal to one another, and the diameter bisects it, that is, divides it into two equal parts.

N. B.--A parallelogram is a four-sided figure, of which the opposite sides are parallel: and the diameter is the straight line joining two of its opposite angles.

Let ACDB be a parallelogram, of which BC is a diameter: the opposite sides and angles of the figure shall be equal to one another; and the diameter BC shall bisect it.

Because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal* to one another: #29. 1. and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal*

#29. 1.

to one another: wherefore the two triangles ABC, CBD have two angles ABC, BCA in the one, equal to two angles BCD, CBD in the other, each to each, and one side BC, which is adjacent to their equal angles, common to the two triangles; therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other, viz. the side AB to the side #26. 1. CD, and AC to BD, and the angle BAC equal to the angle BDC: and because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, the whole angle ABD is equal to the whole angle ACD: and the angle BAC has been shown to be equal to the angle BDC: therefore the opposite sides and angles of parallelograms are equal to one another. Also, their diameter bisects them: for AB being equal to CD, and BC common, the two AB, BC are equal to the two DC, CB, each to each; and the angle ABC has been proved equal to the angle BCD; therefore the triangle ABC is

+2 Ax.

*4. 1.

equal to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. Q. E. D.

PROP. XXXV. THEOR.

Parallelograms upon the same base, and between the same parallels, are equal to one another.

Let the parallelograms ABCD, EBCF be upon the same base BC, and between the same parallels AF, BC: the parallelogram ABCD shall be equal to the parallelogram EBCF.

If the sides AD, DF of the parallelograms ABCD, DBCF, opposite to the base BC, be terminated in the same point D; it is plain that each of the parallelograms is double* of the triangle BDC; and *34. 1. that they are therefore equal† to

+6 Ax.

one another.

But if the sides AD, EF, opposite to the base BC of the parallelograms ABCD, EBCF, be not terminated

in the same point; then, B because ABCD is a paralle

*34 1. *1 Ax.

+34.1.

*29. 1.

B

logram, AD is equal* to BC; for a like reason EP is equal to BC; wherefore AD is equal to EF; and DE is common; therefore the whole, or *2 or 3 Ax. the remainder, AE is equal to the whole or the remainder, DF: AB also is equal† to DC; therefore the two EA, AB are equal to the two FD, DC, each to each; and the exterior angle FDC is equal to the interior EAB: therefore the base EB is equal to the base FC, and the triangle EAB equal to the triangle FDC. Take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB, and the remainders* are equal; that is, the parallelogram ABCD is equal to the parallelogram EBCF. Therefore parallelograms upon the same base, &c. Q. E. v.

*4. 1.

*3 Ax.

PROP. XXXVI. THEOR.

Parallelograms upon equal bases, and between the same parallels, are equal to one another.

Let ABCD, EFGH be parallelograms upon equal bases BC, FG and between the same parallels

AH, BG: the

parallelogram

ABCD shall be equal to EFGH.
Draw BE, CH; then because

A

DE

tHyp: *34. 1. +1 Ax.

+Hyp.

#33. 1.

BC is equal to FG, and FG to* EH, BC is equal tot EH: and they aret parallels, and joined towards the same parts by the straight lines BE, CH: but straight lines which join the extremities of equal and parallel straight lines towards the same parts, are themselves equal and parallel; therefore EB, HC are both equal and parallel; and therefore EBCH is at parallelogram; and it is #35. 1. equal to ABCD, because they are upon the same base BC, and between the same parallels BC, AH: for the like reason, the parallelogram EFGH is equal to the same

*Def. 34. 1.

+1 Ax.

EBCH: therefore the parallelogram ABCD is equalt to EFGH. Wherefore parallelograms, &c. Q. E. D.

PROP. XXXVII. THEOR.

Triangles upon the same base, and between the same
parallels, are equal to one another.

Let the triangles ABC, DBC be upon the same base BC, and between the same parallels AD,

BC: the triangle ABC shall be equal E to the triangle DBC.

12 Post.

#31. 1.

Producet AD both ways to the points E, F, and through B draw* BE parallel

to CA; and through C draw CF

+ Def. 34. 1. #35. 1.

B

parallel to BD: therefore each of the figures EBCA, DBCF is at parallelogram: and EBCA is equal* to DBCF, because they are upon the same base BC, and between the same parallels BC, EF; and the triangle ABC is the half of the parallelogram EBCA, because the diameter AB bisects it: and the triangle DBC is the half of the parallelogram DBCF, because the diameter DC bisects it: but the halves #7 Ax. of equal things are* equal; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D.

34. 1.

This proposition is unnecessary, for it is included in that which follows, and is not subsidiary to the demonstration of the more

general case. The student will readily see that, as the truths of geometry are entirely independent of the size and positions of the figures treated of, they remain the same, however the latter circumstances may vary. It is shown in the next proposition that triangles between the same parallels are always equal, provided only their bases are equal, and without any restriction as to the part of the parallel occupied by one base in reference to the other; the bases may coincide or not. It may be remarked that the 35th, like this, is only a particular case of the proposition next in order, but that cannot be suppressed, because it helps to prove the general case.

PROP. XXXVIII. THEOR.

Triangles upon equal bases, and between the same parallels, are equal to one another.

Let the triangles ABC, DEF be upon equal bases BC EF, and between the same parallels BF, AD: the triangle ABC shall be equal to the triangle DEF.

#31. 1.

Producet AD both ways to the points G, H, †2 Post. and through B draw BG parallel to CA, and through F draw FH parallel to ED: then each of the figures GBCA,

+Def. 34. 1. DEFH, is at parallelogram: and they are

B

*36. 1. equal to one another, because they are upon equal bases BC, EF, and between the same parallels BF, GH; and the triangle ABC is the half of the parallelogram GBCA, because the diameter AB bisects* it; and the triangle DEF is the half of the parallelogram DEFH, because the diameter DF bisects it: but the halves of equal things are equal: therefore the triangle ABC is equal to the triangle DEF. Wherefore triangles, &c. Q. E. D.

#34. 1.

*7 Ax.