angle ACB is equal to the angle CBD; and because the straight Book I. 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 c 27. 1. lines, &c. Q. E. D.

PROP. XXXIV. THEOR.

THE opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them, that is, divides them 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 are equal to one another; and the diameter BC bisects it.

a to one

C

B

D

a 29. 1.

Because AB is parallel to CD, A and BC meets them, the alternate angles ABC, BCD are equal a to one another; and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are equal another; wherefore the two triangles ABC, CBD have two angles ABC, BCA in one, equal to two angles BCD, CBD in the other, each to each, and one side BC common to the two triangles, which is adjacent to their equal angles; therefore their other sides shall be equal, each to each, and the third angle of the one to the third angle of the otherb, viz. the b 26. 1. side AB to the side 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 is

Book I. equal to the angle BCD; therefore the triangle ABC is equal to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. Q. E. D.

c4. 1.

See note.

See the 2d

and 3d fi. gures.j

a 34. 1.

PROP. XXXV. THEOR.

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

A

D

F

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 a of the triangle BDC; and they are therefore equal to one another.

C.

But, if the sides AD, EF, opposite B to the base BC of the parallelograms ABCD, EBCF, be not terminated in the same point; then, because ABCD is a parallelogram, AD is equal a to BC; for the same reason EF is equal to BC; wherefore AD is equal to EF; and DE is common; therefore the whole, or the remainc 2. or 3. der AE, is equal to the whole, or the remainder DF; AB also is equal to DC; and the two EA, AB are therefore equal to the

b 1. Ax.

Ax.

d 19. 1. e 4. 1.

f3. Ax.

d

e

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; the remainders therefore are equal f, that is, the parallelogram ABCD is equal to the parallelogram EBCF. Therefore, parallelograms upon the same base, &c. Q. E. D.

PARALLELOGRAMS upon equal bases, and

between the same parallels, are equal to one another.

Book I.

cause BC is equal to FG, and FG to a EH, BC is equal to EH; a 34. 1. and they are parallels, and joined towards the same parts by the straight lines BE, CH: but straight lines which join equal and parallel straight lines towards the same parts, are themselves equal and parallel b; therefore EB, CH are both equal and pa- b 33. 1. rallel, and EBCH is a parallelogram; and it is equal to ABCD, c 35. 1: because it is upon the same base BC, and between the same parallels BC, AD: for the like reason, the parallelogram EFGH is equal to the same EBCH: therefore also the parallelogram ABCD is equal 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 E

AD, BC: the triangle ABC is equal to the triangle DBC.

Produce AD both ways to the points E, F, and through B draw BE parallel to CA ; and through C draw CF parallel to BD therefore each of the figures EBCA, DBCF

B

A D

C

F

a 31. 1.

is a parallelogram; and EBCA is equal to DBCF, because b 35. 1. they are upon the same base BC, and between the same parallels

BC, EF; and the triangle ABC is the half of the parallelo-
F

c 34. 1.

Book I gram 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 of equal things are d 7. Ax. equal d; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D.

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 is equal to the triangle DEF.

Produce AD both ways to the points G, H, and through B a 31. 1. draw BG parallela to CA, and through F draw FH parallel to

b 36. 1.

ED: then each of
the figures GBCA,
DEFH is a parallel-
ogram; and they
are equal to one
another, becausethey
are upon equal bases
BC, EF, and be-

tween the same pa

c 34. 1. rallels 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 d7. Ax. equal; therefore the triangle ABC is equal to the triangle DEF. Wherefore triangles, &c. Q. E. D.

PROP. XXXIX. THEOR.

EQUAL triangles upon the same base, and upon the same side of it, are between the same parallels.

Let the equal triangles ABC, DBC be upon the same ba BC, and upon the same side of it; they are between the parallels.

Join AD; AD is parallel to BC: for, if it is not, thr a 31. 1. point A drawa AE parallel to BC, and join EC:

E

D

r b 37. 1.

ABC is equal to the triangle EBC, because it is upon the same Book I. base BC, and between the same paral- "A lels BC, AE: but the triangle ABC is equal to the triangle BDC; therefore also the triangle BDC is equal to the triangle EBC, the greater to the less, which is impossible: therefore AE is not parallel to BC. In the same manner, it can be demonstrated that no other line but AD is parallel to BC; AD is therefore parallel to it. Wherefore, equal triangles upon, &c. Q. E. D.

B

PROP. XL. THEOR.

EQUAL triangles upon equal bases, in the same straight line, and towards the same parts, are between the same parallels.

Let the equal triangles ABC, DEF be upon equal bases BC, EF, in the same straight

line BF, and towards the same parts; they are between the same parallels.

B

A

D

C.

E

F

Join AD; AD is parallel to BC; for, if it is not, through A draw a AG parallel to BF, and join GF the triangle ABC is equal b to the triangle GEF, because they are upon equal bases BC, EF, and between the same parallels BF, AG: but the triangle ABC is equal to the triangle DEF; therefore also the triangle DEF is equal to the triangle GEF, the greater to the less, which is impossible: therefore AG is not parallel to BF: and in the same manner it can be demonstrated that there is no other parallel to it but AD; AD is therefore parallel to BF. Wherefore, equal triangles, &c. Q. E. D.

PROP. XLI. THEOR.

IF a parallelogram and triangle be upon the same base, and between the same parallels; the parallelo gram shall be double of the triangle.

a 31. 1.

b 28. 1.