parallel to CE, and BD falls upon them, the exterior angle ECD is equal to the interior and opposite angle ABC; but the angle ACE was shown to be equal to the angle BAC; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; to these angles add the angle ACB, and the angles ACD, ACB are equal to the three angles CBA, BAC, ACB ; but the angles ACD, ACB are equal (13. 1.) to two right angles; therefore also the angles CBA, BAC, ACB are equal to two right angles. Wherefore, if a side of a triangle, &c. Q. E. D.

COR. 1. All the interior angles of any rectilineal figure are equal to twice as many right angles as the figure has sides, wanting four right angles.

E

D

For any rectilineal figure ABCDE can be divided into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles. And, by the preceding proposition, all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure; and the same angles are equal to the angles of the figure, together with the angles at the point F, which is the common vertex of the triangles: that is, (2. Cor. 15. 1.) together with four right angles. Therefore, twice as many right angles as the figure has sides, are equal to all the angles of the figure, together with four right angles, that is, the angles of the figure are equal to twice as many right angles as the figure has sides, wanting four.

A

B

COR. 2. All the exterior angles of any rectilineal figure are together equal to four right angles.

Because every interior

angle ABC, with its adja-
cent exterior ABD, is e-
qual (13. 1.) to two right
angles; therefore all the
interior, together with all
the exterior angles of the
figure, are equal to twice
as many right angles as
there are sides of the fi- D
gure; that is, by the fore-

going corollary, they are equal to all the interior

angles of the figure, to

gether with four right angles; therefore all the exterior angles are equal to four right angles.

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 are also equal and parallel.

C

B

D

Join BC; and because AB is pa- A rallel to CD, and BC meets them, the alternate angles ABC, BCD are equal (29. 1.); 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; and the angle ABC is equal to the angle BCD; therefore the base AC is equal (4. 1.) to the base BD, and the triangle ABC to the triangle BCD, and the other angles to the other angles (4. 1.) 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 (27.1.) to BD; and it was shown to be equal to it. Therefore, straight lines, &c. Q. E.D.

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 are equal to one another; and the diameter BC bisects it.

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

A

C

B

D

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 the side BC, which is adjacent to these 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 (26. 1.), viz. the side AB to the side CD, and AC to BD, and the angle le BAĆ 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 a parallelogram are equal to one another; also, its diameter bisects it; for AB being equal to CD, and BC common, the two AB, BC are equal to the two DC, CB, each to each; now the angle ABC is equal to the angle BCD; therefore the triangle ABC is equal (4.1.) to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. Therefore, &c. Q. E.'D.

PROP. XXXV. THEOR.

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

(SEE THE 20 AND 3d FIGURES.)

Let the parallelograms ABCD, EBCF be upon the same base BC, and between the same parallels AF, BC; the parallelogram ABCD is 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 (34. 1.) of the triangle BDC; and they are therefore equal to one another.

A

B

F

But, if the sides AD, EF, opposite 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 (34. 1.) to BC; for the same reason EF is equal to BC; wherefore AD is equal (1. Ax.) to EF; and DE is common; therefore the whale, or the remainder, AE is equal (2. or 3. Ax.) to the whole, or the remainder DF; now AB is also equal to DC; therefore the two EA, AB are equal to the two

FD, DC, each to each; but the exterior angle FDC is equal (29. 1.) to the interior EAB, wherefore the base EB is equal to the base FC,

and the triangle EAB (4. 1.) to the triangle FDC. Take the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB; the remainders will then be equal (3. Ax.), 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.

A

DE

H

Let ABCD, EFGH be parallelograms upon equal bases BC, FG and between the same parallels AH, BG; the parallelogram ABCD is equal to EFGH.

C

F

G

Join BE, CH; and because BC is equal to FG, and FG to (34. 1.) EH.BC is equal to EH; and they B 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 (33.1.); therefore EB, CH are both equal and parallel, and EBCH is a parallelogram; and it is equal (35. 1.) to ABCD, because it is upon the same base BC, and between the same parallels BG, AH; 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 AD, BC: The triangle ABC, is equal to the triangle DBC.

Produce AD both ways to the points E, F, and through B draw 731. 1.) BE parallel to CA; and through C draw CF parallel to BD: Therefore, each of the figures EBCA,DBCF is a paral

E

B

A D

F

lelogram; and DBCA is equal (35. 1.) to DBCF, because they are upon the same base BC, and between the same parallels BC, EF;

but the triangle ABC is the half of the parallelogram EBCA, because the diameter AB bisects (34. 1.) it; and the triangle DBC is the half of the parallelogram DBCF, because the diameter DC bisects it; and the halves of equal things are equal. (7. Ax.); 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.

A

D

H

Produce AD both ways to the points G, H, and through B draw BG parallel (31. 1.) to CA, and through F draw FH parallel to ED: Then each of the figures GBCA, DEFH is a parallelogram; and they are equal to (36. 1.) one another, because they are upon equal bases BC, EF and between the same parallels BF, GH; and the triangle ABC is the half

B

C

F

(34. 1.) of the parallelogram GBCA, because the diameter AB bisects it; and the triangle DEF is the half (34. 1.) of the parallelogram DEFH, because the diameter DF bisects it; But the halves of equal things are equal (7. Ax); 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 base BC, and upon the same side of it; they are between the same parallels.

D

E

Join AD; AD is parallel to BC; for, if it is not, through the point A draw (31. 1.) AE parallel to BC, and join EC; The triangle ABC, is equal (37. 1.) to the triangle EBC, because it is upon the same base BC, and between the same parallels BC, AE; But the triangle ABC is equal to the A 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 may 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