D; the exterior angle ACD is equal to the two interior an opposite angles CAB ABC, and the three interior angles of the triangle, viz. ABC, BCA, CAB, are together equal to two right angles.

Through the point C draw CE parallel (31. 1.) to the straight line AB; and because AB is parallel to CE, and AC meets them, the alternate angles BAC, ACE are equal (29. 1.)

B

E

Again, because AB is 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 shewn to be equal to the angie BAC; therefore the whole exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; to these equals 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, together with four right angles, are equal to twice as many right angles as the figure has sides.

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 all the angles of the figure, together with four right angles, are equal to twice as many right angies as the figure has sides.

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

A

Because every interior angle ABC, with its adjacent exterior ABD, is equal (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 figure; that is, by the foregoing D corollary, they are equal to all the interior angles figure, together with four right angles; therefore all the exterior angles are equal to four right angles.

Proposition XXXIII. Theorem.

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

D

Join BC; and because AB is parallel 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 wo 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. J.) 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 shewn to be equal to it. fore, straight lines, &c.

Q. E. D.

There

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 ABCD 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

B

angles ACB, CBD are equal (29. 1.) to one 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 other, (26. 1.) viz. the side (26 1.) AB to the side CD, and AC to BD, nd 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 ngle ACD: And the angle BAC has been shewn to be equal to the ingle BDC therefore the opposite sides and angles of parallelograins 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 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. Q. E. D.

Proposition XXXV. Theorem.

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 (34. 1.) of the triangle BDC; and they are therefore equal to 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 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 coinmon; therefore the whole, or the remainder, AE is equal (2. or 3. Ar.) to the whole, or the remainder DF; AB also is equal to DC; and the two EA, AB are therefore equal to the

two FD, DC, each to each; and the exterior angle FDC is equal (29. 1.) to the interior EAB, therefore the base EB is equal to the base FC, and the triangle EAB equal (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 therefore are equal, (3. Ax.) that is, the parallelogram ABCD is equal to the parallelogram EBCF. Therefore parallelograms upon the same base, &c. Q. E.D.

Proposition XXXVI. Theorem.

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

Join BE, CH; and because BC is equal to FG, and FG to

(34. 1.) EH, BC is equal to EH; 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, (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 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.

Proposition. XXXVII. Theorem.

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.

E

B

C

F

Produce AD both ways to the points E, F, and through B draw (31.1.) BE parallel to CA; and through C draw CF parallel to BD: Therefore each of the figures EBCA, DBCF is a parallelogram; and EBCA is equal (35. 1.) 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 (34. 1.)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 equal; (7. Ax.) therefore, the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D.

Proposition XXXVIII. Theorem.

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 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.)

G

D

H

AN

B

one another, because they are upon equal bases BC, EF, and are between the same parallels BF, GH; and the triangle ABC is the half (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. Ar.) therefore the triangle ABC is equal to the trianglo DEF. therefore triangles, &c. Q. E. D.

f 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. 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 triangle BDC; therefore also the triangle BDC, is equal to the triangle EBC, the greater to the less, which is impossible: There

E

fore 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, &c. Q. E. D.

Proposition XL. Theorem..

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.

Join AD; AD is parallel to BC: For, if it is not; throughA draw(31.1.) AG parallel to BF, and join GF: The

B

triangle ABC is equal (37. 1.) to the triangle GEF, because they are upon equal bases BC, EF, and between the same parallels BF, ÁG: 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.

Proposition XLI. Theorem.

If a parallelogram and triangle be upon the same base, and between the same parallels; the parallelogiam shall be double of the triangle. Let the parallelogram ABCD and the triangle EBC be upon the same base BC, and between the same parallels BC, AE; the parallelogram ABCD is double of the triangle EBC