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 the two triangles ABC, CBD have

B

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 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 (2 Ax.) 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 equal to the angle BCD; therefore the triangle ABC is equal (4. 1.) to the triangle BCD, and the diameter BC divides the parallellogram ACDB into two equal parts. Q. E. D.

PROPOSITION 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.-See the 2nd and 3rd figures. If the sides AD, EF of the parallelograms ABCD, EBCF, 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 (6 Ax.) 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;

D E

D

For the same reason EF is equal to BC; wherefore AD is equal (1. Ax.) to EF; and DE is common; therefore the whole, or the remainder AE, is equal (2 or 3 Ax.) 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.

B

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, etc.

Q. E. D.

PROPOSITION 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 is equal to EFGH.

DE

H

Join BE, CH: and because BC is equal (Hyp.) to FG, and FG to (34. 1.) EH, BC is equal to EH (1 Ax.); 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 B EB, CH are both equal and parallel, and EBCH is a (Def. 34. 1.) parallelogram; and it is equal (35. 1.) to ABCD, because it is upon the same base BC, and between the same parallels BC, AD:

C

F

For the like reason, the parallelogram EFGH is equal to the same EBCH:

Therefore also the parallelogram ABCD is equal (1 Ax.) to EFGH. Wherefore parallelograms, etc. Q. E. D.

PROPOSITION 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.

D

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 (Def. 34. 1.) parallelogram; and EBCA is equal (35. 1.) to DBCF, because they are upon the same base BC, and between the same parallels BC, EF;

B

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

PROPOSITION 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.

G

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 (Def. 34. 1.) parallelogram; and they are equal (36. 1.) to one another, because they are upon equal bases BC, EF, and between the same parallels BF, GH;

B

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 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 etc. Q. E. D.

PROPOSITION 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.

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:

B

E

But the triangle ABC is equal (Hyp.) to the triangle BDC, therefore also the triangle BDC is equal (1 Ax.) 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, etc.

Q. E. D.

PROPOSITION 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.

Join AD; AD is parallel to BC: for, if it is not, through A draw (31. 1.) AG parallel to BF, and join GF: the triangle ABC is equal (38. 1.) 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 (Hyp.) to the triangle DEF; therefore also the

E

triangle DEF is equal (1 Ax.) 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, etc.

Q. E. D.

PROPOSITION XLI.

THEOR. If a parallelogram and a triangle be upon the same base, and between the same parallels; the parallelogram shall be double of the triangle.

Let the paralellogram 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.

Join AC; then the triangle ABC is equal (37. 1.) to the triangle EBC, because they are upon the same base BC, and between the same parallels BC, AE.

But the parallelogram ABCD is double (34. 1.) of the triangle ABC, because the diameter AC divides it into two equal parts; wherefore ABCD is also double (1 Ax.) of the triangle EBC. Therefore, if a parallelogram, etc. Q. E. D.

PROPOSITION XLII.

D

E

PROB. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

Let ABC be the given triangle, and D the given rectilineal angle. It is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D.

F

Bisect (10. 1.) BC in E, join AE, and at the point E in the straight line EC make (23. 1.) the angle CEF equal to D; and through A draw (31.1.) AG parallel to EC, and through C draw CG parallel to EF: therefore FECG is a (Def. 34. 1.) parallelogram.

L

B

E

And because BE is equal (Constr.) to EC, the triangle ABE is likewise equal (38. 1.) to the triangle AEC, since they are upon equal bases BE, EC, and between the same parallels BC, AG; therefore the triangle ABC is double of the triangle AEC.

And the parallelogram FECG is likewise double (41. 1.) of the

triangle AEC, because it is upon the same base, and between the same parallels :

Therefore the parallelogram FECG is equal (6 Ax.) to the triangle ABC, and it has one of its angles CEF equal to the given angle D; wherefore there has been described a parallelogram FECG equal to a given triangle ABC, having one of its angles CEF equal to the given angle D. Which was to be done.

PROPOSITION XLIII.

THEOR. The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another.

Let ABCD be a parallelogram, of which the diameter is AC, and EH, FG the parallelograms about AC, that is, through which AC passes, and BK, KD the other parallelograms which make up the whole figure ABCD, which are therefore called the complements. The complement BK is equal to the complement KD.

H

D

Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal (34. 1.) to the triangle ADC: and because EKHA is a parallelogram, the diameter of which is AK, the triangle AEK is equal to the triangle AHK:

For the same reason, the triangle KGC is equal to the triangle KFC:

B

C

K

F

Then, because the triangle AEK is equal to the triangle AHK, and the triangle KGC to KFC; the triangle AEK, together with the triangle KGC, is equal (2 Ax.) to the triangle AHK together with the triangle KFC: but, the whole triangle ABC is equal to the whole ADC; therefore the remaining complement BK is equal (3 Ax.) to the remaining complement KD. Wherefore the complements, etc.

Q. E. D.

PROPOSITION XLIV.

PROB. To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

Let AB be the given straight line, C the given triangle, and D the given rectilineal angle. It is required to apply to the straight line AB a parallelogram equal to the triangle C, and having an angle equal to D.

EBG equal to the angle D, so that BE be in the same straight line with AB, and produce FG to H, and through A draw (31. 1.) AH parallel to BG or EF, and join HB.

Then, because the straight line HF falls upon the parallels AH, EF, the angles AHF, HFE are together equal (29. 1.) to two right angles; wherefore the angles BHF, HFE are less than two right angles; but