*

is the half of the parallelogram DEFH, because the diameter DF bisects it: but the halves of *7 Ax. equal things are 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 base BC, and upon the same side of it: they shall be between the same parallels.

Join AD; AD shall be parallel to BC. For * 31. 1. if it is not, through the point A draw* AE parallel to BC, and join EC. The triangle ABC is equal to the triangle EBC, be

* 37. 1.

cause they are upon the same base BC, and between the same parallels BC, AE: but the triangle +Hyp. ABC is equal to the triangle DBC; therefore also the triangle

+1 Ax. DBC 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.

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 shall be between the B
same parallels.

CE

Join AD: AD shall be parallel to BC. For, if it is not, through A draw* AG parallel to BF, *31. 1. and join GF. The triangle ABC is equal to the *38. 1. 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 +Hyp. triangle DEF; therefore also the triangle DEF is equal to the triangle GEF, the greater to †1 Ax. 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 a triangle be upon the same base, and between the same parallels; the parallelogram 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 shall be double of the triangle EBC.

*

Join AC: then the triangle ABC is equal to the triangle EBC, because they are upon the same base BC, and between the

B

D F

37.1.

same parallels BC, AE: but the parallelogram ABCD is double of the triangle ABC, because the diameter AC divides it into two equal * *34. 1. parts; wherefore ABCD is also double of the triangle EBC. Therefore, if a parallelogram, &c. Q. E. D.

PROP. XLII. 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

10. 1.

*23.1.

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.

Bisect* BC in E, join AE, and at the point E in the straight line EC make the angle CEF *31. 1. equal to D; and through A draw * AFG parallel to EC, and through C draw CG parallel to EF: therefore FECG is at parallelogram. And because BE is equal to EC, the triangle ABE is

† Def.

34 1.
+ Const.

*

*38. 1. equal to the triangle AEC,

AF G

since they are upon equal BEC
bases BE, EC, and between

the same parallels BC, AG; therefore the triangle ABC is double of the triangle AEC: but *41. 1. the parallelogram FECG is likewise double* of the triangle AEC, because they are upon the same base EC, and between the same parallels EC, AG: therefore the parallelogram FECG is +6 Ax. equal to the triangle ABC; and it has one of +Const. its angles CEF equal† to the given angle D: wherefore a parallelogram FECG has been described equal to the given triangle ABC, having one of its angles CEF equal to the given angle D. Which was to be done.

PROP. XLIII. THEOR.

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

E

A H

K

F

Let ABCD be a parallelogram, of which the diameter is AC; and EH, GF parallelograms about AC, that is, through which AC passes and BK, KD the other parallelograms which make up the whole figure ABCD, which are

B G

therefore called the complements. The complement BK shall be equal to the complement KD.

*

Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the * 31. 1. triangle ADC. Again, because EKHA is a parallelogram, the diameter of which is AK, the triangle AEK is equal† to the triangle AHK; †3⁄41⁄2 1. and for the same reason, the triangle KGC is equal to the triangle KFC. Therefore, 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 was proved equal to the whole ADC; therefore the remaining complement BK is equal to the remaining +3 Ax. complement KD. Wherefore the complements, &c. Q. E. D.

PROP. 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, and 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.

Make the parallelogram BEFG equal to the triangle C, and having

the angle EBG

equal to the angle D, so that BE be

H A

in the same straight line with AB; and produce FG to H; and through A draw* AH parallel *31. 1. to BG or EF, and join HB. Then, because the straight line HF falls upon the parallels AH,

*29. 1. EF, the angles AHF, HFE, are together equal

*

to two right angles; wherefore the angles BHF, HFE are less than two right angles; but straight lines which with another straight line make the interior angles upon the same side less *12 Ax. than two right angles, do meet* if produced far enough; therefore HB, FE shall meet if produced; let them meet in K, and through K draw KL, parallel to EA or FH, and produce HA, GB to the points L, M: Then HLKF is a parallelogram, of which the diameter is HK, and AG, ME are parallelograms about HK; and LB, BF are the complements: therefore *43.1. LB is equal to BF: but BF is equal to the 1 Ax. triangle C; wherefore LB is equal to the trian*15. 1. gle C: and because the angle GBE is equal to + Const. the angle ABM, and likewise † to the angle D; +1 Ax. the angle ABM is equal to the angle D.

+ Const.

*42. 1.

*

*

Therefore, to the straight line AB the parallelogram LB is applied, equal to the triangle C, and having the angle ABM equal to the angle D. Which was to be done.

PROP. XLV. PROB.

To describe a parallergram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle.

Let ABCD be the given rectilineal figure, and E the given rectilineal angle. It is required to describe a parallelogram equal to ABCD, and having an angle equal to E..

Join DB; and describe* the parallelogram FH equal to the triangle ADB, and having the angle FKH equal to the angle E; and to the *44. 1. straight line GH apply* the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E: the figure FKML shall be the parallelogram required.

+ Const.

Because the angle E is equal † to each of the †1 Ax, angles FKH, GHM, the angle FKH is equal† to GHM: add to each of these the angle KHG;