Book I. and between the fame parallels BC, AH: For the like reafon, the parallelogram EFGH is equal to the fame EBCH: Therefore alfo the parallelogram ABCD is equal to EFGH. Wherefore, parallelograms, &c. Q. E. D.

TR

RIANGLES upon the fame bafe, and between the fame parallels, are equal to one another.

E

A D

F

Let the triangles ABC, DBC be upon the fame 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

a 31. 1. through B draw a BE på

rallel to CA; and through

B

b

C draw CF parallel to BD: Therefore, each of the figures b 35. 1. EBCA, DB F is a parallelogram; and EBCA is equal to DBCF, because they are upon the fame base BC, and between the fame parallels BC, EF; and the triangle ABC is the half of the parallelogram EBCA, because the diameter € 34. 1. AB bifects it; and the triangle DBC is the half of the parallelogram DBCF, because the diameter DC bifects it: But d7. Ax the halves of equal things are equal; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D.

TRI

C

RIANGLES upon equal bafes, and between the fame parallels, are equal to one another.

Let the triangles ABC, DEF be upon equal bases BC, EF, and between the fame parallels BF, AD: The triangle ABC is equal to the triangle DEF.

Produce

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

rellel to ED: Then each of the figures GBCA, DEFH is a parallelogram; and they are equal to b

one another, because

they are upon equal

bafes BC, EF, and B

between the fame parallels BF, GH; and the triangle ABC is the half of the parallelogram GBCA, because the 34. 8. diameter AB bifects it; and the triangle DEF is the half of the parallelogram DEFH, because the diameter DF bifects it: But the halves of equal things are equal d'; therefore the d7. Ax. triangle ABC is equal to the triangle DEF. Wherefore triangles, &c. Q. E. D.

E

PRO B. XXXIX. THEOR.

AL triangles upon the same base, and upon the fame fide of it, are between the fame

parallels.

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

Join AD; AD is parallel to BC; for, if it is not, through

the point A draw a AE parallel to BC, and join EC: The a 31. 1. triangle ABC is equal to the triangle EBC, because it is up- b 37. 1. on the fame base BC, and between

the fame parallels BC, AE: But the triangle ABC is equal to the triangle BDC; therefore alfo the triangle. BDC is equal to the triangle EBC, the greater to the lefs, which is impoffible: Therefore AE is not parallel to BC. In the fame manner, it

B

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.

D 4

PROP.

Book I.

2.31. T.

PROP. XL. THEO R.

QUAL triangles upon equal bafes, in the fame ftraight line, and towards the fame parts, are between the fame parallels.

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

ftraight line BF, and to-
wards the fame parts;
they are between the
fame parallels.

B

D

G

C

I

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 b 38. 1. equal b to the triangle GEF, because they are upon equal bafes BC, EF, and between the fame parallels BF, AG: But the triangle ABC is equal to the triangle DEF; therefore alfo the triangle DEF is equal to the triangle GEF, the greater to the lefs, which is impoffible: Therefore AG is not parallel to BF: And in the fame 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 lane parellels; the pa

F a parallelogram and a triangle be upon the fame

rallelogram is double of the triangle.

Let the parallelogram ABCD and the triangle EBC be upon the fame bafe BC, and between the fame parallels BC,

AE;

AE; the parallelogram ABCD is double of the triangle EBC.

Join AC; then the triangle ABC is equal a to the triangle EBC, because they are upon the fame bafe BC, and between the fame parallels BC, AE. But the parallelogram ABCD is double b of the triangle ABC, because the

B

diameter AC divides it into two equal parts; wherefore ABCD is alfo double of the triangle EBC. Therefore, if a parallelogram, &c. Q. E. D.

Book I.

a 37. I.

b 34. I.

PROP. XLII. PRO B.

O defcribe a parallelogram that shall be equal Tto 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 defcribe a parallelogram that fhall be equal to the given triangle ABC, and have one of its angles equal to D.

F G

Bifect a BC in E, join AE, and at the point E in the a 10. I. ftraight line EC make b the angle CEF equal to D; and b 23. 1. through A draw AG parallel to EC, and through G draw © 31. 1. CG parallel to EF: Therefore FECG is a parallelogram: And becaufe BE is equal to EC, the triangle ABE is likwife equal d to the triangle AEČ, fince they are upon equal bafes BE, EC, and between the fame parallels BC, AG; therefore the triangle ABC

B

d 38. r.

is double of the triangle AEC. And the parallelogram FECG is likewife double e of the triangle AEC, because it is up- e 41. 1. on the fame base, and between the fame parallels: Therefore the parallelogram FECG is equal to the triangle ABC, and it has one of its angles CEF equal to the given angle D:

Wherefore

Book I. Wherefore there has been defcribed 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.

a 34. I.

THE

PROP. XLIII. THE OR.

1

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

A

D

K

F

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

:

E

B

Becaufe ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal a to the triangle ADC: And, because EKHA is a parallelogram, and AK its diameter, the triangle AEK is equal to the triangle AHK For the fame reason, the triangle KGC is equal to the triangle KFC. Then, because the triangle AEK is equal to the triangle AHK, and the triangle KGC to the triangle KFC; the triangle AEK, together with the triangle KGC is equal 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 to the remaining complement KD. Wherefore, the complements, &c. Q. E. D.

PROP.