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Book I.

See N.

TRA

PROP. XXXVII. THEOR.

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

B

A D

C

F

Let the triangles ABC, DBC be upon the fame base BC, and between the fame parallels E AD, BC: The triangle ABC is equal to the triangle DBC. Produce AD both ways to the points E, F, and make EA, DF each of them equal to BC, and join BE, CF; and becaufe EA is equal and parallel to BC, the ftraight lines BE, CA which join them are equal and parallel, and therefore EBCA is a parallelogram: For the fame reafon, DBCF is a parallelogram: and EBCA is equal to DBCF, because they are upon the fame bafe BC, and between the fame parallels BC, EF; and the triangle ABC is the half of the parallelogram EBCA, because the diameter AB 34. bifects it; and the triangle DBC is the half of the parallelogram DBCF, because the diameter DC bifects it: But the 7. Ax. halves of equal things are equal; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D.

a 33. 1.

I.

b35. 1.

d

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b

TR

PROP. XXXVIII. THEOR.

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

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

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Produce AD both ways to the points G, H, and make AG equal to BC, and DH to EF, and join BG, FH; and it may be demonftrated, as in the preceding, G

a

that each of the figures GBCA, DEFH is a parallelogram; and 36. 1. they are equal to one another, because they are upon equal bafes BC, EF, and between the fame parallels BF, GH: and

34. 1. the triangle ABC is the half

B

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of the parallelogram GBCA, because the diameter AB bifects it; and the triangle DEF is the half of the parallelogram

DEFH,

DEFH, because the diameter DF bifects it: But the halves Book I. of equal things are equal; therefore the triangle ABC is equal to the triangle DEF. Wherefore triangles, &c. Q. E. D.

C 7. Ax.

PROP. XXXIX. THEOR.

E

QUAL triangles upon the fame bafe, and upon See N. the fame fide of it, are between the fame

parallels.

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

d

Join AD; AD is parallel to BC; for, if it is not, through the point A draw a AE parallel to BC, and because AB meets a 31. 1, the parallels BC, AE, the interior angles CBA, BAE are equal b to two right angles; therefore the angles DBA, BAE are lefs than two right angles; and because AB meets BD, AE, and makes the interior angles DBA, BAE together less than two right angles, BD, AE fhall meet one another; let them meet c*12. Ax, in E, and join EC: The triangle ABC is equal to the triangle d 37.1. EBC, because it is upon the fame base BC, and between the fame parallels BC, AE: But the triangle ABC is equal to the triangle BDC; therefore allo 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 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.

A

B

E

D

C

COR. Hence it is manifeft, that the ftraight line, which meets another ftraight line, fhall alfo, if produced, meet any straight line parallel to that other.

E

PROP. XL. THEOR.

QUAL triangles upon equal bafes, in the fame See N. 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 towards the fame parts; they are between the fame parallels.

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

A draw AG parallel to BF; and because ED meets BF, it a 31. 1.

fhall

BOOK I. fhall :1fo meet b AG, which is parallel to BF; let it meet it

bCor.39.1. in G, and join CF: The triangle ABC is equal to the triangle GEF, because they are

c38. 1. upon equal bafes BC, EF, and between the fame parallels BF, AG: But the triangle ABC is equal to the triangle DEF;

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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 same manner, it can be demonftrated, that there is no other parallel to it but AD; AD is therefore parallel to BE. Wherefore equal triangles, &c. Q. E. D.

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PROP. XLI. THEOR.

F a parallelogram and a triangle be upon the fame bafe, and between the fame parallels; the parallelogram fhall be 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; the parallelogram ABCD is double of the triangle EBC. Join AC; then the triangle ABC A

a 37. 1. is equal a to the triangle EBC, be

cause they are upon the fame base
BC, and between the fame parallels
BC, AE. But the parallelogram

b34. 1. ABCD is doubled of the triangle
ABC, becaufe the diameter AC di-
vides it into two equal parts; where-

2.10. I.

B

D

I

C

fore ABCD is alfo double of the triangle EBC. Therefore, if a parallelogram, &c. Q. E. D.

PROP. XLII. PROB.

To defcribe 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.

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Bifect BC in E, join AE, and at the point E in the straight

b

d

A

F

G

b 23. 1.

C 31. I. dCor.39.1.

line EC make the angle CEF equal to D, and through C draw Book I. CG parallel to EF, and through A draw AG parallel to BC, meeting EF, CG in F, G: Therefore FECG is a parallelogram: And becaufe BE is equal to EC, the triangle ABE is likewife equal to the triangle AEC, fince they are upon equal bafes BE, EC, and between the fame parallels BC, AG; B therefore the triangle ABC is double of

e

D

e 38. I.

E

C

the triangle AEC: And the parallelogram FECG is likewife double f of the triangle AEC, becaufe it is upon the fame bafe, f 41. 1. 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 there has been defcribed a parallelogram FECG equal to a given triangle ABC, having one of its augles CEF equal to the given angle D. Which was to be done.

THE

PROP. XLIII. THEOR.

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

A

H

E

K

F

Let ABCD be a parallelogram, of which the diameter is AC, and EH, FG the parallelograms about AC, that is, through which AC paffes, 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.

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Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle a 34. i. ADC: And, because EKHA is a parallelogram, the diameter of which is AK, 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 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.

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BOOK I.

a 42. I.

T

PROP. XLIV. PROB.

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

Let AB be the given ftraight 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 paral

lelogram BEFG equal
to the triangle C, and
having the angle EBG
equal to the angle D,
fo that BE be in the

fame ftraight line with

AB, and produce FE,

FG to K, H; and

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e

c

b 31. 1. through A draw ↳ AH parallel to BG or EF, meeting FH in cCor.39.1. H; and join HB; it fhall, if produced, meet FK, which is parallel to GB; 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 the parallelograms about HK; and LB, BF are the complements; therefore LB is equal to BF: But BF is equal to the triangle C; wherefore LB is equal to the triangle Ĉ: And because the angle GBE is equal to the angle ABM, and likewife to the angle D; the angle ABM is equal to the angle D: Therefore the parallelogram LB is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to the angle D. Which was to be done.

e 43. 1.

f 15. 1.

T

PROP. XLV. PROB.

See N. O defcribe a parallelogram equal to a given recti-. lineal figure, and having an angle equal to a given rectilineal angle.

a 42..

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

Join DB, and defcribe a the parallelogram FH equal to the triangle ADB, and having the angle HKF equal to the angle b 44. 1. E; and produce KH to M; and to the ftraight line GH apply

the

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