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20r3 Ax. +34. 1.

nated in the same point; then, because ABCD is a parallelogram, AD is equal to BC; for the same 34. 1. reason EF is equal to BC; wherefore AD is equal to * 1 Ax. EF; and DE is common; therefore the whole, or the remainder, AE is equal to the whole, or the remainder DF: AB also is equal+ to DC; therefore the two EA, AB are equal to the two FD, DC, each to each; and the exterior angle FDC is equal to the interior EAB: therefore the base EB is equal to the base FC, and the triangle EAB equal* to the triangle FDC. Take

[graphic]

B

DE

B

EDF

29.1.

4. 1.

the triangle FDC from the trapezium ABCF, and from the same trapezium take the triangle EAB, and the remainders are equal; that is, the parallelogram *3 Ax. ABCD is equal to the parallelogram EBCF. Therefore parallelograms upon the same base, &c. Q. E. D.

PROP. 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 shall be equal to EFGH.

E

H

+1 Ax.

33.1.

Join BE, CH; and because BC, is equal+ to FG, and + Hyp. FG to* EH, BC is equal to+ EH; and they are+ paral-34.1. lels, and joined towards the same parts by the straight + Hyp. lines BE, CH: but straight lines which join the extremities of equal and parallel straight lines towards the same parts, are themselves* equal and parallel; therefore EB, HC, are both equal and parallel; and therefore EBCH is at parallelogram; and it is equal* to ABCD, because they are upon the same base BC, and between the same parallels BC, AH: for the like reason, the parallelogram EFGH is equal to the same EBCH: therefore the parallelogram ABCD is equal+ +1 Ax to EFGH. Wherefore parallelograms, &c.

[graphic]

PROP. XXXVII. THEOR.

Q. E. D.

Triangles upon the same base, and between the same parallels, are equal to one another.

1.

Def. 34.

35. 1.

+ 2 Post.

* 31. 1.

+ Def. 34.

1.

35. 1.

* 34. 1.

7 Ax.

+ 2 Post.

* 31.1.

Let the triangles ABC, DBC, be upon the same base BC, and between the same

parallels AD, BC: the triangle E
ABC shall be equal to the triangle
DBC.

Produce+ AD both ways to the
points E, F, and through B draw*

BE parallel to CA; and through C draw CF parallel to BD: therefore each of the figures EBCA, DBCF is at parallelogram: and EBCA is equal 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* 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; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D.

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PROP. 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 shall be equal to the triangle DEF. Produce+ AD both ways to the points G, H, and through B draw BG parallel to CA, and through F draw FH parallel to ED: then each of the figures + Def. 34. GBCA, DEFH, is at parallelogram: and they are equal one another, because they are upon equal bases BC, EF, and between the same parallels BF, GH; and the triangle ABC is the half of the parallelogram GBCA,

1.

* 36. 1.

* 34. 1.

* 7 Ax.

to

C

B

CE

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

B

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 same parallels.

*

31. 1.

* 37. 1.

[graphic]

↑ Hyp.

† 1 A*.

Join AD; AD shall be parallel to BC. For, if it is not, through A draw* AG parallel to BF, and join 31. 1. GF. The triangle ABC is equal to the triangle * 38.1. GEF, because they are upon equal bases BC, EF, and between the same parallels BF, AG: but the triangle ABC is equal to the triangle DEF; therefore also the + Hyp. triangle DEF is equal to the triangle GEF, the greater +1 Ax 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.

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

D

* 37.1.

* 34. 1.

* 10. 1.

* 23. 1. *31.1. + Def. 34.

1.

† Constr.

* 38. 1.

* 41. 1.

† 6 Ax.
t Constr.

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.

*

DE

Join AC: then the triangle ABC is equal 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 of the triangle ABC, because the diameter AC divides it into two equal 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 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 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 equal 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, B E of the triangle AEC: but the parallel

*

A T

ogram 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 equal † to the triangle ABC; and it has one of 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.

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

E

A H

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 thewhole figure ABCD, which are therefore called the complements, BK shall be equal to the complement KD.

BG

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The complement

*

Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle * 34. 1. ADC. Again, because EKHA is a parallelogram, the diameter of which is AK, the triangle AEK is equal to the triangle AHK; and for the same reason, † 34.1. 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 to the triangle AHK together with the triangle +2 Ax. KFC: but the whole triangle ABC was proved equal to the whole ADC; therefore the remaining complement BK, is equal + to the remaining complement KD. +3 Ax. Wherefore the complements, &c. Q. E. D.

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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

F E

K

42. 1.

C, and having the angle EBG

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equal to the angle D, so that C

B

BE be in the same straight

H

line with AB; and produce

FG to H; and through A draw* AH parallel to BG 31. 1. or EF, and join HB. Then, because the straight line

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