Page images
PDF
EPUB

€ 34. 1.

с

Book I 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 equald; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D.

7 Ax.

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 is equal to the triangle DEF.

Produce AD both ways to the points G, H, and through * 31. 1. B draw BG parallela to CA, and through F draw FH parallel to ED: Then G

each of the figures
GBCA, DEFH, is
a parallelogram;
and they are equal

36. 1. tob one another, be

H

[blocks in formation]

EF, and between the same parallels BF, GH; and the 34. 1. triangle ABC is the half 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 7 Ax. equald; 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 same side of it, are between the same parallels.

the

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 31. 1. the point A drawa AE parallel to BC, and join EC: The

triangle ABC is equal to the triangle EBC, because it is upon the same base BC, and between the same parallels BC, AE: But the triangle ABC is equal to the triangle BDC, therefore also the triangle BDC; is equal to the triangle EBC, the greater to the less, which is impos

E

sible: 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. È. 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 to

wards the same parts; they are between the same parallels.

Join AD; AD is parallel to BC: For, if it

is not, through A drawa

B

D

b

Book 1.

37. 1.

51. 1.

AG parallel to BF, and join GF: The triangle ABC is equal to the triangle GEF, because they are upon equal 38. 1. bases BC, EF, and between the same parallels BF, AG: But the triangle ABC is equal to the triangle DEF; there fore also the triangle DEF is equal 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, &c. Q. E. D.

PROP. XLI. THEOR.

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

BOOK I.

Let the parallelogram ABCD and the triangle EBC be upon the same base BC, and between the same parallels BC, AE; the parallelogram ABCD A

is double of the triangle EBC.

Join AC; then the triangle ABC

a 37. 1. is equal to the triangle EBC, because they are upon the same base BC, and between the same parallels BC, AE. But the parallelogram b 34. 1. ABCD is doubleb of the triangle ABC, because the diameter AC di

B

DE

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

a

10. 1.

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.

Bisecta BC in E, join AE, and at the point E in the 23. 1. straight line EC makeb the angle CEF equal to D; and 31. 1. through A drawe AG parallel

A F

G

to EC, and through C draw CGC parallel to EF: Therefore FECG is a parallelogram: And because BE is equal to EC, the triangle 38. 1. ABE is likewise equal to the triangle-AEC, since they are upon equal bases BE, EC, B E and between the same paral

lels BC, AG; therefore the triangle ABC is double of the triangle AEC. And the parallelogram FECG is likewise 41. 1. double of the triangle AEC, because it is upon the same base, and between the same 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 described a parallelogram FECG BOOK L equal to a 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.

H

[ocr errors]

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 parallelo- E grams which make up the whole figure ABCD, which are therefore called the complements. The complement BK is equal to the complement KD.

B

Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ADC: And, a 54. 1. because EKHA is a parallelogram, the diameter of which is AK, the triangle AEK is equal to the triangle AHK : By the same reason, the triangle KGC is equal to the triangle KFC: Then, because the triangle AÈK 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.

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.

[blocks in formation]

12 Ax.

BE be in the same straight line with AB, and produce FG 31. 1. to H; and through A drawb 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 29. 1. 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 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 the parallelograms about HK; and LB, BF are the complements: therefore LB is 43. 1. equale to BF; But BF is equal to the triangle C; where

fore LB is equal to the triangle C; and because the angle £15. 1. GBE is equal to the angle ABM, and likewise 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.

PROP. XLV. PROB.

See N. To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle.

42. 1.

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 HKF equal to the

« PreviousContinue »