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Book I.

I.

Let the parallelogram ABCD and the triangle EBC be upori the fame base BC, and between the fame parallels BC, AE; the parallelogram ABCD is double of the tri

angle EBC.

Join AC; then the triangle ABC is

a

a. 37. 1. equal to the triangle EBC, because they are upon the fame bafe BC, and between the fame parallels BC, AE.

b. 34.. but the parallelogram ABCD is double b of the triangle ABC, because the diameter AC divides it into two equal

a. 1o. f.

A

B

DE

parts; wherefore ABCD is also double of the triangle EBC. there= fore if a parallelogram, &c. Q. E. D.

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PROP. XLII. PROB.

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

c

Bifect BC in E, join AE, and at the point E in the straight b. 23. 1. line EC make the angle CEF equal to D; and thro' A draw AG parallel to EC, and thro' C draw c

C. 31. I.

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A F

G

N

BEC

therefore the triangle ABC is dou- B

D

ble of the triangle AEC. and the parallelogram FECG is likewife e. 41. 1. double of the triangle AEC, because it is upon 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 there has been described a paralle→

logram

logram FECG equal to a given triangle ABC, having one of its an- Book I. gles 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.

A H

D

K

E

F

Let ABCD be a parallelogram, of which the diameter is AC, and EH, FG the parallelograms about AC, that is, thro' which AC paes, and BK, KD the other parallelograms which make up the whole figure ABCD, which are therefore called the complements. the

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the complement KD.

C

Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ADC. and because EKHA is a a. 34. r. parallelogram, the diameter of which is AK, the triangle AEK is equal to the triangle AHK, by the fame reafon, 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.

To a

PROP. XLIV. PROB.

O 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 straight line, and C the given triangle, and D the given rectilineal angle. It is required to apply to the ftraight line AB a parallelogram equal to the triangle C, and having an angle equal to D.

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fame Straight line with AB, and produce FG to H; and thro' Ą 6. 31. 1. draw b AH parallel to BG'or EF, and join HB. then because the

traight line HF falls upon the parallels AH, EF, the angles AHF, c. 29. HFE are together equal to two right angles; wherefore the angles

BHF, HFE are leffer than two right angles. but ftraight lines which with an other ftraight line make the interior angles upon the fame d. 1. Ax. file lefs than two right angles, do meet if produced far enough. therefore HB, FE fhall meet, if produced; let them meet in K, and thro' 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 ¢ 43. 1. 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 15. I. C. and because the angle 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.

3. 42. I.

Tod

f

PROP. XLV. PROB.

e

defcribe a parallelogram 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 defcribe a parallelogram equal to ABCD and having an angle equal to E.

Join DB, and defcribe the parallelogram FH equal to the triangle ADB, and having the angle HKF equal to the angle E; and to the b. 44. 1. fuaight line GH apply the parallelogram GM equal to the triangle

b

DBC

DBC having the angle GHM equal to the angle E. and because the Book I. angle E is equal to each of the angles FKH, GHM, the angle FKH is equal to GHM; add to each of these the angle KHG; therefore

the angles FKH,

KHG are equal to the angles KHG, GHM. but FKH, KHG are equal to two right angles;

therefore

alfo KHG, GHM

are equal to two

C

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CKH

M

C. 29. I.

d. 14. 5.

right angles. and because at the point H in the straight line GH, the two straight lines KH, HM upon the oppofite fides of it make the adjacent angles equal to two right angles, KH is in the fame ftraight line with HM. and because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal; add to each of thefe the angle HGL; therefore the angles MHG, HGL are equal to the angles HGF, HGL. but the angles MHG, HGL are equal to two right angles; wherefore alfo the angles HGF, HGL are equal to two right angles, and FG is therefore in the same straight line with GL. and because KF is parallel to HG, and HG to ML; KF is parallel to ML. and KM, FL are paral- e. 30. s. lels; wherefore KFLM is a parallelogram. and because the triangle ABD is equal to the parallelogram HF, and the triangle DBC to the parallelogram GM; the whole rectilineal figure ABCD is equal to the whole parallelogram KFLM. therefore the parallelogram KFLM has been defcribed equal to the given rectilineal figure ABCD, having the angle FKM equal to the given angle E. Which was to be done.

e

COR. From this it is manifest how to a given straight line to apply a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure, viz. by applying to the given straight line, a parallelogram equal to the b. 44. x. firft triangle ABD, and having an angle equal to the given angle.

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PROP. XLVI. PROB.

TO describe a square upon a given straight line.

Let AB be the given straight line; it is required to describe a fquare upon AB.

a

From the point A draw 2 AC at right angles to AB; and make b AD equal to AB, and thro' the point D draw DE parallel to it, and thro' B draw BE parallel to AD, therefore ADEB is a parallelogram; whence AB is equal to DE, and AD to BĘ, but BA is equal to AD; therefore the four

d

ftraight lines BA, AD, DE, EB are e-
qual to one another, and the parallelo-
gram ADEB is equilateral. likewife all D
its angles are right angles; because the
ftraight line AD meeting the parallels
AB, DE, the angles BAD, ADE are ẹ-

e

A

E

B

29. 1. qual to two right angles; but BAD is a right angle, therefore alfo ADE is a right angle. but the oppofite angles of parallelograms are equal; therefore each of the oppofite angics ABE, BED is a right angle; wherefore the figure ADEB is rectangular. and it has been demonftrated that it is equilateral; it is therefore a fquare, and it is described upon the given straight line AB. Which was to be done.

46. I,

COR. Hence every parallelogram that has one right angle has all its angles right angles.

PROP. XLVII. THEOR.

IN any right angled triangle, the fquare which is defcribed upon the fide fubtending the right angle, is equal to the fquares defcribed upon the fides which contain the right angle.

Let ABC be a right angled triangle having the right angle BAC; the fquare defcribed upon the fide BC, is equal to the fquares defcribed upon BA, AC.

a

On BC defcribe the fquare BDEC, and on BA, AC the squares

GB,

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