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Book 1. Let the parallelogram ABCD and the triangle EBC be upon m the fame base BC, and between the fame parallels BC, AE ;

the parallelogram ABCD is double of
the triangle EBC.

DE Join AC; then the triangle ABC a 37. I. is equal to the triangle EBC, be

cause they are upon the same base
BC, and betw.en the same parallels
BC, AE.

But the parallelogram b 34. I. ABCD is double b of the triangle

ABC, because the diameter AC di-
vides it into two equal parts; where- B
fore ABCD is alto double of the tri.
angle EBC. Therefore, if a parallelogram, &c. Q. E. D.

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a 1o. 1. 1 23 I. C 3I. I.

TO

"O describe a parallelogram that shall be equal to a

given triangle, and have one of its anglos equal to a given rectilineal angle,

Let ABC be the given triangle, and D the given rectilineal angle It is required to defcr be a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D.

Bila • BC in E, join AE, and at the point E in the straight
line EC makt :hr-angle CEF equal to D; and through A draw
CAG parallel to LC, and through
C draw CGS parallel to EF :

AF G
Therefore FECG is a parallelo-
gram: , And because BE is equal
to EC, the triangle ABE is like.
wile equal to the triangle AEC,
fince they are upon equal bases

D
BE, EC, and tetween the time
paralels LC, AG; thi sefore the
triangle ABC is roub.e of the B E C С
triangle AEC : Aid the para!
lelogic m PECG is tokrw.le double e of the triangle AEC, be
c?uie it is upon the fame base, and between the same pa.
Falles : The elore the parallelogram FECG is equal to the
trungle ABC, and it has one of its angles CEF equal to the
given angle D: Wherefore there has been described a paralle-

logram

d 38. 1.

€ 41. I.

logram FECG equal to a given triangle ABC, having one of Book I. its angles CEF equal to the given angle D. Which was to be done.

PROP XLIII. THE O R. "HE complements of the parallelograms which are

about the diameter of any parallelogram, are equal to one another

Lek ABCD be a parallelogram, of which the diameter is AC, and EH, FG the paral

A H

D lelograms about AC, that is, through which AC pafes, and

K BK, KD che other parallelo. E

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

B G

C Because ABCD is a paral. lelogram, and AC its diameter, the triangle ABC is equal a to a 34. I. the triangle ADC : And, because EKH A is a parallelogram,

the diameter of which is AK, the triangle AEK is equal to the tri. angle AHK: By the same reason, the triangle KGC is equal to the criangle KFC: Chen, because the triangle AEK is equal to the triangle AHK, and the triangle KOC 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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a given straight line to apply a parallelogram,

which thall 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. le is required to apply to the straight line AB a parallelogram equal to the triangle C, and having an angle cqual to D.

Make

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Book t.
Make the

FE
parallelogram

K a 42. I.

BEFG equal
to the triangle
C, and having

D
theangleEBG

G

M equal to the

B В
angleD, fothat

C
BE be in the
fame straight

Η Α L line with AB, b 31. 1. and produce FG to H; and thro' A draw 6 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 toge C 29. . ther equal to two right angles; wherefore the angles BHF,

HFE are lesser than two right angles : But straight lines which

with another straight line make the interior angles upon the d 13. An. same fide 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 paraljelogram, of which the diameter

is HK, and AG, ME are the parallelograms about HK ; and LB, BF are the complements; 43. 1.

therefore LB is equal to BF: But BF is equal to the triangle C

wherefore LB is equal to the triangleC: And because the angle f 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.

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O describe a parallelogram equal to a given re&ili

neal figure, and having an angle equal to a given tectilineal angle.

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

Join DB, and describe a the parallelogram FH equal to the triangle ADB, and having the angle HKF equal to the angle E; and to the straight line GH apply the parallelogram GM equal

2 42. 1.

b 44. I.

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angles A

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

F G L KHG GHM ; but FKH, KHG

E are equal to two

c 29. I. right angles ; therefore allo KHG, GHM are equal to two right angles; and

B CK HM because at the point H in the straight line GH, the two straight lines KH, HM, upon the opposte sides of it make the adjacent angles equal to two right angles, KH is in the same straight line d with HM;d 14. I. and because the straight line HG meets the parallels KM, FG, the alternate angles MHG, HGF are equal; Add to each of these, 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 also the angles HGF, HGL are equal to two right angles, and FG is therefore in the fame Itraight line with GL: And because KF is parallel to HG, and HG to ML ; KF is parallel to ML: And KM, FL - 30. t. are parallels; 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 described equal to the given rectilineal figure ABCD, having the angle FKM equal to the given angle E. Which was to be done.

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 Itraight line, a parallelo- b 44. 1. gram equal to the first triangle ABD, and having an angle co qual to the given angle.

PROP.

Book Ti

PRO P. XLVI.

PRO B.

To describe a square upon a given straight line.

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

From the point A draw a AC at right angles to AB; and makeb AD equal to AB, and through the point D draw DE parallel to AB, and through B draw BE parallel to AD; therefore ADEB is a parallelogram; whence AB is equal to DE, and AD to BE: But BA is equal to AD; C therefore the four straight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEBD

E is equilateral, likewise all its angles are right angles; because the straight line AD meeting the parallels AB, DE, the angles BAD, ADE are e. qual to two right angles ; but BAD is a right angle; therefore also ADE is a right angle; but the opposite A

B angles of parallelograms are equal d ; therefore each of the opposite angles ABE, BED is a right angle; wherefore the figure ADEB is rectangular, and it has been demonstrated that it is equilateral; it is therefore a square, and it is described upon the given straight line AB : Which was to be done.

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

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In any right angled triangle, the square which

is de scribed upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.

Let ABC be a right angled triangle having the right angle
SAC; the square described upon the fide BC is equal to the
squares described upon BA, AC.
On BC describe the square BDEC,and onBA, AC the squares

GB,

a 46. 1.

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