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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 is double of the tri

A D E angle EBC.

Join AC; then the triangle ABC is 2. 37. 1. equal * to the triangle EBC, because

they are upon the same base BC, and

between the fame parallels BC, AE. b. 34. 1. but the parallelogram ABCD is double b

of the triangle ABC, because the dia-
meter AC divides it into two equal

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

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To describe a parallelogram that shall be equal to a

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

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Let ABC be the given triangle, and 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.

Biftet * BC in E, join AE, and at the point E in the straight b. 23. 1. line EC make b the angle CEF equal to D; and thro' A draw AG 6. 31.5. parallel to EC, and thro' C draw c

AF G
CG parallel to EF. therefore
FECG is a parallelogram. and be-

cause BE is equal to EC, the trid. 38. 1. angle ABE is likewise equal to the triangle AEC, since they are

D
upon equal bases BE, EC and be-
tween the same parallels BC, AG;
therefore the triangle ABC is dou: B E

ble of the triangle AEC. and the parallelogram FECG is likewise 6. 41. 1. double e 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 10 the given angle D, wherefore there has been described a paralle

logram

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

Let ABCD be a parallelogram, of which the diameter is AC, and EH, FG the parallelo

A H

D grams about AC, that is, thro' which AC palles, and BK, KD

K the other parallelograms which

E

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

B the complement KD.

G Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ADC. and 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 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.

a. 34. ri

PROP. XLIV. PROB. Тоа

O 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 De C 3

Make

M

Bock I. Make a the

F

K parallelogram a. 42. 1. BEFG equal to the triangle C

D and having the

G angle EBG e

B
qual to the an-

C
gle D, so that
BE be in the

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

itraight line HF falls upon the parallels AH, EF, the angles AHF, ç. 29. 1. HFE are together equal to two right angles; wherefore the angles

BAF, HFE are lesser than two right angles. but straight lines which

with an other straight line make the interior angles upon the fame d. 12. Ax. fille less than two right angles, do meet if produced far enough.

therefore HB, FE Mall 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 dia

meter is HK, and AG, ME are the parallelograms about HK; and ç 43. . L, 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 15.3. C. and because the angle GBE is equal f 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.
O describe a parallelogram equal to a given rectili-

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

Let ABCD be the given recilineal figure, and E the given rectie lineal angle. It is required to describe a parallelogram equal to ABCD and having an angle ecual to E.

Join LB, and cccribe the parallelogram FH equal to the triangle

ADB, and having the angle HET equal to the angle E; and to the b. 44. 1. fuzigli linc CH arriyb the parallelogram GM equal to the triangle

DBC

3. 42. I

А

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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,

D

F GL
KHG are equal to
the angles KHG,
GHM. but FKH,

E
KHG are equal
to two right an-
gles; therefore
also KHG, GHM
are equal to two

B C K HM right angles. and because at the point H in the straight line GH, the two straight lines KH, HM upon the opposite sides of it make the adjacent angles equal to two right angles, KH is in the same

d. 14. . straight a 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 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 same straight line with GL. and because KF is parallel to HG, and HG to ML; KF is parallelo to ML. and KM, FL are paral- e. 30. se 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 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 straight line, a parallelogram equal to the 6.44. 5. first triangle ABD, and having an angle equal to the given angle,

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Book 1,

PROP. XLVI.

PROB.

To describe a square upon a given straight line.

2. II. I.

b. 3. I.

Let AB be the given straight linc; it is required to describc a square upon AB.

From the point A draw a AC at right angles to AB; and make

b AD equal to AB, and thro’ the point D draw DE parallel to it, f. 31. d, and thro’ B draw BE parallel to AD, therefore ADEB is a paralled. 34. 1. logram; whence AB is equal to DE, and AD to BE, but BA is equal to AD; therefore the four

СІ
straight lines BA, AD, DE, EB are e-
qual to one another, and the parallelo-
gram ADEB is equilateral. likewife all D

E
its angles are right angles; because the
straight line AD meeting the parallels

AD, DE, the angles BAD, ADE are e29.3. qual e to two right angles; but BAD is

a right angle, therefore also ADE is a
right angle. but the opposite angles of A

В parallelograms are equal ~; 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 angles right angles.

THEOR.

PROP. XLVII.
IN any right angled triangle, the square which is de-

scribed upon the side subrending the right angle, is equal to the squares described upon the sides which con, tain the right angle.

Let ABC be a right angled triangle having the right angle BAC; the square described upon the side BC, is equal to the squares defcribed upon BA, AC. On BC describe the square BDEC, and on BA, AC the squares

GB,

2. 46. 1,

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