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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 A

LE is double of the triangle EBC.

Join AC; then the triangle ABC a 37. 1. is equala to the triangle EBC, be

cause 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
vides 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.

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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. a 10. 1. Bisecta BC in E, join AE, and at the point E in the 623. 1. straight line EC makeb the angle CEF equal to D; and 31. 1. through A drawC AG parallel

to EC, and through C draw
CGC parallel to EF: There-
fore FECG is a parallelo-
gram: And because BE is -

equal to EC, the triangle
38. 1. ABE is likewise equals to

the triangle-AEC, since they
are upon equal bases BE,EC,
and between the same paral- D
lels BC, AG; therefore the triangle ABC is double of the

triangle AEC. And the parallelogram FECG is likewise • 41. 1. doublee of the triangle AEC, because it is upon the same

base, and between the same parallels : Therefore the pa- rallelogram FECG is equal to the triangle ABC, and it

has one of its angles CEF equal to the given angle D;

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Book L

wherefore there has been described a parallelogram FECG 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.

Let ABCD be a parallelogram, of which the diameter is AC, and EH, FG, the parallelograms about AC, that is, L H through which AC passes, and BK, KD, the other parallelograms which make up the whole figure ABCD, which are therefore called the complements. The complement BK is equal to the comple- b ment KD.

Because ABCD is a parallelogram, and AC its diameter, the triangle ABC is equal to the triangle ADC: And, . 34. 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 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.

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.

Book I. Makea the

paralleloa 42. 1. 1

gram BEFG
equal to the
triangle C,
and having
the angle
EBG equal
to the angle

LA
D, so that

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 . 12 Ax, upon the same side less than two right angles, do meetd 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 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.

PROP. XLV. PROB.

See n. To describe a parallelogram equal to a given rec

tilineal 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 describe a paral

lelogram equal to ABCD, and having an angle equal to E. 49. 1. Join DB, and described the parallelogram FH equal to

the triangle ADB, and having the angle HKF equal to the

44. 1.

€ 29. 1.

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angle E; and to the straight line GH apply the parallelo. Book t. gram GM equal to the triangle DBC, having the angle GHM equal to the angle E; and because the 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, G#M; but FAH, KHG are equal to two right an

K H M gles; Therefore also KHG, GHM, are equal to two 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 straight lined with HM, and because the straight line • 14. 1. HG meets the parallels KM, FG, the alternate angles MHG, HGF are equalc: 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 parallele to ML; and KM, . 30. 1. FL 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 applyingb to the given • 41. 1. straight line a parallelogram equal to the first 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 de

scribe a square upon AB. . 11. 1. From the point A drawa AC at right angles to AB; and

03. 1. makeb AD equal to AB, and through the point D draw DE * 31. 1. parallelc to AB, and through B draw BE parallel to AD; 34. 1. therefore ADEB is a parallelogram: whence AB is equald

to DE, and AD to BE: hut BA is
equal to AD; therefore the four
straight lines BA, AD, DE, EB, are
equal to one another, and the paral-
lelogram ADEB is equilateral, like- DH
wise all its angles are right angles;
because the straight line AD meet-

ing the parallels AB, DE, the angles * 29. 1. BAD, ADE are equale to two right

angles : but BAD is a right angle;
therefore also ADE is a right angle; u
but the opposite angles of parallelo- A
grams are equald; 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 equi-
lateral; 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.

B

PROP. XLVII. THEOR. In any riglit-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.

Let ABC he a right-angled triangle having the right angle BAC; the square described upon the side BC is equal to

the squares described upon BA, AC. . * 46. 2. On BC describes the square BDEC, and on BA, AC the

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