EBG equal to

B

C

H

A

L

b 31. I.

C 29. I.

12. Ax.

€ 43. I.

P15. I.

See N.

a 42. 1.

b 44. I.

the angle D,
fo that BE be
in the fame
ftraight line
with AB, and produce FG to H; and thro' A draw b AH paral-
lel 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 equal to two right angles; wherefore the anglesBHF,
HFE are less than two right angles: But ftraight lines which
with another straight line make the interior angles upon the
fame fide less than two right angles, do meet d 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 paral-
lelogram, of which the diameter is HK, and AG, ME are the
parallelograms about HK; and LB, BF are the complements;
therefore LB is equal e to BF: But BF is equal to the triangle
C; wherefore LB is equal to the triangle 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.

T

PROP. XLV. PRO B.

O describe 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 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

to

GL

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 each of thefe the angle KHG; therefore the angles FKH, KHG are equal to the angles KHG, GHM;but FKH, A KHG are equal c to two right angles: Therefore alfo KHG, GHM are equal to two right

angles; and be

D

F

E

C 29.1.

caufe at the point

H in the ftraight

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 straight lined with HM; and because the d 14. 1. ftraight 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 alfo the angles HGF, HGL are equal to two right angles, and FG is therefore in the fame ftraight line with GL and because KF is parallel to HG, and HG to ML; KF is parallel to ML: and KM, FL e 30. 1. 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 b to the given ftraight line a parallelogram e- b 44.5. qual to the first triangle ABD, and having an angle equal to the given angle.

PROP.

Book I.

a II. I.

b 3. I.

C 31. I.

d 34. I.

€ 29. I.

a 46. I.

To

PROP. XLVI. PRO B.

O defcribe a fquare upon a given ftraight line.

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

From the point A draw a AC at right angles to AB; and make b AD equal to AB, and through the point D draw DE parallel to AB, and through B draw BE parallel to AD; therefor ADEB is a parallelogram: whence AB is equal d to DE, and AD to BE: But BA is equal to

AD; therefore the four straight lines
BA, AD, DE, EB are equal to one
another, and the parallelogram ADEBD
is equilateral, likewife all its angles
are right angles; because the straight
line AD meeting the parallels AB,
DE, the angles BAD, ADE are equal
to two right angles; but EAD is a
right angle; therefore alfo ADE is a
right angle; but the oppofite angles A

E

B

of parallelograms are equal d; therefore each of the oppofite angles 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 defcribed upon the given ftraight line AB: Which was to be done.

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 described upon BA, AC.

On BC describe the fquare BDEC, and on BA, AC the

fquares

fquares GB, HC; and through A draw b AL parallel to BD, or Book 1. CE, and join AD,FC; then, because each of the angles BAC,

49

BAG is aright angle, the

make with it at the point A the adjacent angles equal to two right angles; therefore CA is in the fame ftraight lined with AG; for the fame reafon, AB and AH are in the fame ftraight line; and because the angle DBC is equal to the angle FBA, each of them being a right angle, add to each the angle ABC, and the whole angle DBA is

B

G

b 31. I

equale to the whole FBC; and because the two fides AB, BD e 2. Ax. are equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC; therefore the base AD is equal f to f4. T. the bafe FC, and the triangle ABD to the triangle FBC: Now the parallelogram BL is double g of the triangle ABD, becaufe g 41. 1. they are upon the fame base BD, and between the same parallels, BD, AL; and the fquare GB is double of the triangle FBC, because these alfo are upon the fame base FB and between the fame parallels FB, GC. But the doubles of equals are equal h to one another: Therefore the parallelogram BL h 6. A. is equal to the fquare GB: And, in the fame manner, by joining AE, BK, it is demonftrated that the parallelogram CL is equal to the fquare HC: Therefore the whole square BDEC is equal to the two fquares GB, HC; and the fquare BDEC is described upon the straight line BC, and the fquares GB, HC upon BA, AC: Wherefore the fquare upon the fide BC is equal to the fquares upon the fides BA, AC. Therefore, in any right angled triangle, &c. Q. E. D.

PROP. XLVIII. THEOR.

IF F the fquare defcribed upon one of the fides of a triangle, be equal to the fquares defcribed upon the other two fides of it; the angle contained by these two fides is a right angle.

Book I.

a II. I.

b 47. I.

c8. I.

If the fquare described upon BC, one of the fides of the triangle ABC, be equal to the squares upon the other fides BA, AC, the angle BAC is a right angle.

Ꭰ .

From the point A draw a AD at right angles to AC, and make AD equal to BA, and join DC: Then, because DA is equal to AB, the fquare of DA is equal to the fquare of AB: To each of these add the fquare of AC; therefore the squares of DA, AC, are equal to the fquares of BA, AC: But the fquare of DC is equalb A to the fquares of DA, AC, because DAC is a right angle; and the fquare of BC, by hypothefis, is equal to the squares of BA, AC; therefore the fquare of DC is equal to the fquare of BC, and therefore alfo B the fide DC is equal to the fide BC. And

because the side DA is equal to AB, and AC common to the two triangles DAC, BAC, the two DA, AC are equal to the two BA, AC; and the base DC is equal to the base BC; therefore the angle DAC is equal to the angle BAC: But DAC is a right angle; therefore also BAC is a right angle. There fore, if the fquare, &c. Q. E. D.

THE