T

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 ftraight line, and C the given tri-angle, 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 Ď.

Book I.

Make a the parallelogram BEFG equal to the triangle C, a 42. 1

C

having the angle EBG equal to the angle D, and the fide BE in the fame ftraight line with AB: produce FG to H, and through A draw b AH parallel to BG or EF, and join b31. 1. HB. Then because the ftraight line HF falls upon the parailels AH, EF, the angles AHF, HFE, are together equal c c 29. 1 to two right angles; wherefore the angles BHF, 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 if produced far enough: Therefore HB, FE will 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 equal to BF: But BF is e- d 43. 1. qual to the triangle C; wherefore LB is equal to the triangle C; and because the angle GBE is equale to the angle e 15. 1. ABM, and likewife to the angle D; the angle ABM

d

is equal to the angle D: Therefore the parallelogram

LB

Book I. LB is applied to the ftraight line AB, is equal to the triangle C, and has the angle ABM equal to the angle D: Which was to be done.

42. Z.

b 44. 1.

29. 1.

Tre

PROP. XLV. PROB.

O 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 ftraight line GH apply the parallelogram GM equal to the triangle DBC, having the angle GHM equal to the angle E. And becaufe 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, GHM; but FKH, KHG are equal to two right angles; there

c

314. I.

fore alfo KHG, GHM are equal to two right anglès: and becaufe at the point H in the ftraight line GH, the two straight lines KH, HM, upon the oppofite fides of GH, make the adjacent angles equal to two right angles, KH is in the fame ftraight lined with HM. And because the 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

1

But the angles MHG, HGL, are equal to two right angles; Book I. wherefore alfo the angles HGF, HGL are equal to two right angles, and FG is therefore in the fame straight line with GL. And because KF is parallel to HG, and HG to ML, KF is parallele to ML; but KM, FL are parallels; where- e 30. I. fore 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 manifeft how to a given straight line to apply a parallelogram, which fhall 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 b 44. I. parallelogram equal to the firit triangle ABD, and having an angle equal to the given angle.

PROP. XLVI. PRO B.

To defcribe a fquare upon a given straight line.

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

C

C

D

d 34. x.

From the point A draw a AC at right angles to AB; and a 11. 1. make b AD equal to AB, and through the point D draw b 3. 1. DE parallel to AB, and through B draw BE parallel to c 31. I. AD; therefore ADEB is a parallelogram: Whence AB is equal 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 ADEB is equilateral : it is likewife rectangular; for the ftraight line AD meeting the parallels AB, DE, makes the angles BAD, ADE equale to two right angles; but BAD is a right angle; therefore also ADE is a right angle; Al now the oppofite angles of parallelo.

E

B

grams

e 29. x.

Book I. grams are equald; therefore each of the oppofite angler 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 square, and it is defcribed upon the gi ven straight line AB: Which was to be done.

a 46. I.

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

IN

PROP. XLVII. THEOR.

N 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 described upon the fide BC is equal to the fquares defcribed upon BA, AC.

On BC describe a the square BDEC, and on BA, AC the b 31. 1. fquares GB, HC; and through A draw b AL parallel to BD or CE, and join AD, FC; then, because each of the angles BAC, BAG is a right

c 25. def. angle c, the two ftraight

lines AC, AG upon the
oppofite fides
make with it at the point
A the adjacent angles e-
qual to two right angles;
therefore CA is in the
fame ftraight lined with
AG; for the fame reafon,
AB and AH are in the
Now
fame ftraight line.

because the angle DBC is

angle, adding to each the

e 2. Ax. angle ABC, the whole angle DBA will be equale to the whole FBC; and because the two fides AB, BD, are equal to the two FB, BC, each to each, and the angle DBA equal to

the

f

f 4. I.

g 41. I.

the angle FBC, therefore the bafe AD is equal to the bafe Book I. FC, and the triangle ABD to the triangle FBC. But the parallelogram BL is double g of the triangle ABD, because they are upon the fame bafe BD, and between the fame parallels, BD, AL; and the fquare GB is double of the triangle BFC, because these alfo are upon the fame base FB, and between the same parallels FB, GC. Now the doubles of equals are equal h to one another; therefore the parallelogram h 6. Ax. BL is equal to the square GB: And, in the same manner, by joining AE, BK, it is demonstrated that the parallelogram CL is equal to the fquare HC. Therefore, the whole fquare BDEC is equal to the two fquares GB, HC; and the fquare BDEC is described upon the straight line BC, and the fquares CB, 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.

IT

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 thefe two fides is a right angle.

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

From the point A draw a AD at right angles to AC, and a 11. f. make AD equal to BA, and join DC. Then, because DA is

equal to AB, the square of DA is equal
to the fquare of AB: To each of thefe
add the fquare of AC; therefore the
fquares of DA, AC, are equal to the
fquares of BA, AC. But the square of
DC is equal to the fquares of DA,
AC, because DAC is a right angle;
and the fquare of BC, by hypothefis, is
equal to the fquares of BA, AC; there- B

b

D

b. 47. I.

fore, the fquare of DC is equal to the fquare of BC: and

therefore