b. 31. I.

C. 29. I.

c

d

straight line with AB, and produce FG to H; and thro' A draw '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 equal to two right angles; wherefore the angles BHF, HFE are lefs than two right angles. but straight lines which with another straight line make the interior angles upon the d. 12. Ax. fame fide lefs than two right angles, do meet if produced far enough. therefore HB, FE fhall 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 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 equal to the triangle C; wherefore LB is equal to the triangle C. and because the angle 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.

e. 43. I.

f. 15. 1.

f

a. 42. I.

b. 44. I.

T°

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 describe a parallelogram equal 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

to the triangle DBC having the angle GHM equal to the angle E. Book I. 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

equal to two right angles. and because at the point H in the straight line GH, the two straight lines KH, HM upon the oppofite fides

d

C

of it make the adjacent angles equal to two right angles, KH is in the same straight line with HM. and because the straight line d. 14. 1. 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 parallel to ML. and KM, e. 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 defcribed equal to the given rectilineal figure ABCD, having the angle FKM equal to the given angle E. Which was to be done.

e

COR. From this it is manifeft how to a given ftraight 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,

b

viz. by applying to the given straight line, a parallelogram equal b. 44. Ia to the first triangle ABD, and having an angle equal to the given

angle.

C4

Book I.

A. II. I.

b.

3. I.

C. 31. I.

d. 34. I.

PROP. XLVI. PROB.

To defcribe a fquare upon a given straight line.

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 make b AD equal to AB, and thro' the point D draw DE parallel © to it, and thro' 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; therefore the four

ftraight lines BA, AD, DE, EB are equal to one another, and the parallelogram ADEB is equilateral. likewife all D its angles are right angles; because the ftraight line AD meeting the parallels AB, DE, the angles BAD, ADE are *. 29. 1. equal to two right angles; but BAD is a right angle, therefore alfo ADE is

2. 46. I.

A

B

a right angle. but the oppofite angles
of parallelograms are equal; therefore each of the oppofite angles
ABE, BED is a right angle; wherefore the figure ADEB is rec✶
tangular. and it has been demonftrated 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.

IN

N any right angled triangle, the fquare which is described upon the fide fubtending the right angle, is equal to the fquares described 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 described upon BA, AC.

On BC defcribe the fquare BDEC, and on BA, AC the fquares

GB, HC; and thro' A draw b AL parallel to BD or CE, and join
AD, FC. then because each of the angles BAC, BAG is a right
angle, the two straight lines
AC, AG upon the oppofite

G

Book I.

b. 31. I.

C. 30. Def.

H

K

d. 14. I.

D

equal to the whole FBC. and because the two fides AB, BD are e. 2. Az, equal to the two FB, BC, each to each, and the angle DBA equal to the angle FBC; therefore the bafe AD is equal to the base FC, f. 4.1. and the triangle ABD to the triangle FBC. now the parallelogram BL is double of the triangle ABD, because they are upon the g. 41. I. same base BD, and between the fame parallels BD, AL; and the fquare GB is double of the triangle FBC, because these also are upon the fame base FB, and between the fame parallels FB, GC. but the doubles of equals are equal to one another. therefore the h. 6. Ax. parallelogram BL is equal to the fquare GR. 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 fquare BDEC is equal to the two squares GB, HC. and the square BDEC is described upon the ftraight 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.

IF

PROP. XLVIII. THEOR.

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

—

Book I. If the fquare defcribed upon BC one of the fides of the trian gle ABC be equal to the squares upon the other fides BA, AC; the angle BAC is a right angle.

a. II. I.

D

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 square of AB; to each of these add the fquare of AC, therefore the fquares of DA, AC are equal to the fquares of BA, AC. but the

5. 47. I. fquare of DC is equal to the fquares of A

C. 8. I.

b

DA, AC, because DAC is a right angle;
and the fquare of BC, by Hypothesis, is
equal to the fquares of BA, AC; therefore,
the square of DC is equal to the fquare of B

BC; and therefore also the side DC is equal to the fide BC. and
because the fide 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 bafe BC; therefore the an-
gle DAC is equal to the angle BAC. but DAC is a right angle,
therefore alfo BAC is a right angle. Therefore if the fquare, &c
Q. E. D.