C

But the angles MHG, HGL are equal to two right angles; Book I. wherefore also 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 e to ML; but KM, FL are parallels; where- e 30. 1. 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 ftraight 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. 1. parallelogram equal to the first 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 de

fcribe a fquare upon AB.

C

C

E

31.1.

d 34. I

From the point A draw a AC at right angles to AB; and a II. I. make b AD equal to AB, and through the point D draw b 3. I. DE parallel to AB, and through B draw BE parallel to AD; therefore ADEB is a parallelogram: Whence AB is equal d 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 pa- D rallelogram ADEB is equilateral : it is likewife rectangular; for the ftraight line AD meeting the parallels AB, DE, makes the angles BAD, ADE equal e to two right angles; but BAD is a right angle; therefore alfo ADE is a right angle; A now the oppofite angles of parallelo

B

grams

€ 29. I.

BookI. grams are equald; 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 straight line AB: Which was to be done.

a 46. I.

b 31. I.

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

IN

PRO P. XLVII. THE OR.

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

On BC describe a the fquare BDEC, and on BA, AC the 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 30. def. angle c, the two ftraight
lines AC, AG upon the
oppofite fides of AB, F
make with it at the point
A the adjacent angles e-
qual to two right angles;
therefore CA is in the
d 14.1. fame ftraight line d with

H

K

B

AG; for the fame reason,
AB and AH are in the
fame straight line; and be-
cause the angle DBC is e-
qual to the angle FBA,

each of them being a

right angle, add to each

e 2. Ax. the angle ABC, and the whole angle DBA is 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 4. I.

g 41. I.

the angle FBC; therefore the bafe AD is equalf to the bafe Book I. FC, and the triangle ABD to the triangle FBC: Now the parallelogram BL is double g of the triangle ABD, because they are upon the fame base BD, and between the fame parallels, BD, AL; and the fquare GB is double of the triangie FBC, because these alfo are upon the fame base FB, and between the fame parallels FB, GC. But the doubles of cquals are equal to one another; therefore the parallelogram h 6 Ax. BL 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 fquare BDEC is equal to the two fquares GB, HC; and the fquare BDEC is defcribed 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.

PROP. XLVIII. THEOR.

F the square described upon one of the fides of a

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. 1. 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
fquares of DA, AC, are equal to the
fquares of BA, AC: But the fquare of
DC is equal b 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

D

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

therefore

b 47. 1.

Book I. therefore also the fide 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, and the base DC likewise equal to the base BC, the angle 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.

c 8. I.

ELE

ELEMENTS

GEOMETRY.

BOOK II.

DEFINITIONS.

I..

EVERY right angled parallelogram, or rectangle, is faid to

A

E

Ꭰ .

In every parallelogram, any of the parallelograms about a
diameter, together with the
two complements, is cal-
led a Gnomon. Thus
'the parallelogram HG,
together with the com-
plements AF, FC, is

the

gnomon, which is

F

H

K

more briefly expreffed

by the letters AGK, or

EHC, which are at the

Book II.