Book I. It is required to make a parallolegram equal to a given right lined figure ABCD, having an angle in it equal to the right lined angle E: Join DB, and make the parallelogram FH equal to the triangle ABD *; the angle FKH equal to E. Upon the right line GH make GM equal to DCB, and the angle GHM equal to Eb; then FM is the parallelogram equal to ABCD.

2 42.

b

44.

C 29.

For, because FH is a parallelogram, the angles FKH, GHK, are equal to two right angles; but the angles GHM, FKH, are each equal to the angle E; therefore equal to one another. Add GHK to both, then the angles GHM, GHK, are equal to FKH, GHK, that is, equal to two right angles; therefore KHM is a right line. For the fame reafon FGL is a right lined; but FK, LM, are each parallel to GH; therefore pae 30. and rallel to one another. Wherefore FM is a parallelogram equal to the right lined figure ABCD, and an angle FKM equal to E. Wherefore, &c.

d 14.

· conftruct.

b 3.

C 31.

d 34.

COR. Hence a parallelogram may be made equal to a given right lined figure of any number of fides; for a parallelogram can be made equal to any triangle upon any given right line.

It is required to defcribe a fquare upon the given right line AB.

d

e

From the point A, in the given right line AB, draw the perpendicular AC2; cut off AD equal to AB b; through D draw DE parallel to AB, and BE parallel to AD; then ADEB is a parallelogram, the oppofite fides of which are equal a; that is, e by conftr. DE equal to AB, and BE to AD; but AD is equal to AB ; therefore the four fides are equal to one another. But the angles ADE, BAD, are equal to two right angles; and BAD is a right angle; therefore ADE is likewise a right angle; but the oppofite angles of every parallelogram are equal": g Def. 30. Therefore ADEB is a fquares. Wherefore, &c.

£ 29.

IN

PRO P. XLVII. T HE OR.

every right angled triangle the square described upon the fide fubtending the right angle is equal to the squares of the fides containing the right angle.

Let