BOOK II. DEFINITIONS. 1. A rectangle is a parallelogram whofe angles are all right angles. 2. A fquare is a rectangle, whose fides are all equal tơ each other. 3. Every rectangle is faid to be contained by any two of the right lines which contain one of the right angles. 4. If two right lines be drawn through any point in the diagonal of a parallelogram, parallel to its opposite fides, the figures which are interfected by the diagonal are called parallelograms about the diagonal. 5. And the other two parallelograms, which are not interfected by the diagonal, are called complements to the parallelograms which are about the diagonal. 6. In every parallelogram, either of the two parallelograms about the diagonal, together with the two complements, is called a gnomon. 7. The altitude of any figure is a perpendicular drawn. from the vertical angle to the base. PROP. I PROBLEM. UPON a given right line to defcribe a fquare. C Let AB be the given right line; it is required to describe a fquare upon it. Make AD, BC, each perpendicular and equal to AB (I. 11 and 3.), and join DC; then will AC be the fquare required. For, fince the angles DAB, ABC are right angles (by Conft.), AD will be parallel to BC (I. 22 Cor.) And because AD, BC are equal and parallel, AB, DC will, alfo, be equal and parallel (I. 30.) But AD, BC are each equal to AB (by Conft.); whence AD, AB, BC and CD are all equal to each other. The The figure AC, therefore, is an equilateral parallelogram; and it has, likewife, all its angles right angles. For the angle DAB is equal to the angle DCB, and the angle ABC to the angle ADC (I. 30.) But the angles DAB, ABC are right angles (by Const.) ; confequently the angles DCB, ADC are, alfo, right angles. The figure AC, therefore, being both equilateral and rectangular, is a fquare; and it is defcribed upon the line AB, as was to be done. PRO P. II. THEOREM. Rectangles and Squares contained under equal lines are equal to each other. Let BD, FH be two rectangles, having the fides AB, BC equal to the fides EF, FG, each to each; then will the rectangle BD be equal to the rectangle FH. For draw the diagonals AC, EG: Then, fince the two fides AB, BC are equal to the two fides EF, FG, each to each (by Hyp.), and the angle B is equal to the angle F (I. 8.), the triangle ABC will be equal to the triangle EFG (1.4.) But the diagonal of every parallelogram divides it into two equal parts (I. 30.); whence the halves being equal; the wholes will also be equal. The rectangle BD is, therefore, equal to the rectangle FH; and in the fame manner it may be proved when thẻ figures are fquares. Q.E.D. PROP. III. THEOREM. The fides and diagonals of equal fquares are equal to each other. Let BD, FH, be two equal fquares; then will the fide AB be equal to the fide EF, and the diagonal AC to the diagonal EG. For if AB, EF be not equal, one of them must be greater than the other; let AB be the greater, and make BL, BK each equal to EF or FG (I. 3.); and join LK. Then, becaufe BL is equal to FE, BK to FG, and the angle LBK to the angle EFG, being each of them right angles, the triangle BLK will be equal to the triangle FEG (I. 4.) But the triangle FEG is equal to the triangle BAC, being each of them the halves of the equal fquares FH, BD (I. 30.); whence the triangle BLK is alfo equal to the triangle BAC, the lefs to the greater, which is abfurd. The fide AB, therefore is not greater than the fide EF; and in the fame manner it may be proved that it cannot be less; confequently they are equal to each other. And because AB is equal to EF, BC to FG, and the angle ABC to the angle EFG (I. 8.), the fide, Ac' will alfo be equal to the fide EG (I. 4.) E Q. E, D. PROP. PROP. IV. THEOREM. The fquare of a greater line is greater than the square of a lefs; and the greater square has the greater fide. Let the right line AB be greater than the right line EF; then will BD, the square of AB, be greater than FH, the fquare of EF. For fince AB is greater than EF, and BC than FG (by Hyp.), take вK, a part of BA, equal to EF, and BL, a part of BC, equal to FG (I. 3.); and join KL. Then, because BK is equal to FE, BL to FG, and the angle KBL to the angle EFG (I. 8.), the triangle BLK will be equal to the triangle FGE (I. 4.) But the triangle BCA is greater than the triangle BLK, whence it is alfo greater than the triangle FGE, And fince the fquare BD is double the triangle BCA, and the fquare FH is double the triangle FGE (I. 30.), the fquare BD will also be greater than the square -FH. Again, let the fquare BD be greater than the fquare FH; then will the side AB be greater than the side EF. For if AB be not greater than EF, it must be either equal to it, or lefs; but it cannot be equal to it, for then the |