BOOK II. DEFINITIONS. 1. EVERY right-angled parallelogram is said to be contained by any two of the straight lines which contain one of the right angles. II. In every parallelogram, any of the parallelograms about a diameter, together with the two complements, is called a Gnomon. Thus the parallelogram HG, together with the complements AF, FC, is a gnomon, which is more briefly expressed by the letters AGK, or EHC, which are at the opposite angles of the parallelograms which make the gnomon. PROP. I. THEOR. If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line. Let A and BC be two straight lines, and let BC be divided into any parts in the points D, E: the rectangle contained by the straight lines A, BC shall be equal to the rectangle contained by A, BD, together with that contained by A, DE, and that contained by A, EC. From the point B draw BF at right angles to BC, and make BG equal to A; through G draw GH parallel to BC, and through D, E, C, draw DK, EL, CH parallel to BG: Then the rectangle BH is equal to the rectangles BK, DL, EH: And BH is contained by A, BC, for it is contained by GB, BC, of which GB is equal to A ; and BK is contained by A, BD, for it is contained by GB, BD, of which GB is equal to A; and DL is contained by A, DE, because DK, that is BG, is equal to A; and, in like manner, EH is contained by A, EC: Therefore the rectangle contained by A, BC, is equal to the several rectangles contained by A, BD, and by A, DE, and by A, EC. Wherefore, If there be two straight lines &c. Q.E.D. B LH PROP. II. THEOR. If a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts are together equal to the square of the whole line. Let the straight line AB be divided into any two parts in the point C: the rectangle* AB, BC, together with the rectangle AB, AC, shall be equal to the square of AB. Upon AB describe the square ADEB (1. 46), and N.B. To avoid repeating the word contained too frequently, the rectangle contained by two straight lines AB, BC, will in future be called, simply, the rectangle AB, BC. through C draw CF, parallel to AD or BE: Then AE is equal to the rectangles AF, CE: And AE is the square of AB; and AF is the rectangle contained by AB, AC, for it is contained by AD, AC, of which AD is equal to AB; and CE is contained by AB, BC, for BE is equal to AB: Therefore the rectangle AB, AC, together with the rectangle AB, BC, is equal to the square of AB. Wherefore, If a straight line &c. Q. E.D. D F P PROP. III. THEOR. If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts is equal to the rectangle contained by the two parts, together with the square of the aforesaid part. Let the straight line AB be divided into any two parts in the point C: the rectangle AB, BC shall be equal to the rectangle AC, CB, together with the square of BC. Upon BC describe the square CDEB, and produce ED to F, and through A draw AF parallel to CD or BE: A C Then the rectangle AE is equal to the rectangles AD, CE: And AE is the rectangle contained by AB, BC, for it is contained by AB, BE, of which BE is equal to BC; and AD is contained by AC, CB, for CD is equal to CB; and CE is the square of BC: Therefore the rectangle AB, BC is equal to the rectangle AC, CB, together with the square of BC. Wherefore, If a straight line &c. Q. E.D. F D PROP. IV. THEOR. If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts. Let the straight line AB be divided into any two parts in C: the square of AB shall be equal to the squares of AC, CB, together with twice the rectangle contained by AC, CB. A F D Upon AB describe the square ADEB, and join BD, and through C draw CGF parallel to AD or BE, and through G draw HK parallel to AB or DE: Then, because CF is parallel to AD, and BD falls upon them, the exterior angle CGB is equal to the interior and opposite angle ADB (1. 29): but the angle ADB is equal to the angle ABD, because AD is equal to AB, being sides of a square; therefore the angle CGB is equal to the angle CBG, and therefore also the side CG is equal to the side CB: But CG is equal to BK, and CB to GK; therefore the figure BCGK is equilateral : It is also rectangular; for, because BK is parallel to CG, and BC meets them, the angles KBC, BCG are equal to two right angles; and KBC is a right angle, therefore also BCG is a right angle, and therefore also the angles CGK, GKB opposite to these, are right angles, and BCGK is rectangular: And it has been shown to be equilateral; therefore it is a square, and it is upon the side CB: For the like reason HF is a square, and it is upon the side HG, which is equal to AC: Therefore HF, CK are the squares of AC, CB: And because the complement AG is equal to the complement GE (1. 43), and that AG is the rectangle contained by AC, CB, for CG is equal to CB, therefore H D G C B G K GE is also equal to the rectangle AC, CB, and AG, GE are equal to twice the rectangle AC, CB; and HF, CK are the squares of AC, CB; therefore the four figures HF, CK, AG, GE are together equal to the squares of AC, CB, and twice the rectangle AC, CB: But HF, CK, AG, GE make up the whole figure ADEB, which is the square of AB; therefore the square of AB is equal to the squares of AC, CB, and twice the rectangle AC, CB. Wherefore, If a straight line, &c. Q.E.D. COR. From the demonstration, it is manifest that the parallelograms about the diameter of a square are likewise squares. PROP. V. THEOR. If a straight line be divided into two equal parts, and also into two unequal parts, the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line. Let the straight line AB be divided into two equal parts at the point C, and into two unequal parts at the point D: the rectangle AD, DB, together with the square of CD, shall be equal to the square of CB. Upon CB describe the square CEFB, join BE, and through D draw DHG parallel to CE or BF; through H draw KLM parallel to CB or EF, and through A draw AK parallel to CL or BM. Then, because the complement CH is equal to the complement HF, to each of these add DM; therefore the whole CM is equal to the whole DF: But CM is equal to AL, because AC is equal to CB; therefore also AL is equal to DF: To each of these add CH; therefore the whole AH is equal A C D B K GF |