EUCLID'S ELEMENTS OF GEOMETRY. Book II. DEFINITIONS. I. Every right-angled parallelogram, or rectangle, 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 the gnomon, which is more briefly expressed by the letters 4GK, or EHC, which are at the opposite angles of the parallelograms which make the gnomon." PROP. I.-THEOREM. 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, is 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 (I. 11.) BF at right angles to BC, and make BG equal (1.3.) to 4; and through G draw (I. 31.) GH parallel to BC; and through D, E, C, draw (I. 31.) DK, EL, CH, parallel to BG; then the rectangle BH is equal to the rectangles BK, DL, EH; and 1. BH is contained by A, BC, for it is contained by GB, BC, and GB is equal (Constr.) to 4; and 2. BK is contained by A, BD, for it is contained by GB, BD, of which GB is equal to ▲; and DL is contained by A, DE, 3. because DK, that is, BG, (I. 34.) is equal to ; and in like manner 4. The rectangle EH is contained by A, EC. Therefore the rectangle contained by A, BC, is equal to the several rectangles contained by A, BD, by A, DE, and by A, EC. Wherefore, if there be two straight lines, &c. Q.E.D. PROP. II.—THEOREM. 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 contained by AB, BC, together with the rectangle* AB, AC, shall be equal to the square of AB. Upon AB describe (I. 46.) the square ADEB, and through C draw N.B.--To avoid repeating the word contained too frequently, the rectangle contained by two straight lines AB, AC, is sometimes simply called the rectangle AB, AC. (I. 31.) CF parallel to AD or BE; then AE is equal to the rectangles AF, CE; and AE is the square of AB; and 1. AF is the rectangle contained by BA, AC, for it is contained by DA, AC, of which AD is equal (Def. 30.) to AB; and 2. CE is contained by AB, BC, for BE is equal to AB; therefore 3. The rectangle contained by AB, AC, together with the rectangle AB, BC, is equal to the square of AB. If, therefore, a straight line, &c. Q.E.D. 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, is equal to the rectangle AC, CB, together with the square of BC. C Upon BC describe (I. 46.) the square CDEB, and produce ED to F; and through draw (I. 31.) AF parallel to CD or BE; then (Constr.) The rectangle AE is equal to the rectangles AD, CE, 1. and 2. AE is the rectangle contained by AB, BC, for it is contained by AB, BE, of which BE is equal (Def. 30.) to BC; and 3. AD is contained by AC, CB, for CD is equal to CB; and CE is the square of BC; therefore 4. The rectangle AB, BC, is equal to the rectangle AC, CB, together with the square of BC. If, therefore, a straight line, &c. Q.E.D. PROP. IV.-THEOREM. 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 is equal to the squares of AC, CB, together with twice the rectangle contained by AC, CB. Upon AB describe (I. 46.) the square ADEB, and join BD; and through draw (I. 31.) CGF parallel to AD or BE; and through G draw HK parallel to AB or DE. And because CF is parallel to AĎ, and BD falls upon them, (I. 29.) 1. The exterior angle BGC is equal to the interior and opposite angle ADB; but (I. 5.) 2. ADB is equal to the angle ABD, because BA is equal to AD, being sides of a square ; wherefore 3. The angle CGB is equal to the angle GBC; and therefore (I. 6.) 4. The side BC is equal to the side CG: but CB is equal (I. 34.) also to GK, and CG to BK; wherefore 5. The figure CGKB is equilateral. It is likewise rectangular; for, since CG is parallel to BK, and CB meets them, therefore (I. 29.) 6. The angles KBC, GCB, are equal to two right angles ; and KBC is (Def. 30.) a right angle; wherefore 7. GCB is a right angle; and therefore also (I. 34.) the opposite angles CGK, GKB, are right angles, and 8. The figure CGKB is rectangular. But it is also equilateral, as was demonstrated; wherefore and it is upon the side CB. For the same reason, also, and it is upon the side HG, which is equal (I. 34.) to AC; therefore 11. HF, CK, are the squares of AC, CB. And because the complement 4G is equal (I. 43.) to the complement GE, and that 12. AG is the rectangle contained by AC, CB, for GC is equal (Def. 30.) to CB; therefore also wherefore 13. 14. AG, GE, are equal to twice the rectangle AC, CB: and HF, CK, are the squares of AC, CB; wherefore 15. The four figures HF, CK, AG, GE, are equal to the squares of AC, CB, and to twice the rectangle AC, CB: but HF, CK, AG, GE, make up the whole figure ADEB, which is the square of AB: therefore 16. 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.-THEOREM. 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 in the point C, and into two unequal parts at the point D; the rectangle AD, DB, together with the square of CD, is equal to the square of CB. Upon CB describe (I. 46.) the square CEFB, join BE, and through D draw (I. 31.) DHG parallel to CE or BF; and through H draw KLM parallel to CB or EF; and also through 4 draw 4K parallel to CL or BM. And because the complement CH is equal (I. 43.) to the complement HF, to each of these add DM; therefore 1. The whole CM is equal to the whole DF; but (I. 36.) CM is equal to AL, because AC is equal (Hyp.) to CB; therefore also 2. AL is equal to DF. To each of these add CH, and 3. The whole AH is equal to DF and CH. |