B G Proposition 1. 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 number of parts at the points D, E; the rectangle contained by the straight lines A, BC, will 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 (I. 11) BF at right angles to BC; and make (I. 1) BG equal to A; through G draw (I. 29) GH parallel to BC; and through D, E and C draw (I. 29) DK, EL and CH parallel to BG. B D E Then the rectangle BH is equal to F the rectangles BK, DL, EH. But BH-BG.BC- A. BC, because A is equal to BG. Also, BK-BG.BD=A. BD; DL-DK.DE-A.DE; EH EL. EC-A.EC; A.BC- A.BD+A.DE+ A . EC. H C H Proposition 2. 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; then AB.AC+AB. CB = AB2. On AB describe (I. 41) the square AE, and draw (I. 29) through C, CF parallel to AD. Then but and and AF+CE-AE; AF-AD.AC-AB. AC, CECF. CB = AB. CB, AE-AB2; AB. AC+AB. CB = AB'. Proposition 3. Theorem.-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. C F D B E Then but also, and AE-AD+CE; AE-AB. BE-AB. BC; AD AC. CD = AC. CB, = CE=CB2. AB.BC- AC. CB+ CB2. Proposition 4. 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 two parts. Let the straight line AB be divided into any two parts in C; then will AB2 = AC2+ BC2+2A C. CB. A H C B D Upon AB describe (I. 41) the square ABED; join BD; through C draw (I. 29) CF parallel to AD, and through G draw (I. 29) HGK parallel to AB. Because CF is parallel to AD, and BD falls upon them, the exterior angle BGC is equal (I. 27) to the interior and opposite angle ADB; but ADB is equal (I. 6) to ABD. Therefore CGB is equal to CBG, and therefore the side CG is equal to (I. 23) the side CB; but CB is equal (I. 32) to GK; therefore the figure CGKB is equilateral. It is also rectangular, for the angle CBK being a right angle, the other angles are also right angles (I. 41, Cor.). Wherefore it is a square upon the side CB. For the same reason HF is a square upon the side HG, which is equal to AC. The complement A G is equal (I. 40) to the complement GE. Also, AG-AC. CG-AC. CB; AG+ GE=2A C. CB. K F E Also, .'. But HF AC, and CK= CB2; HF+CK+AG+ GE=AC2+ CB3+2A C. CB. HF+CK+AG+ GE-AE-AB2; Corollary 1.-Hence it is manifest that the parallelograms about the diagonal of a square are also squares. Corollary 2.-The square on a line is four times the square on half the line. = For let C be the middle point of AB, then AB2 = AC2 + CB2+2A C. CB=4A C2. x Proposition 5. Theorem.—The square on the difference of two lines, is equal to the sum of their squares diminished by twice the rectangle of the lines. Let AB, BC be two lines: place BC on BA from B toward A; then AC is their difference; then AC2 = AB2 + BC2 – 2AB.BC. On AB describe the square ABED; join DB, and through C draw CF parallel to BE; through G draw HK parallel to AB; then AE-AB2, CK CB and HF = AC2. A H D C B K F E Because (I. 40) AG = GE, add to both CK; then AK= CE, and AK+ CE=2AK=2AB.BK =2AB.BC. Now, if from AE+ CK, AK+ CE be taken away, there will remain the square HF; that is, HF = AE + CK - (AK + CE), or AC2 = AB2 + BC2 - 2AB. BC. Proposition 6. Theorem.—The rectangle contained by the sum and difference of two lines is equal to the difference of their squares. Let AB, BC be two lines; then (AB+BC) (AB-BC) = AB2 – BC2. From BA cut off (I. 1) BD equal to BC, then AC- AB+BC and AD-AB-BC. On AB describe the square AEFB; join AF, and draw DG parallel to AE; through H draw KHL parallel to AC, and through C draw CL parallel to AK. Then EB = AB2 and GM = DB2. Now, because DB = BC (I. 34), DM = BL; but (I. 40) DM = KG, therefore BL = KG ; add to both AM, and the gnomon BKG = AL. Now the gnomon BKG is the difference between the squares EB and GM, or AB2 – DB2; and the rectangle AL-AC. AK-AC.AD =(AB+BC) (AB-BC); therefore (AB+BC) (AB-BC) AB2 - DB2. Corollary.—If a line be bisected and produced, the rectangle of the whole line thus produced and the part produced, together with the square of the line between the points of section, is equal to the square of the line made up of the half and the part produced. For, CD being bisected in B and produced in A, we have, adding DB2 to the last equals, AC.AD+DB2 = AB2. Proposition 7. Theorem. In obtuse-angled triangles the square of the side subtending the obtuse angle, is greater than the sum of the squares of the other two sides, by twice the rectangle contained by one of |