BK is equal to BC. [II. 4, Cor.] Therefore the gnomon AKF, together with the square CK, is equal to twice the rectangle AB, BC. To each of these equals add HF, which is equal to the square on AC. [II. 4, Cor., and I. 34.] Therefore the gnomon AKF, together with the squares CK, HF, is equal to twice the rectangle AB, BC, together with the square on AC. But the gnomom AKF together with the squares CK, HF, make up the whole figure ADEB and CK, which are the squares on AB and BC. Therefore the squares on AB, BC, are equal to twice the rectangle AB, BC, together with the square on AC. Therefore, if a straight line, &c. Q.E.D. If AB and BC be considered as two lines, AC will be their difference, and it is proved that the square on AC is less than the squares on AB and BC by twice the rectangle AB, BC. Therefore the square on the difference of two lines is equal to the squares on the lines diminished by twice their rectangle. The corresponding proposition in Algebra is similarly expressed, and is thus represented in symbols, (a - b)2=a2+ B2 – 2ab. PROPOSITION 8. THEOREM. If a straight line be divided into any two parts, four times the rectangle contained by the whole line and one of the parts together with the square on the other part, is equal to the square on the straight line which is made up of the whole and that part. Let the straight line AB be divided into any 1. two parts at the point C: four times the rectangle AB, BC, together with the square on AC, shall be equal to the square on the straight line made up of AB and BC together. 92 may K z Х P Р R AEFD; and construct BD, L BD [Post. 2, and I. 3.] on Produce AB to D, so that BD be equal to CB; M figures such as in preceding proposi Then because 3. CB is equal to [Const.] and that to KN, [I. 34.] therefore GK is equal to KN. PAx. 1.) For the same reason PR is equal to RO.. of the parallelogram C0; [I. 43.] therefore also BN is BN, CK, GR, RN are equal to one another, and so the quadruple of one of them CK. Again, because CB is equal to BD, [Const.] and that BD is equal to BK, [II. 4, Cor.] that is to CG, [I. 34.] and that CB is equal to GK, (I. 34.] that is to GP; [II. 4, Cor.] therefore CG is equal to GP. (Ax. 1.] And because CG is equal to GP, and PR to RO, the rectangle AG is equal to the rectangle MP, and the rectangle PL to the rectangle RF. [I. 36.] But MP is equal to PL, because they are the complements of the parallelogram ML; [I. 43.] therefore also AG is equal to RF. [Ax. 1.] Therefore the four rectangles AG, MP, PL, RF are equal to one another, and so the four are quadruple of one of them AG. And it was demonstrated that the four CK, BN, GR and RN are quadruple of CK; therefore the eight rectangles which make up the gnomon AOH are quadruple four are of AK. And because AK is the rectangle contained by AB, BC, for BK is equal to BC; therefore four times the rectangle AB, BC is quadruple of AK. But the gnomon AOH was demonstrated to be quadruple of AK. Therefore four times the rectangle AB, BC is equal to the gnomon AOH. [Ax. 1.] To each of these add XH, which is equal to the square on AC. [II. 4, Cor., and I. 34.] Therefore four times the rectangle AB, BC, together with the square on AC, is equal to the gnomon AOH and the square XH. But the gnomon AOH and the square XH make up the figure AEFD, which is the square on AD. Therefore four times the rectangle AB, BC, together with the square on AC, is equal to the square on AD, that is to the square on the line made of AB and BC together. Therefore, if a straight line, &c. Q.E.D. If AB contains a lineal units and BC b lineal units, shew that this proposition may be expressed in symbols, thus : (a+b)2 – (a-6)=4ab. If AB and BC be considered as two lines, AD is their sum and AC their difference. Therefore four times the rectangle contained by two lines is less than the square on the sum of the two lines by the square on their difference. PROPOSITION 9. THEOREM. If a straight line be divided into two equal, and also into two unequal parts, the squares on the two unequal parts are together double of the square on half the line and of the square on the line between the points of section. 1. 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 squares on AD, DB shall be together double of the squares on AC, CD. 2. From the point C draw CE at right angles to AB, [I. 11.] and make it equal to AC or CB, [I. 3.) and join EĀ, EB; through D draw DF parallel to CE, and through F draw FG parallel to BA; [I. 31.] and join AF. Then, because 3. AC is equal to CE, Const.) the angle EAC is equal to the angle AEC. [I. 5.) And because the angle ACE is a right angle, [Const.] the two other angles Ā с AEC, EAC are together equal to one right angle; [I. 32.] and they are equal to one another; therefore each of them is half a right angle. For the same reason each of the angles CEB, EBC is haif a right angle. Therefore the whole angle AEB is a right angle. And because the angle GEF is half a right angle, and the angle EGF a right angle, for it is equal to the interior and opposite angle ECB; [I. 29.) therefore the remaining angle EFG is half a right angle. Therefore the angle GEF is equal to the angle EFG, and the side EG is equal to the side GF. [I. 6.] Again, because the angle at B is half a right angle, and the angle FDB a right angle, for it is equal to the interior and opposite angle ECB; [I. 29.] therefore the remaining angle BFD is half a right angle. Therefore the angle at B is equal to the angle BFD, and the side DF is equal to the side DB. [I. 6.] And because AC is equal to CE, [Const.] the square on AC is equal to the square on CE; therefore the squares on AC, CE are double of the square on AC. But the square on AE is equal to the squares on AC, CE, because the angle ACE is a right angle; [I. 47.] therefore the square on AE is double of the square on AC. Again, because EG is equal to GF, [Const.] the square on EG is equal to the square on GÈ; therefore the squares on EG, GF are double of the square on GF. But the square on EF is equal to the squares on EG, GF, because the angle EGF is a right angle; [I. 47.] therefore the square on EF is double of the square on GF. And GF is equal to CD; [I. 34.] therefore the square on EF is double of the square on CD. But it has been demonstrated that the square on AE is also double of the square on AC. Therefore the squares on AE, EF are double of the squares on AC, CD. But the square on AF is equal to the squares on AE, EF, because the angle AEF is a right angle. [I. 47.] Therefore the square on AF is double of the squares on AC, CD. But the squares on AD, DF are equal to the square on AF, because the angle ADF-is a right angle. [I. 47.] Therefore the squares on AD, DF are double of the squares on AC, CD. And DF'is equal to DB; therefore the squares on AD, DB are double of the squares on AC, CD. Therefore, if a straight line, &c. Q.E.D. PROPOSITION 10. THEOREM. If a straight line be bisected, and produced to any point, the square on the whole line thus produced, and the square on the part of it produced, are together double of the square on half the line bisected and of the square on the line made up of the half and the part produced. 1. Let the straight line AB be bisected at C, and produced to D: the squares on AD, DB shall be together double of the squares on AC, CD. From the point C draw CE at right angles to AB, and make it equal to AC or CB; [I. 11, 1. 3.] and join AE, EB; through E draw EF parallel to AB, and through D draw DF parallel to CE. [I. 31.] Then because EC is parallel to FD, therefore EB is not parallel to FD. [Ax. 12.] Hence EB and FD will meet if produced. Let them meet in G. 3. Then because AC is equal to CE, [Const.] the angle CEA is equal to the angle EAC; [T. 5.] and the angle ACE is a right angle; therefore each of the angles CEA, EAC is half a right angle. For the |