Through D draw DHG parallel to CE or BF (I. 31). allel to CL or BM. -- PROOF. Then the complement CH is equal to the complement HF (I. 43). To each of these add DM; therefore the whole CM is equal to the whole DF (Ax. 2). For But CM is equal to AL (I. 36), because AC is equal to CB AL=CM (Hyp.); Therefore also AL is equal to DF (Ax. 1). =DF. To each of these add CH; therefore the whole AH is AH= equal to DF and CH (Ax. 2). But AH is contained by AD and BD, since DH is equal to DB (II. 4, cor.), And DF, together with CH, is the gnomon CMG; DF+CH. =AD DB. Therefore the gnomon CMG is equal to the rectangleÁD, DB... CMG. To each of these equals add LG, which is equal to the Add to square on CD (II. 4, cor., and I ⚫ 34); Therefore the gnomon CMG, together with LG, is equal to the rectangle AD, DB, together with the square on CD. But the gnomon CMG and LG make up the whole figure CEFB, which is the square on CB; each LG or CD2. Therefore the rectangle AD, DB, together with the square .. CB2 on CD, is equal to the square on CB. Therefore, if a straight line, &c. Q. E. D. COROLLARY.-From this proposition it is manifest that the difference of the squares on two unequal lines AC, CD is equal to the rectangle contained by their sum and difference. Proposition 6.-Theorem. If a straight line be bisected, and produced to any point, the rectangle contained by the whole line thus produced and the part of it produced, together with the square on half the line bisected, is equal to the square on the straight line which is made up of the half and the part produced. Let the straight line AB be bisected in C, and produced to D; =AD DB +CD2, AD DB + CB2 = CD2, For AL CH = HF. ... AM or AD DB =CMG. Add to each LG or CB2, ...AD DB + CB2 = CD2, AB2+BC2 =2 AB BC + AC2. The rectangle AD, DB, together with the square on CB, shall be equal to the square on CD. CONSTRUCTION. Upon CD describe the square CEFD (I. 46), and join DE. Through B draw BHG parallel to CE or DF (I. 31). Through H draw KLM parallel k to AD or EF. And through A draw AK par allel to CL or DM. PROOF.-Because AC is equal C B L H M E to CB (Hyp.), the rectangle AL is equal to CH (I. 36). To each of these add CM; therefore the whole AM is equal to the gnomon CMG (Ax. 2). But AM is the rectangle contained by AD and DB, since DM is equal to DB (II. 4, cor.); Therefore the gnomon CMG is equal to the rectangle AD, DB (Ax. 1). Add to each of these LG, which is equal to the square on CB (II. 4, cor., and I. 34); Therefore the rectangle AD, DB, together with the square on CB, is equal to the gnomon CMG and the figure LG. But the gnomon CMG and LG make up the whole figure CEFD, which is the square on CD; Therefore the rectangle AD, DB, together with the square on CB, is equal to the square on CD. Therefore, if a straight line, &c. Q. E. D. Proposition 7.-Theorem. If a straight line be divided into any two parts, the squares on the whole line and on one of the parts are equal to twice the rectangle contained by the whole and that part, together with the square on the other part. Let the straight line AB be divided into any two parts in the point C; The squares on AB and BC shall be equal to twice the rectangle AB, BC, together with the square on AC, CONSTRUCTION.-Upon AB describe the square ADEB (I. 46), and join BD. Through C draw CGF parallel to AD or BE (I. 31). Through G draw HGK parallel to AB or DE (I. 31). PROOF. Then AG is equal to GE (I. 43). To each of these add CK; therefore the whole AK is equal to the whole CE; Therefore AK and CE are double of AK. But AK and CE are the gnomon AKF, together with the square CK; Therefore the gnomon AKF, together with the square is double of AK. CK, But twice the rectangle AB, BC is also double of AK, for 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. For ...AKF + CK= 2AB BC. To each of these equals add HF, which is equal to the Add HF or square on AC (II. 4, cor., and I. 34); Therefore the gnomon AKF, together with the squares CK and HF, is equal to twice the rectangle AB, BC, together with the square on AC. AC2 to each equal. But the gnomon AKF, together with the squares CK and.. AB2 HF, make up the whole figure ADEB and CK, the squares on AB and BC; which are + BC2 Therefore the squares on AB and BC are equal to twice the rectangle AB, BC, together with the square on AC. Therefore, if a straight line, &c. Q. E. D. 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 line and the first mentioned part. Let the straight line AB be divided into any two parts in the point C; 4 AB BC +AC2= Four times the rectangle AB, BC, together with the square (AB+BC)2, on AC, shall be equal to the square on the straight line For BN=CK =GR=RN ... the four And the rectangles AG, MP, PL, RF, are equal to each other, and are together= 4 AG. made up of AB and BC together. CONSTRUCTION.-Produce AB to D, so that BD may be equal to CB (Post. 2, and I. 3). Upon AD describe the (I. 46), AEFD square And construct two figures such as in the preceding propositions, PROOF.-Because CB is equal to BD (Const.), CB to GK, and BD to KN (Ax. 1), For the same reason PR is equal to RO. G And because CB is equal to BD, and GK to KN, Therefore the rectangle CK is equal to BN, and GR to RN (I. 36). But CK is equal to RN, because they are the complements of the parallelogram CO (I. 43); Therefore also BN is equal to GR (Ax. 1). Therefore the four rectangles BN, CK, GR, RN are equal to one another, and so the four are quadruple of one of them, CK. Again, because CB is equal to BD (Const.); And that BD is equal to BK, that is CG (II. 4, Cor., and I. 34); And that CB is equal to GK, that is GP (I. 34, and 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 MP, and PL to RF (I. 36). But MP is equal to PL, because they are complements of the parallelogram ML (I. 43), and 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 of AK. And because AK is the rectangle contained by AB and 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 Hence 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; adding 4 AB BC + AC2 Therefore four times the rectangle AB, BC, together with the square on AC, is equal to the square on AD, that is, on = AF = the line made up of AB and BC together. Therefore, if a straight line, &c. Q. E. D. 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. (AB+BC)2. Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts in the point D; The squares on AD and DB shall be together double of AD2+DB2 the squares on AC and CD. CONSTRUCTION. 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 EA, EB. Through D draw DF parallel to CE (I. 31). Through F draw FG parallel to BA (I. 31), and join AF. PROOF.-Because AC is equal to CE (Const.), the angle EAC is equal to the angle AEC (I. 5). =2(AC2+ CD2). |