And CE is the square on BC. Therefore the rectangle AB, BC is equal to the square on Proposition 4.-Theorem. If a straight line be divided into any two parts, the square on the whole line is equal to the sum of the squares on 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 on AB shall be equal to the squares on AC and AR? = CB, together with twice the rectangle contained by AC AC+CB2 and CB. 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 H G equal to the interior and opposite angle ADB (I. 29). Because AB is equal to AD, being sides of a square, the angle ADB is equal to the angle ABD (I. 5); +2AC CB. Therefore the angle CGB is equal to the angle CBG Show first (Ax. 1); Therefore the side BC is equal to the side CG (I. 6). It is likewise rectangular. For since CG is parallel to BK, and CB meets them, the angles KBC and GCB are together equal to two right angles (I. 29). But KBC is a right angle (Const.), therefore GCB is a right angle (Ax. 3). Therefore also the angles CGK, GKB, opposite to these, are right angles (I. 34). that CK is a square = CB2 £o also Therefore CGKB is rectangular; and it has been proved equilateral; therefore it is a square; and it is upon the side CB. For the same reason HF is also a square, and it is on HF AC the side HG, which is equal to AC (I. 34). And =2AC CB ... whole figure or AD DB +CD2 =CB2 Therefore HF and CK are the squares on AC and CB. And because the complement AG is equal to the complement GE (I. 43), And that AG is the rectangle contained by AC and CG, that is, by AC and CB, Therefore GE is also equal to the rectangle AC, CB; Therefore AG, GE are together equal to twice the rectangle AC, CB; And HF, CK are the squares on AC and CB. Therefore the four figures HF, CK, AG, GE are equal to the squares on AC and CB, together with twice the rectangle AC, CB. But HF, CK, AG, GE, make up the whole figure ADEB, which is the square on AB; Therefore the square on AB is equal to the squares on AC and CB and twice the rectangle AC CB. Therefore, if a straight line, &c. Q. E. D. COROLLARY.-From this demonstration it follows that the parallelograms about the diameter of a square are likewise squares. Proposition 5.-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 on the line between the points of section, is equal to the square on half the line. Let the straight line AB be bisected in C, and divided unequally in D; The rectangle AD, BD, together with the square on CD, shall be equal to the square on CB. CONSTRUCTION.-Upon CB describe the square CEFB (I. 46), and join BE. Through D draw DHG parallel to CE or BF (I. 31). 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). D B L H M G For =DF. 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). To each of these add CH; therefore the whole AH is. AH= cqual 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+CII. =AD DB. Therefore the gnomon CMG is equal to the rectangleAD, 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 f CD2. AD DB + CB2 = CD2. For AL CII = HF. ... AM or AD DB =CMG. Add to each LG or CB2. ...AD D3 + CB2 = CD2. AB2+BC2 =2AB 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 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 AÇ. But AK and CE are the gnomon AKF, together with the square CK; For AK=CE. Therefore the gnomon AKF, together with the square CK,... AKF is double of AK. 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. + 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 .. AB? HF, make up the whole figure ADEB and CK, which are the squares on AB and BC; 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; E + BC2 |