therefore the four figures HF, CK, AG, GE, are equal to the squares of AC, CB, together with twice the rectangle AC, CB; but HF, CK, AG, GE, make up the whole figure ADEB, which is the square of AB; therefore 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; then 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. (References-Prop. I. 31, 34, 36, 43, 46; II. 4, Cor.) Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts at the point D. Then the rectangle AD, DB, together with the square of CD, shall be equal to the square of CB. Upon CB describe the square CEFB, (1. 46) and join BE; through D draw DHG parallel to CE or BF; (1. 31) through H draw KLM parallel to CB or EF; and through A draw AK parallel to CL or BM. DEMONSTRATION Then, because the complement CH is equal to the complement HF, (1. 43) to each of these add DM; therefore the whole CM is equal to the whole DF; but CM is equal to AL, (1. 36) because AC is equal to CB; (hyp.) therefore also, AL is equal to DF; to each of these equals add CH, therefore the whole AH is equal to DF and CH. But AH is contained by AD, DB, since DH is equal to DB; and DF, together with CH, is the gnomon CMG; therefore the gnomon CMG is equal to the rectangle AD, DB; (ax. 1) to each of these equals add LG, which is equal to the square of CD, (11. 4, Cor.) since CD is equal to LH; therefore the gnomon CMG, together with LG, is equal to the rectangle AD, DB, together with the square of CD; but the gnomon CMG and LG make up the whole figure CEFB, which is the square of CB; therefore the rectangle AD, DB, together with the square of CD, is equal to the square of CB. Wherefore, if a straight line, &c. Q. E. D. Cor. From this proposition it is manifest that the difference of the squares of two unequal lines AC, CD, is equal to the rectangle contained by their sum and difference. PROP. VI.-THEOREM. If a straight line be bisected, and produced to any point; then the rectangle contained by the whole line thus produced, and the part of it produced, together with the square of half the line bisected, is equal to the square of the straight line which is made up of the half and the part produced. (References-Prop. I. 31, 36, 43; II. 4, Cor.) Let the straight line AB be bisected in C, and produced to D. CONSTRUCTION Upon CD describe the square CEFD, (1. 46) and join DE; through B draw BHG parallel to CE or DF; (1. 31) through H draw KLM parallel to AD or EF; and through A draw AK parallel to CL or DM, Because AC is equal to CB, (hyp.) therefore the rectangle AL is equal to CH, (1. 36) but CH is equal to HF; (1. 43) therefore also AL is equal to HF; to each of these add CM; therefore the whole AM is equal to the gnomon CMG. But AM is the rectangle contained by AD, DB, since DM is equal to DB; (11. 4, Cor.) therefore the gnomon CMG is equal to the rectangle AD, DB; add to each of these LG, which is equal to the square of CB; therefore the rectangle AD, DB, together with the square of 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 of CD; therefore the rectangle AD, DB, together with the square of CB, is equal to the square of CD. Wherefore, if a straight line, &c. Q. E. D. PROP. VII.- THEOREM. If a straight line be divided into any two parts; then the squares of the whole line and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. (References - Prop. I. 34, 43, 46; 11. 4. Cor.) Let the straight line AB be divided into any two parts in the point C, Then the squares of AB, BC, shall be equal to twice the rectangle AB, BC, together with the square of AC. Upon AB describe the square ADEB, (1. 46) and join BD; · DEMONSTRATION Then, because AG is equal to GE, (1. 43) add to each of them CK; therefore the whole AK is equal to the whole CE; and therefore AK, CE, are double of AK; but AK, CE, are the gnomon AKF, together with the square CK, therefore the gnomon AKF, together with the square CK, is double of AK; but twice the rectangle AB, BC, is 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; to each of these equals add HF, which is equal to the square of AC; therefore the gnomon AKF, together with the squares CK, HF, is equal to twice the rectangle AB, BC, and the square of AC; but the gnomon AKF, together with the squares CK, HF, make up the whole figures ADEB and CK, which are the squares of AB and BC; therefore the squares of AB and BC are equal to twice the rectangle AB, BC, together with the square of AC. Wherefore, if a straight line, &c. Q. E. D. PROP. VIII.-THEOREM. If a straight line be divided into any two parts; then four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line which is made up of the whole and that part. (References Prop. 1. 3, 31, 34, 36, 43, 46; п. 4, Cor.) Let the straight line AB be divided into any two parts in the point C. Then four times the rectangle AB, BC, together with the square of AC, shall be equal to the square of the straight line made up of AB and BC together. Produce AB to D, so that BD be equal to CB; (I. 3) upon AD describe the square AEFD; (1. 46) DEMONSTRATION Then, because CR is equal to BD, (constr.) and that CB is equal to GK, and BD to KN; (1. 34) therefore GK is equal to KN; for the same reason, PR is equal to RO; and because CB is equal to BD, and GK to KN, therefore the rectangle CK is equal to BN, and GR to RN; (1. 36) but CK is equal to RN, because they are the complements of the parallelogram CO; (1. 43) therefore also BN is equal to GR, and therefore the four rectangles BN, CK, GR, RN, are equal to one another, and so are quadruple of one of them, CK. Again, because CB is equal to BD, and that BD is equal to BK, that is CG; (1. 34) and because CB is equal to GK, that is GP; therefore CG is equal to GP; and because CG is equal to GP, and PR to RO, the rectangle AG is equal to MP, and PL to RF; |