c. 29. I. d. S.I. Book II. Upon AB describe the square ADEB, and join BD, and thro mc draw 6 CGF parallel to AD or BE, and thro'G draw HK parallel a. 46, s. to AB or DE, and because CF is parallel to AD, and BD falls upon b. 31. 1. them, the exterior angle BGC is equal to the interior and oppo site angle ADB ; but ADB is equal d to the angle ABD, because BA is equal to AD, being sides of a square; wherefore the angle CGB is equal to the angle GBC, and A с B C. 6. 1. therefore the side B C is equal to the f. 34. 1. fide CG. but CB is equal also f to GK, and CG to BK; wherefore the figure H K K CGKB is equilateral. it is likewise rectangular; for CG is parallel to BK, and CB meets them, the angles KBC, GCB are therefore equal to two right angles ; and KBC is a right angle, D F E wherefore GCB is a right angle ; and therefore also the anglese CGK, GKB opposite to these are right angles, and CGKB is rectangular. but it is also equilateral, as was demonstrated; wherefore it is a square, and it is upon the side CB. for the same reason HF also is a square, and it is upon the side HG which is equal to AC. there fore HF, CK are the squares of AC, CB. and because the comple8.43.1. ment AG is equal to the complement GE, and that AG is the rectangle contained by AC, CB, for GC is equal to CB; therefore GE is also equal to the rectangle AC, CB; wherefore AG, GE are equal to twice the rectangle AC, CB. and HF, CK are the squares of AC, CB; wherefore the four figures HF, CK, AG, GE are equal to the squares of AC, CB and to 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 parallelo: grams about the diameter of a square are likewise squares. PROP. Book II, PROP. V. THEOR. also into two unequal parts; 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. Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts at the point D; the rectangle AD, DB together with the square of CD, is equal to the square of CB. Upon CB describe the square CEFB, join BE, and thro’D draw a: 46. I. DHG parallel to CE or BF; and thro' H draw KLM parallel to b. 31. 1. CB or EF; and also thro' A draw AK parallel to CL or BM. and because the complement CH is equal to the complement HF, to c. 43. 1. each of these add DM, A С D B therefore the whole CM is equal to the whole DF; L H but CM is equal to AL, KY M d. 36. I. because AC is equal to CB; therefore also AL is equal to DF, to each of these add CH, and the whole AH is E equal to DF and CH. but AH is the rectangle contained by AD, DB, for DH is equal to DB; and DF together with CH is the c. Cor. 4. gnomon CMG; therefore the gnomon CMG is equal to the rectangle AD, DB. to each of these add LG, which is equal to the square of CD, therefore the gnomon CMG together with LG is equal to the rectangle AD, DB together with the square of CD. but the gromon 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 traight line, &c. Q. E. D, PROP IF Book II. PROP. VI. THEOR. 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 of half of the line bisected, is equal to the square of 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 the point D; the rectangle AD, DB together with the square of CB, is equal to the square of CD. a. 46. 1. Upon CD describe the square CEFD, join DE, and thro' B b. 31. 1. draw 6 BHG parallel to CE or DF, and thro' H draw KLM parallel to AD or EF, and also thro' A draw AK parallel to CL or DM. and C. 36.1. because AC is equal to CB, the rectangle AL is equal to CH; bat d. 43. 1. CH is equal to HF; there A С B D fore also AL is equal to HF. to each of these add CM, L Н. therefore the whole AM is e M qual to the gnomon CMG. K and AM is the rectangle con tained by AD, DB, for DM is e. Cor. 4. 7. equal to DB. therefore the E gnomon CMG is equal to G F 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. THEOR. a straight line be divided into any two parts, 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. Let the straight line AB be divided into any two parts in the I point Ć; the squares of AB, B C are equal to twice the rectangle Book II. AB, BC together with the square of AC. Upon AB describe the square ADEB, and construct the figure 2.46.1. as in the preceeding Propofitions, and because AG is equal 6 to GE, 6.43. 1, add to each of them CK; the whole AK is therefore equal to the whole CE; therefore AK, CE are double of AK. but AK, CE are the A С B K C. Cor.4.11 D to twice the rectangle AB, BC. to F E THEOR: PROP. VIII. times the rectangle contained by the whole line, and up of the whole and that part. Let the straight line A B be divided into any two parts in the point C; four times the rectangle AB, BC, together with the square of AC, is 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, and upon AD describe the square AEFD; and construct two figures such as in the preceeding. Because CB is equal to BD, and that CB is equal to a. 34. 1. GK, and BD to KN; therefore GK is equal to KN. for the same D reafon Book II. reason PR is equal to RO. and because CB is equal to BD, and MGK to KN, the rectangle CK is equal to BN, and GR to RN, but b. 36. 1. CK is equal to RN, because they are the complements of the pa€ 43. 1. rallelogram CO; therefore also BN is equal to GR. and the four rectangles BN, CK, GR, RN, are therefore equal to one another, and so are quadruple of one of them CK. again, because CB is equal d. Cor.4.2. to BD, and that BD is equal a to BK, that is to CG; and CB equal C B D M N P RO C. 43. 1. and PL to RF. but MP is equal to PL, because they are the com- H LF equal to the gnomon AOH. to each of these add XH, which is Cor. 4.2. equal d to the square of AC; therefore four times the rectangle AB, BC together with the square of AC is equal to the gnomon AOH and the square XH. but the gnomon AOH and XH make up the figure AEFD which is the square of AD. therefore four times the rectangle AB, BC together with the square of AC is equal to the square of AD, that is, of AB and BC added together in one Straight line. Wherefore if a straight line, &c. Q. E. D. PROP |