B DEC * 11. 1. 3. 1. * 31. 1. contained by the straight lines A, BC shall be equal to the rectangle contained by A, BD, together with that contained by A, DE, and that contained by A, EC. KLH From the point B draw* BF at right angles to BC, and make BG equal * to A; and through G draw* GH parallel to BC; and through D, E, C, draw* DK, EL, CH pa- * 31. 1. rallel to BG. Then the rectangle BH is equal to the rectangles BK, DL, EH: but BH is contained by A, BC, for it is contained by GB, BC, and GB is equal t to A; and BK is contained by A, BD, for it is + Constr. contained by GB, BD, of which GB is equal to A; and DL is contained by A, DE, because DK, that is * BG, * 34.1. is equal to A; and in like manner the rectangle EH is contained by A, EC: therefore the rectangle contained by A, BC, is equal to the several rectangles contained by A, BD, and by A, DE, and by A, EC. Wherefore, if there be two straight lines, &c. Q. E. D. PROP. II. THEOR. If a straight line be divided into any two parts, the rect angles contained by the whole and each of the parts, are together equal to the square of the whole line. Let the straight line AB be divided into any two parts in the point C: the rectangle contained by AB, BC, together with the rectangle | AB, AC shall be equal to the square of AB. Upon A B describe * the square ADEB, and through C draw * CF, parallel to AD or BE. Then AE is equal to the rectangles AF, CE: but AE is the square of AB; and AF is the rectangle contained by BA, AC; for it is contained by DA, AC, of which AD is equal + to AB: and CE is contained + 30 Def. by AB, BC, for BE is equal to AB: therefore the rectangle Aç, contained by AB, AC, together with the rect C B * 46.1. * 31. 1. D F E # N.B.To avoid repeating the word contained too frequently, the rectangle contained by two straight lines AB, AC is sometimes simply called the rectangle AB, AC. angle AB, BC, is equal to the square of AB. If therefore a straight line, &c. Q. E. D. PROP. III. THEOR. 46. 1. с B If a straight line be divided into any two parts, the rect angle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part. Let the straight line AB be divided into any two parts in the point C: the rectangle AB, BC shall be equal to the rectangle AC, CB, together with the square of BC. Upon BC describe * the square CDEB, and produce ED to F, and through A draw * AF parallel to CD or BE. Then the rectangle AE is equal to the rectangles AD, CE: but AE is the rectangle contained by AB, BC, for it is contained by AB, BE, of which BE is equal + to BC; and AD is contained by AC, CB, for CD is equal to CB; and DB is the square of BC: therefore the rectangle AB, BC, is equal to the rectangle AC, CB, together with the squarc of BC. If therefore á straight line, &c. Q. E. D. 31.1. E † 30 Def. PROP. IV. THEOR. If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the trvo 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 of AB shall be equal to the squares of AC, CB, and to twice the rectangle contained by AC, CB. Upon AB describe * the square ADEB, and join BD, and through C draw* CGF parallel to AD or BE, and through G draw HK parallel to AB or DE. And because CF is parallel + to AD, and BD falls upon them, the exterior angle BGC is equal * to the interior and opposite angle ADB; but ADB is equal * to the angle ABD, becausc BA is equal to ADĚ, being sides of a * 46. 1. * 31. 1. + Constr. * 29.1. ** 5. 1. † 30 Der. С В * 6. 1. K K 34. 1. F E square; wherefore the angle CGB. is equal t to the angle CBG; and therefore + 1 Ax. the side BC is equal * to the side CG: but CB is equal * also to GK, and CG H to BK; wherefore the figure CGKB is + equilateral: it is likewise rectangular; D for, since CG is parallel to BK, and CB meets them, therefore the angles KBC, GCB are equal + to two right angles: but KBC is a + right angle; † 29. 1. wherefore GCB is a † right angle: and therefore also | s0 Def. the angles * CGK, GKB, opposite to these, are right 134. 1. angles; and therefore CGKB is rectangular: but it is and 1 Ax. also equilateral, as was demonstrated; wherefore it is a t square, and it is upon the side CB: for the same + 50 Def. reason HF also is a square, and it is upon the side HG, which is equal + to AC: therefore HF, CK are the † 34. 1. squares of AC, CB: and because the complement AG is equal * to the complement GE, and that AG is the * 43. 1. rectangle contained by AC, CB, for GC is equal t to † 30 Def. CB; therefore GE is also equalt to the rectangle AC, † 1 Ax. 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 t to the squares of AC, † 1 Ax. CB, and twice the rectangle AC, CB. Wherefore, if a straight line, &c. Q. E. D. Cor. From the demonstration, it is manifest, that parallelograms about the diameter of a square are like wise squares. PROP. V. THEOR. into two unequal parts; the rectangle contained by the 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, shall be equal to the square of CB. Upon CB describe * the square CEFB, join BE, and * 46. 1. through D draw * DHG parallel to CË or BF; and * 31. 1. through H draw KLM parallel to CB or EF; and also * 43. 1. * 36. 1. С DB M К. GF through A draw AK parallel to CL or BM. And be- HF, to each of these add DM; therefore the whole CM + 2 Ax. is equalt to the whole DF: but CM is equal * to AL, + Hyp. because AC is equalt to CB; therefore also AL is equalt to DF: to each of these add CH, + 2 Ax. and the whole AH is equalt to DF contained by AD, DB, for DH is * Cor. 4. 2. equal * to DB; and DF together & 30 Def. with CH is the gnomon CMG; therefore the gnomon CMG is equal + to the rectangle * Cor. 4. 2. AD, DB: to each of these add LG, which is equal * to and 34. 1. the square of CD; therefore the gnomon CMG, toge ther with LG is equal+ to the rectangle AD, DB toge- From this proposition it is manifest, that the differ- +1 Ax. + 2 Ax. ference. PROP. VI. THEOR. If a straight line be bisected, and produced to any point ; the rectangle contained by the whole line thus produced, Let the straight line AB be bisected in C, and pro- Upon CD describe * the square CEFD, join DE, G F * 46. 1. * 31.1. C B D + Hyp. M is equal + to the gnomon CMG: but AM is the rect- + 2 Ax. PROP. VII. THEOR. A If a straight line be divided into any two parts, the squares of the whole line, and of one of the parts, are equal to Let the straight line AB be divided into any two Upon AB describe * the square ADEB, and con- * 46.1. struct the figure as in the preceding propositions. And because AG is equal * to GE, add to each of them CK; * 43. 1. therefore the whole AK is equal to the whole CE; 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; therefore the gnomon AKF, together with the square CK, is equal + & 30 Def. to twice the rectangle AB, BC: to each of these equals add HF, which is equal t to the square of AC: there- † Cor. 4. 2. fore the gnomon AKF, together with the squares CK, HF, is equal + to twice the rectangle, AB, BC, and the + 2 Ax. square of AC: but the gnomon AKF, together with the squares CK, HF, make up the whole figure 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. H + 1 Ax. D * Cor. 4.2. & 34. 1. |