From the point A draw (11. 1.) AD at right angles to AC, and make AD equal to BA, and join DC : then because DA is equal to AB, the square of DA is equal to the square of D AB: to each of these add the square of AC : therefore the squares of DA, AC are equal to the squares of BA, AC: but the square of DC is equal (47. 1.) to the squares of DA, AC, because DAC 4 is a right angle; and the square of BC, by hypothesis, is equal to the squares of BA, AC; therefore the square of DC is equal to the square of BC; and therefore also the side DC is equal to the side BC. And because the side DA is equal B C to AB, and AC common to the two triangles DAC, BAC, the two DA, AC are equal to the two BA, AC: and the base DC is equal to the base BC: therefore the angle DAC is equal (8.1.) to the angle BAC: but DAC is a right angle; therefore also BAC is a right angle. Therefore, if the square, &c. Q. E. D. THE E L E MENT'S OF EU C LID. BOOK II. I EVERY right angled parallelogram is said to be contained by any two of the straight lines which contain one of the right angles. II. In every parallelogram, any of the parallelograms about a diameter, together with the two complements is E called a gnomon. ‘Thus the paral-A —— D “lelogram HG, together with the com- PROP. I. THEOR. If there be two straight lines, one of which is divides into any number of parts; the rectangle contained by the two straight lines, is equal to the rectangles contained by the undivided line, and the several parts of the divided line. Let A and BC be two straight lines; and let BC be divided into any parts in the points D, E: the rectangle contained by the straight lines A, BC is equal to the rectangle B D E C contained by A, BD, together with that ---T— contained by A, DE, and that contained by A, EC. From the point B draw (11. 1.) BF at right angles to BC, and make BG equal (3. 1.) to A; and through G| G draw (31. 1.) GH parallel to BC; and | K 1. H through D, E, C draw (31. 1.) DK, A EL, CH parallel to BG: then the rectangle BH is equal to the rect- F angles BK, DL, EH ; and BH is contained by A, BC, for it is contained by GB, BC, and GB is equal to A.; and BK is contained by A, BD, for it is contained by GB, BD, of which GB is equal to A; and DL is contained by A, DE, because DK, that is (34. 1.) BG 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 also 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 rectangles 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 A C B two parts in the point C; the rectangle contained sby AB, BC, together with the rectangle+ AB, AC, shall be equal to the square of AB. Upon AB describe (46. 1.) the square ADEB, and through C draw (31. 1) CF, parallel to AD or BE; then AE is equal to the rectangles AF, | CE: and AE is the square of AB: and AF is the rectangle contained by BA, AC; for it is contain- 1—--ed by DA, AC, of which AD is equal to AB; and D CE is contained by AB, BC, for BE is equal to AB; therefore the rectangle contained by AB, AC, together with the rectangle AB, BC, is equal to the square of AB. If therefore a straight line, &c. Q. E. D. PROP. III. THEOR. If a straight line be divided into any two parts, the rectangle 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 foresaid part. Let the straight line AB be divided into two parts in the point C; the rectangle AB, BC is equal to the rectangle AC, CB together with the square of BC. Upon BC describe (46. 1.) the square A C B 'CDEB, and produce ED' to F, and T' CD or BE; then the rectangle AE is to BC; and DB is the square of BC; F D E equal to the rectangles AD, CE; and : AE is the rectangle contained by AB, BC, for it is contained by AB, BE, of * N. B. To avoid repeating the word contained too frequently, the rectangle cono by two straight lines AB, AC is sometimes simply called the rectangle AB, AC. therefore the rectangle AB, BC, is equal to the rectangle AC, CB, together with the square of BC. If therefore a straight line, &c. Q. E. D. 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 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 of AB is equal to the squares of AC, CB, and to twice the rectangle contained by AC, CB. Upon AB describe (46. 1.) the square ADEB, and join BD, and through C draw (31. 1.) 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 (29. 1.) to the interior and opposite angle ADB; but ADB is equal (5. 1.) to the angle ABD, because BA is equal to AD, being sides of a square; wherefore the angle CGB is equal to the A C B angle GBC; and therefore the side BC is equal (6. 1.) to the side CG ; but CB is wherefore the figure CGKB is equilateral; ~f~--— K 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; where- —--—" fore GCB is a right angle: and therefore D F E also the angles (34.1.) 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 C.B.: for the same reason HF also is a square, and it is upon the side HG, which is equal to AC : therefore HF, CK are the squares of AC, CB; and because the complement AG is equal (43. 1.) 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 parallelograms about the diameter of a square are likewise squares. PROP. W. THEOR. 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 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 (46. 1.) the square CEFB, join BE, and through D draw (31. 1.) DHG parallel to CE or BF; and through H draw KLM parallel to CB or EF; and also through A draw AK parallel to CL or BM; and because the complement CH is equal (43. 1.) to the complement HF, to each of these add DM; therefore the whole CM is equal A C D B to the whole DF; but CM is equal (36. 1.) to AL, because AC I, is equal to CB; therefore also K AL is equal to DF. To each of M these add CH, and the whole AH is equal to DF and CH: but AH is the rectangle contained by AD, DB, for DH is equal (Cor. 4. 2.) to DB, and E G F DF together with CH is the gnomon CMG ; therefore the gnomon CMG is equal to the rectangle AD, DB : to each of these add LG, which is equal (43. 1.) 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 gnomon CMG and LG makes 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. 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. THEOR. lf 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 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. Upon CD describe (46. 1.) the square of CEFD, join DE, and through B draw (31. 1.) BHG parallel to CE or DF, and through |