G F CGB is equal to the angle GBC; and therefore the side bc is equal (1. 6.) to the side cg : But CB is equal (i. 34.) also to Gk, and co to BK; wherefore the H K figure 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; D and KBC is a right angle; wherefore GCB is a right angle; and therefore also the angles (1. 34.) 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 AG, which is equal to Ac: therefore HF, CK are the squares of AC, CB; and because the complement AG is equal (1. 43.) to the complement GE, and that AG is the rectangle contained by A C, 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. V. 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 co describe (1. 46.) the square CEFB, join be, and through a draw (1. 31.) DHG parallel to CE or bF; and through a draw KLM parallel to CB or EF; and also through A draw Ak parallel cL or BM; and because the complement cu is equal (1. 43.) to the complement , to each of these add DM; therefore the whole CM A С DB is equal to the whole DF; but cm is L H equal (1. 36.) to Al, because Ac is K M м equal to CB; therefore also al is equal to DF. To each of these add ch, and the whole Ah is equal to DF E G F and ch: But as is the rectangle contained by AD, BD, for du is equal (11. 4. Cor.) to DB; and 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 (II. 4. Cor.) to the square of cd, therefore the gnomon GMG, 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. THEOR. 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 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 Upon co describe (1. 46.) the square CEFd, join DE, and through B draw (1. 31.) BHG parallel to, CE or DF, and through a draw klm parallel to AD or EF, and also through A draw ak parallel to cl or DM: and because ac is equal to CB, the B D rectangle Al is equal (1. 36.) to ch; but ch is equal (1. 43.) to HF; M therefore also al is equal to AF: To each of these add cm; therefore the whole Am is equal to the gnomon G F CMG: And Am is the rectangle con A C L H K E tained by AD, DB, for DM is equal (11. 4. Cor.) to DB: 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. THEOR. H If a straight line be divided into any two parts, the squares of the whole line, and 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 point c; the squares of AB, BC are equal to twice the rectangle AB, BC, together with the square of AC. Upon A B describe (1. 46.) the square ADE B, and construct the figure as in the preceding propositions: And because AG is equal (1. 43.) to GE, add to each of them ck; the whole AK is therefore equal to the whole CE; A с B therefore AK, CE are double of AK; but AK, CE are the gnomon ARF, together with the square CK, therefore the gnomon K AKF, together with the square ck, is double of AK: But twice the rectangle A B, BC is double of Ak, for ek is equal (II. 4. Cor.) to Bc: Therefore the gnomon D F 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 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 A B, BC, together with the squares of AC. Wherefore, if a straight line, &c. Q. E. D. E PROP. VIII. THEOR. 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 of the other part, is equal to the square of the straight line wħich is made up of the whole and that part. Let the straight line AB be divided into any two parts in the point c; four times the rectangle A B, BC, together with the square of ac, is equal to the square of the straight line made up of a B and BC together. Produce AB to D, so that BD be equal to CB, and upon A D describe the square A EFD; and construct two figures such as in the preceding. Because cB is equal to BD, and that cB is equal (1. 34.) to gk, and BD to KN; 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, the rectangle ck is equal (1. 36.) to bn, and go to RN: But ck is equal (1. 43.) to RN, because they are the complements of the parallelogram co; therefore also en 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 ce is equal to BD, and that BD is equal (11. 4. Cor.) to BK, that is, to cg; and co equal C B D to Gk, that (11. 4. Cor.) is, to GP; therefore cg is equal to GP: and be- M N cause cg is equal to GP, and PR to RO, the rectangle ag is equal to MP, and X PL to RF: But mp is equal (1. 43.) toy PL, because they are the complements of the parallelogram ML; wherefore AG is equal also to RF; Therefore the È H L F four rectangles AG, MP, PL, RF are equal to one another, and so are quadruple of one of them AG; and it was demonstrated, that the four CK, BN, GR, and in are quadruple of ck: Therefore the eight rectangles which contain the gnomon AOH, are quadruple of AK : And because ak is the rectangle contained by AB, BC, for bk is equal to BC, therefore four times the rectangle A B, BC is quadruple of AK: But the gnomon A0 H was demonstrated to be quadruple of AK; therefore four times the rectangle AB, BC is equal to the gnomon Aon. To each of these add x H, which A K P R is equal (11. 4. Cor.) to the square of Ac; Therefore four times the rectangle AB, BC, together with the square of A C, 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. IX. THEOR. If a straight line be divided into two equal, and also into two unequal, parts; the squares of the two unequal parts are together double of the square of half the line, and of the square of the line between the points of section. Let the straight line AB be divided at the point c into two equal, and at d into two unequal parts : The squares of AD, DB are together double of the squares of AC, CD. From the point c draw (1. 11.) CE at right angles to AB, and make it equal to AC or CB, and join EA, EB; through D draw (1. 31.) DF parallel to CE, and through F draw FG parallel to AB; and join AF: Then, because Ac is equal to CE, the angle EAC is equal (1. 5.) to the angle A EC; and because the angle ACE is a right angle, the two others AEC, EAC together make one right angle (1. 32.); and they are equal to one another; each of them, E therefore, is half of a right angle. For the same reason each of the angles CEB, EBC is half a right angle; and therefore the whole AEB is a right angle: And a because the angle GEF is half a right A argle, and EGF a right angle, for it is equal (1. 29.) to the interior and opposite angle ECB, the remaining angle EFG is half a right angle; therefore, the angle GEF is equal to the angle EFG, and the side Eg equal (1. 6.) to the side GF : Again, because the angle at B is half a right angle and FDB a right angle, for it is equal (1. 29.) to the interior and opposite angle ECB, the remaining angle BFD is half a right angle; therefore the angle at B is equal to the angle BFD, and the side DF to (1. 6.) the side DB: And because AC is equal to CE, the square of Ac is equal to the square of CE; therefore the squares of AC, CE are double of the square of AC: But |