C B H H iK F 4 Th. The square of a straight line (AB) is equal to the squares of any two parts (AC, CB) into which it may be divided, and two rectangles of the parts Constr. Upon AB describe the square AE (a); A join BD; draw CF parallel to AD, meeting BD in G; draw HGK parallel to AB (6). Argument. The angles CGB, ADB, as exterior and interior, are equal (c); and ADB equals ABD, from the isosceles (d); therefore, ABD equals CGB (e), and the sides CG, CB are equal (f), and their opposites BK, GK are equal (8). CK is o therefore a square (h) on CB. For similar reasons, HF is a square on HG, which is equal to AC. It is also plain that AG is the rectangle of AC and CG, or CB; also, that GE is the rectangle of GF and GK, which are equal to AC and CB. Now the two squares and two rectangles make up the square of AB. Wherefore, the square, &c. Q. E. D. Recite (a) p. 46; (6) p. 31; (c) p. 29; (d) p. 5; (e) ax. 1; (f) p. 6; (8) p. 34-all of b.i: (h) def. 30 of b. 7, and 1 of b. 2. Cor. The rectangles about the diameter of a square are squares. Scholium. A line or other magnitude divided into two parts, is called a binomial, the properties of which are very remarkable. D3 1-1 5 Th. If a straight line (AB), be divided into two equal parts (in C), and two unequal parts (in D), the square of half the line is equal to the rectangle of the unequal parts and the square of the line between the sectional points. Upon CB, half the line, describe the A square CF (a); join BE; through D, draw DG parallel to BF (b), meeting BE in H; M к through H, draw ML parallel to AB, and produce it to meet AK drawn parallel to CE. The square CF equals the gnomon CMG and the square LG. But the gnomon is equal to the rectangle AH; for the complements CH, HF are equal (c); and with DM added to each, CM equals DF; but CM equals AL, (d); therefore DF equals AL, (e). Again, AH is the rectangle of the unequal parts AD, DB; for DH equals DB ($): LG is also the square of LH, or CD, the line between the sectional points. Therefore, the square of CB, is equal to the rectangle of AD, DB and the square of CD. Wherefore, if a straight line, &c. Q. E. D. Recite (a) p. 46; (6) p. 31; (c) p. 43; (d) p. 36; (e) ax. 1-all of b. 1 ; (f) cor. p. 4 of b. 2. E C F 6 Th. If a straight line (AB) be bisected (in C), and produced to any point (D); the square of the line (CD), composed of the half and the part produced, is equal to the rectangle of the whole line AD and its produced part, (BD), with the square of (CB), half the line bisected. H M M с F Upon CD describe the square CF (a); A BD join DE; through B, draw BG parallel to DF, and meeting the diameter in H; through H, draw ML parallel to AB, and produce it K to meet AK drawn parallel to CE (6). The square CF is equal to the gnomon CMG and the square LG: this is evident. But the gnomon is equal to the rectangle AM: for the rectangles AL and CH are equal (c); also CH and HF (d); therefore AL equals HF (e). Hence the square CF equals the rectangle AM and the square LG. But AM is the rectangle of the whole line AD and its produced part BD: for DM or BH equals BD (f). Also, LG is the square of LH, or CB, half the bisected line. Wherefore, the square of CD equals the square of CB and the rectangle of AD and DB. Q. E. D. Recite (a) p. 46; (b) p. 31; (c) p. 36 ; (d) p. 43; (e) ax. 1; (f) p. 6, all of b. 1, and cor. p. 4, of b. 2. A 7 Th. The two squares described upon a straight line (AB) and a part of that line (BC) are equal to two rectangles of the line and that part, together with the square of the other part (AC). Constr. Upon AB describe the square AE (a), join BD; through C draw CF parallel to BE, and cutting the diameter in G; through G draw HK parallel to AB (6). The squares AE, CK are equal to the rectangles AK, CE and the square HF. This is evi. dent. Now AE is the square of AB, and CK is the F E square of BC. Also, AK is the rectangle of AB and BK, or BC; and CE is the rectangle of BC and BE, or BA; and HF is the square of HG, or AC. Wherefore, the two squares described, &c. Q. E. D. Recite (a) p. 46 of b. 1 ; (6) p. 31 of b. 1. Cor. Hence the sum of the squares of any two lines, is equal to twice the rectangle of the two with the square of their difference CBD 8 Th. If a straight line (AB), be divided into any two parts (in C), four rectangles of the whole line and one part (CB), together with the square of the other part (AC), are equal to the square of the line (AD), composed of the whole and the one part. Produce AB, so that BD be equal to CB; upon X! The complements AK, KF are equal; as also the complements MP, PL .(c); to these latter equals add the equals GR, BN-MR is therefore equal to PL and BN; and AK, KF, MR, PL and BN-namely, the gnomon AOH (d), are equal to four rectangles of AB and BC: to this gnomon add XH, which is the square of XP, or AC. All these, therefore, coincide, and fill the same space, with the figure AEFD, which is the square of AD, composed of AB and BC. Wherefore, if a straight line be divided, &c. Q. E. D. (b) p: 31, 1; (c) p. 43, 1; (d) def. 2, 2. Cor. 1. Four rectangles of any two lines, together with the square of their difference, are equal to the square of their sum; for, in this case, AD is the sum and AC the difference of AB, BC Cor. 2. The square of any line is equal to four times the squaro of its half. HL 9 Th. If a straight line (AB) be equally divided in C) and unequally in D); the squares of the two unequal parts (AD, DB), are double of the squares of half the line (AC) and of the lino (CD) between the points of section. Constr. From C draw CE at right angles to AB (a) and equal to CA, or CB; join EA, EB; draw DF parallel to CE (6); join XF, and draw FG parallel to AB (6): Argument. The sides CE, CA, CB being made equal, their opposite angles are equal (c); and on Ā account of the right angles at C, the two angles at E make one right angle (d). Again, the parallels CE, DF make the exterior angle DFB equal to the interior CEF, or its equal B (e): wherefore DF equals DB (f). Hence the side AF subtends two right angles, one at D and the other at E; and the square of AF equals the squares of the sides contair. ing each of those angles (g): therefore the squares of AE, EF are equal to those of ADD, F, or AD, DB (h). But the square of AE equals the squares of AĆ, ČE, or twice that of AC; and the square of EF equals the squares of GE, GF, or twice that of GF, or its equal CD. Wherefore also, the squares of AD, DB are double the squares of AC and CD. Therefore, if a straight line, &c. Q. E. D (d) p. 32; (e) p. 29; (f) p. 6; B D 10 Th. If a straight line (AB) be bisected (in C), and produced to any point (D); the square of this whole line (AD), and the square of its produced part (BD), are equal to two squares of (AC) half the bisected line, and two squares of (CD) the line composed of the half and the production. From C draw CE at right angles to AB (a) and equal to CA, or CB; make EF equal and parallel to CD-then will FD be equal and parallel to CE (6); join EA, EB, and producing EB, FD, they will meet in G, because the angles F and GEF are less than two right angles C; join also AG. Because the triangles ACE, BCE are right angled, the ci right angle is equal to half the sum of the angles in each (d), and because they are isosceles the angles opposite to equal sides are equal to each other (e), and each is halt a right angle; therefore, AEB is a right angle. Also, since CBE is half a right angle, the alternate angle FEG is the same (f ), and so is also the vertical opposite angle DBG (8). Again, because ECD is a right angle, the alternate angle CDG is the same: therefore DGB is half a right angle; and the sides DG and DB are equal; also the sides FG and FE (k). AG therefore subtends two right angles; 'namely, AEG, ADG; and its square is equal to that of any two sides containing those angles (i): Wherefore, the squares of AE and EG are equal to the squares of AD and DG=B, each pair being equal to the square of AG (k). But the square of AE equals the squares of AC, CE, or two squares of AC; and the square of EG equals the squares EF, FG, or two squares of EF=CD. Therefore, the squares of AD, BD are equal to two squares of AC and two of CD. Q. E. D. Recite (a) p. 11; (b) p. 31, 34; (c) p. 29; (d) p. 32; (e) p. 5; (f) p. 27; (8) p. 15; (h) p. 6; (i) p. 47; (k) ax. 1-all of b. 1. 11 P. To divide a given straight line (AB) into two parts in H), so that the rectangle of the whole line and one of the parts (HB) shall equal the square of the other part (AH). F K Constr. Upon AB describe the square AD (a) B Argument. The straight line AC being bisected in E and produced to F, the rectangle CFXFA and the square of AE are together equal to the square of EF (à), or its equal EB, or the squares of AE and C AB (e). Take the square of AE from both; then the rectangle CFXFA—that is, FK, is equal to the square of AB, viz. AD (Š). Take also from the rectangle and square the part AK, which is common; the remainders are equal; namely, AG to HD ($). But AG is the square of AH; and HD is the rectangle of HB XBD, or HB XAB. The straight line AB is therefore divided as required. Q. E. D. Recite (a) p. 46 ; (6) p. 10; (c) p. 3 of b. 1; (d) p. 6, of b. 2; (e) p. 47; (f) ax. 3 of b. l. 12 Th. In any obtuse angled triangle, if a perpendicular be drawn from one of the acute angles to the opposite side produced; the square of the side subtending the obtuse angle equals the two squares of the sides containing it, and two rectangles of the produced side and its production. Let the triangle ABC be obtuse angled at C; produce BC to meet a perpendicular from A, in the point D. The production is CD, and AB subtends the obtuse angle. Now AB2=AD2 +BD? (a). Therefore, omitting CD', which is both positive and negative, AB2=AC2+ BC2+2BCXCD (c). Wherefore, in any obtuse angled triangle, &c. Q. E. D. (c) ax. 1. 13 Th. In any triangle (ABC), the square of the side (AC) subtending one of the acute angles (B) is less than the squares of the two containing sides (AB, BC), by two rectangles of one of them (BC) and a segment of the same intercepted between the said angular point and a perpendicular (AD) drawn to that side from the opposite angle (A). |