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 A CBD CK 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 F E HL and BC. 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 square of its half. 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 line (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 AF, and draw FG parallel to AB (6). Argument. The sides CE, CA, CB being made equal, their opposite angles are equal (c); and on A account of the right angles at C, the two angles at E make one right angle (d). E C F CD B 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 contain. 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 AC, CE, 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 Therefore, if a straight line, &c. of AC and CD. Q. E. D Recite (a) p. 11; (b) p. 31; (d) p. 32; (c) p. 5; (e) p. 29; (f) p. 6; (g) p. 47; (h) ax. 1-all of b. 1. 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. A C B D F E Because the triangles ACE, BCE are right angled, the c 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 half 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 (g). 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 (h). 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-BD, 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. 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 Constr. Upon AB describe the square AD (a) Argument. The straight line AC being bisected in E and produced to F, the rectangle CFXFA and E the square of AE are together equal to the square of EF (d), or its equal EB, or the squares of AE and C K D 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 (f). Take also from the rectangle and square the part AK, which is common; the remainders are equal; namely, AG to HD (f). But AG is the square of AH; and HD is the rectangle of HBXBD, or HBXAB. The straight line AB is therefore divided as required. Q. E. D. (c) p. 3 of b. 1; (e) p. 47; 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 AB=AD+BD2 (a). And ADAC2-CD2 (a). LOR Also BD2=BC+CD+2BCCD (b). Therefore, omitting CD2, which is both positive and negative, AB=AC+BC2+2BCXCD (c). Wherefore, in any obtuse angled triangle, &c. Recite (a) p. 47 of b. 1; (b) p. 4 of b. 2; (c) ax. 1. BC D Q. E. D. A 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). 1. Let AD meet BC within the triangle. Then BC2+BD2=2BCXBD+CD2 (a): And AB2 =AD+BD2(b): Add these two equations, omitting BD2 from each side. Then AB+BC=AD+CD2+2BCXBD: (c) the last equation: there remains AB+BC2-AC2=2BCXBD. A BDC 2. Let AD meet BC produced: then ACB is an obtuse angle; therefore AB=AC+BC+2BCX A CD (c). To both sides add BC2: then AB2+BC2= AC2+2BC2+2BC×CD (d). But since BD is divided in C, BC2+BCXCD=BCXBD (e); and the doubles are equal; therefore 2BC2+2BCXCD= 2BCXBD (f). Now substitute this in the equation above: it makes AB+BC2-AC2+2BCXBD. 3. If AC be itself the perpendicular on BC; then BC is intercepted between the angular point B and the foot of AC; and AB+BC=AC2+2BCXBC DCB A 14 P. To describe a square that shall be equal to the given rectilineal figure (ABCD). Constr. 1. Join the points A, C, of the given figure; and parallel to AC draw DE (a) to meet the base BC produced in É; join AE. L A F Argument 1. The triangles ACD, B ACE upon the same base AC, and between the parallels AC, DE are equal (b): to each add the triangle ABC: then the quadrilateral ABCD equals the trilateral ABE (c). Constr. 2. Transfer the base BE to GH: place the triangle FGH at the same parallel distance as that of ABE; make FGH a right angle (d); upon GK, half the base GH, describe the rectangle GL (d): then, if the sides of GL are equal, the thing is done; for it is a square, and equal to the given figure. But if the sides are unequal, produce LK so that KM shall equal KG (e); bisect LM in O (f), at which centre describe the arc LPM (g); produce GH to meet the circle in P; join OP, and upon KP describe the square KS (h). Argument 2. The straight line LM being divided equally in O and unequally in K, the rectangle LKXKM, with the square of OK, is equal to the square of OM, or OP (i), or to the squares of OK, KP (k). From these equals reject the square of OK, the rectangle LKX KM is equal to the square of KP (1). But the rectangle LK, KM= KG, equals the triangle FGH, or ABE, or the figure ABCD. Therefore the square of KP, that is, KS, is equal to the same figure; and so, the thing is done which was required. (c) ax. 2; (d) p. 11, 42; (e) pos. 2; (f) p. 10; (g) pos. 3; (h) p. 46-all of b. 1; (i) p. 5, of b. 2; κ` ο. 47, 1; (1) ах. 3. A Th. If one side (BC) of a triangle be bisected (in D), the sum of the squares of the other two sides (AB, AC), is double of the square of that side, and of the square of the line (AD) drawn from the bisectional point to the opposite angle. Draw AE at right angles to BC (a); then AB AE2+BE, and AC=AE+CE2 (b); - adding these equals, therefore, AB+AC2=2AE+BE+ CE2 (c). But since the straight line BC is cut equally in D, and unequally in E, the squares of BE and CE are equal to double squares of BD and DE (d). Therefore AB2+AC2=2BD+2DE2+2AE. Now, the squares of AE, DE equal the square B of AD (e). Therefore AB2+AC2=2BD2+2AD2. Recite (a) p. 12, 1; (d) p. 9, 2; (b) p. 47, 1; (c) ax. 2; (e) p. 47, 1. A DE C Q. E. D. B Th. In any parallelogram (ABCD), the sum of the squares of the diameters (AC, BD) is equal to the sum of the squares of its sides, (AB, AD, CB, CD.) It is obvious from the equality of the opposite sides and angles of the parallelogram (a) of the vertical angles at E (6) of the alternate angles at B, D (c)that BD is bisected in E;-and therefore (d), that AB+AD2=2AE2+2ED2, and that CB2+CD2=2CE2+2EB2. Therefore, the squares of the four sides are equal to four squares of the half of each diameter. But four squares of the half equal the square of the whole (e): therefore the squares of the two diameters equal the squares of the four sides of the parallelogram. Recite (a) p. 34, 1; (d) p. A, 2; A C B |