1. Let AD meet BC within the triangle. Then BC+BD2=2BC×BD+CD2 (a): And AB2 =AD2+BD2(b): Add these two equations, omitting BD from each side. Then AB2+BC2=AD2+CD2+2BCxBD: (c) But AC2 the last equation: there remains AB2+BC2-AC2=2BCX BD. 2. Let AD meet BC produced: then ACB is an obtuse angle; therefore AB2=AC2+BC2+2BCX CD (c). To both sides add BC2: then AB2+BC2= AC2+2BC+2BC×CD (d). But since BD is divided in C, BC2+BCXCD=BCX BD (e); and the doubles are equal; therefore 2BC+2BCXCD= 2BCXBD (ƒ). Now substitute this in the equation above: it makes AB2+BC2=AC2+2BC×BD. 3. If AC be itself the perpendicular on BC; then BC is intercepted between the angular point B and the foot of AC; and AB2+BC2=AC2+2BC×BC (b). Wherefore, in every Recite (a) p. 7,2; (c) p. 12, 2; (e) p. 3, 2; triangle, &c. Q. E. D. (b) p. 47, 1; (d) ax. 2; (ƒ) ax. 6. B 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. 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). H/P E 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 (7). 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. Recite (a) p. 31; k D. 47, 1; (b) p. 37; (e) pos. 2; (c) ax. 2; (f) p. 10; (h) p. 46-all of b. 1; (i) p. 5, of b. 2 ; 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 AB2= AE+BE2, and AC2-AE2+CE (b);- adding these equals, therefore, AB2+AC2=2ÁE2+BE2+ 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=2BD2+2DE2+2AE'. Now, the squares of AE, DE equal the square B DE C of AD (e). Therefore AB+AC2=2BD2+2AD2. Q. E. D. Recite (a) p. 12, 1; (d) p. 9, 2; (b) p. 47, 1; (e) p. 47, 1. (c) ax. 2; 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 (b)—of the alternate angles at B, D (c)— that BD is bisected in E;-and therefore (d), that AB2+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 Q. E. D. (b) p. 15, 1; (c) p. 27, 1; BOOK THIRD. Definitions. 1. Equal circles are those of which the diameters or radii are equal. 2. tangent is a straight line touching a circle, without cutting it, however produced. 3. A circle touches a circle when they meet and do not cut one another. 4. Chords are equidistant from the centre, when the perpendiculars drawn to them from the centre are equal: when the perpendicular is greater the chord is more remote. 5. The angle of a segment is the declination of its chord from the arc. 6. An angle is in a segment when the sides containing it are in the segment: and an angle is said to insist, or stand upon the arc intercepted between the sides containing the angle. 7. Similar segments of a circle are those which contain equal angles. DD Propositions. 1 P. To find the centre of a given circle (ABC). Construction. In the circle draw any chord AB, and bisect it in D (a); through D draw CE at right angles to AB (b); bisect CE in F (a); draw FA, FB (c). D Argument. In the triangle ADF, BDF, the sides AD, BD are made equal, and DF is common: also the angles at D are equal, being right angles (d): there- A fore the sides FA, FB are equal (e); and, being drawn from the bisectional point of CE, they are radii of the circle (ƒ) wherefore the point F is the centre sought. Recite (a) p. 10, 1; (b) p. 11, 1; (d) ax. 10, 1; Corollary. The line (e) p. 4, 1; (c) post. 1; E which bisects another at right angles in a circle, passes through the centre of a circle. 2 Th. If any two points (A, B) be taken in the circumference of a circle (ABC), the straight line (AB) which joins them, shall fall within the circle. Constr. Find the centre D (a), and draw radii from D to A, B, F; also draw DE perpendicular upon AB (b). A Ε ́ Β Argument. In the triangle DAE the exterior angle DEB exceeds the interior DAE (c), or its equal DBE (d); therefore the radius DB, or its equal DF (e), exceeds the side DE (ƒ); and so the part DF exceeds the whole DE, which cannot be admitted (g). Wherefore AB falls not without the circle; and it cannot fall upon the circumference because it is a straight line (h): it therefore falls within the circle. Recite (a) p. 1, 3; Q. E. D. (b) p. 12, 1; (d) p. 5, 1; (e) def. 15, 1; (c) p. 16, 1; (h) def 4, 1. 3 Th. If a diameter (CD) of a circle (ABC) bisect a chord (AB) not passing through the centre, the former will cut the latter at right angles; and the chord so cut shall be bisected (in F). Argument. Find the centre E (a), and join EA, EB. Then, in the triangles EAF, EBF the bases AF, BF are equal, by hyp.; EĂ, EB are equal radii, and EF is common: therefore the adjacent angles AFE, BFE are equal (b); and so CD bisects AB at right angles in F (c). Again, since CD cuts AB at right angles, the angles A at Fare equal (d); and so are the angles at A and B, which are opposite to equal radii (e); therefore the F angles AEF, BEF are equal (ƒ), and likewise the sides AF, FB; that is, the chord AB is bisected in F. Wherefore, if a diameter of a circle, &c. Q. E. D. (b) p. 8, 1; (c) p. 13, 1; (d) ax. 10, 1; (e) p. 5, 1; (f) p. 26, 32, or 4 b. 1. 4 Th. If in a circle (ABCD) two chords (AC, BD) cut each other (in E), which do not both pass through the centre, they do not bisect each other. Argument 1. Let F be the centre; then, if AC pass through it, BD cannot pass through F, by hyp.; therefore AC is not bisected by BD (a). Now 2. If neither of the chords pass through F, draw o FE from the centre to the sectional point. because FE affects to bisect AC, and also BD, the angles FEA, FEB assume to be right angles (b), and equal to each other (c); and so, a part equals the whole, which is impossible (d): therefore neither of the chords is bisected in E. 5 Th. If two circles (ABC, ABG cut each other, they shall not have the same centre. Argument. For, if possible, let E be the common centre; and let C be one of their sectional points; join CE, and draw a straight line EFG, to meet the circles in F and G. Then EC and EF, also EC and EG, are equal radii (a); and so, EF is equal to EG (b)—a part equal to the whole, which cannot be admitted (c). Wherefore E is not the common centre; and E is any point whatever. Hence, if two circles, &c. Recite (a) def. 15, 1; (b) ax. 1; (c) ax. 9. D Q. E. D. 6 Th. If two circles (ABC, CDE) touch each other internally, they shall not have the same centre. Argument. For, if possible, let F be the common centre, and C the point of contact: join FC, and draw a straight line FEB to meet the circles in E and B. Then FC and FE, also FC and FB are equal radii (a); and so, FE and FB are equal (6)-a part equal to the whole, which is impossible (c). Wherefore, F is not the common centre; nor is any other point. Therefore, if two circles, &c. Q. E. D. 7 Th. From any point (F) not the centre, in the diameter (AD) of a circle (ABCD), the greatest straight line drawn to the circumference passes through the centre (E) : and FD, the other part of the diameter, is the least: and other lines drawn from that point diminish as they recede from the greatest towards the least: also from the same point, only two equal straight lines can be drawn, one on each side of the diameter. |