* Let ABC be any triangle, and the angle at B one of its acute angles; and upon BC, one of the sides containing it, let fall the perpendicular* AD from the opposite angle: 12. 1. the square of AC, opposite to the angle B, is less than the squares of CB, BA, by twice the rectangle CB, BD. First, let AD fall within the triangle ABC: and because the straight line CB is divided into two parts in the point D, the squares of CB, BD are equal† to twice the rectangle contained by CB, BD, and the square of DC: To each of these equals add the B D + 7.2. square of AD; therefore the squares of CB, BD, DA are equal to twice the rectangle CB, BD, and the squares of ‡ 2 Ax. AD, DC: But the square of AB is equal to the squares of BD, || 47. 1. DA, because the angle BDA is a right angle; and the square of AC is equal to the squares of AD, DC; therefore the squares of CB, BA are equal to the square of AC, and twice the rectangle CB, BD; that is, the square of AC alone is less than the squares of CB, BA, by twice the rectangle CB, BD. Secondly, let AD fall without the triangle ABC: then, because the angle at D is a right angle, the angle ACB is greater than a right angle; and therefore the square of AB is equal to the squares of AC, CB, and twice the rectangle BC, CD: § 16. 1. ¶ 12. 2. B C D * 2 Ax, To these equals add the square of BC, and the squares of AB, BC are equal* to the square of AC, and twice the square of BC, and twice the rectangle BC, CD: But because BD is divided into two parts in C, the rectangle DB, BC is equal† to the rectangle BC, CD and the t 3. 2. square of BC; and the doubles of these are equal: therefore the squares of AB, BC are equal to the square of AC, and twice the rectangle DB, BC: therefore the square of AC alone is less than the squares of AB, BC, by twice the rectangle DB, BC. Lastly, let the side AC be perpendicular to BC; then is BC the straight line between the perpendicular and the acute angle at B: and it is manifest, that the squares of AB, BC, are equal to the square of AC and twice the square of BC. Therefore, in every triangle, &c. Q. E. D. B A 47. 1. and 2 Ax. F 3 45. 1. + 30 Def. 1 3. 1. | 10. 1. § 5. 2. ¶ 15 Def. * 47. 1. + 3 Ax. Constr. PROPOSITION XIV. PROB. To describe a square that shall be equal to a given rectilineal figure. Let A be the given rectilineal figure: it is required to describe a square that shall be equal to A. Describe a rectangular parallelogram BCDE equal to the rectilineal figure A. If then the sides of it, BE, ED, are equal to one another, it is af square, and what was required is now done: But if they are not B H equal, produce one of them BE to F, and make‡ EF equal to ED, and bisect|| BF in G; and from the centre G, at the distance GB, or GF, describe the semicircle BHF, and produce DE to H, and join GH: Therefore because the straight line BF is divided into two equal parts in the point G, and into two unequal at E, the rectangle BE, EF, together with the square of EG, is equal to the square of GF: But GF is equal to GH; therefore the rectangle BE, EF, together with the square of EG, is equal to the square of GH: But the squares of HE, EG are equal* to the square of GH; therefore the rectangle BE, EF, together with the square of EG, is equal to the squares of HE, EG: Take away the square of EG, which is common to both; and the remaining rectangle BE, EF is equal† to the square of EH: But the rectangle contained by BE, EF is the parallelogram BD, because EF is equal to ED; therefore BD is equal to the square of EH: But BD is equal to the rectilineal figure A; therefore the rectilineal figure A is equal to the square of EH. Wherefore a square has been made equal to the given rectilineal figure, A, viz. the square described upon EH. Which was to be done. BOOK III. DEFINITIONS. I. EQUAL circles are those of which the diameters are equal, or from the centres of which the straight lines to the circumferences are equal. "This is not a definition, but a theorem, the truth of which is evident; for, if the circles be applied to one another, so that their centres coincide, the circles must likewise coincide, since the straight lines from the centres are equal." II. A straight line is said to touch a circle, when it meets the circle, and being pro duced does not cut it. III. Circles are said to touch one another, which meet but do not cut one another. IV. Straight lines are said to be equally distant from the centre of a circle, when the perpendiculars drawn to them from the centre are equal. ས. And the straight line on which the greater perpendicular falls, is said to be farther from the centre. VI. A segment of a circle is the figure con- VII. "The angle of a segment is that which is contained by the straight line and the circumference." VIII. An angle in a segment is the angle contained by two in the circumference of the segment, IX. And an angle is said to insist or stand upon the circumference intercepted between the straight lines that contain the angle. * 10. J. t 11. 1. PROB. To find the centre of a given circle. Let ABC be the given circle; it is required to find its centre. Draw within it any straight line AB, and bisect* it in D; from the point D draw† DC at right angles to AB, and produce it to E, and bisect CE in F: the point F is the centre of the circle ABC. For if it be not, let, if possible, G be the centre, and join GA, GD, GB: then, because DA is equal to DB, and DG common to the two triangles ADG, BDG, the two sides AD, DG are equal to the two BD, DG, each to each; and the base GA is equal to the base GB, because they are drawn from the centre Ga: therefore the angle ADG is equal to the angle GDB: but when a straight line standing upon A D 8. 1. E another straight line makes the adjacent angles equal to one another, each of the angles is a right angle; therefore || 10. Def. 1. the angle GDB is a right angle: But FDB is likewise a § right angle; wherefore the angle § Constr. FDB is equal to the angle GDB, the greater to the less, ¶ 1 Ax. which is impossible: therefore G is not the centre of the circle ABC. In the same manner it can be shown, that no other point but F is the centre; that is, F is the centre of the circle ABC. Which was to be found. COR. From this it is manifest, that if in a circle a straight line bisect another at right angles, the centre of the circle is in the line which bisects the other. PROPOSITION II. THEOR.—If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle. Let ABC be a circle, and A, B any two points in the circumference; the straight line drawn from A to B shall fall within the circle. For if it do not, let it fall, if possible, without, as AEB: find* D the centre of the circle ABC; and join AD, DB, and produce DF, any straight line meeting the circumference AB, to E: Then, because DA is equal† to DB, the angle DAB is equal to the angle DBA ; And because AE, a side of the triangle DAE, is produced to B, the angle DEB is greater || than the angle DAE: B a N.B. Whenever the expression "straight lines from the centre" or "drawn from the centre" occurs, it is to be understood that they are drawn to the circumference. * 1. 3. + 15 Def. 1. 5.1. || 16. 1. |