DEMONSTRATION Then BC is 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. PROP. XIV. - PROBLEM. To describe a square that shall be equal to a given rectilineal figure. (References - Prop. I. 3, 10, 45, 47; 11. 5.) Let A be the given rectilineal figure. It is required to describe a square that shall be equal to A. Describe the rectangular parallelogram BCDE equal to the rectilineal figure A. (1. 45.) If, then, the sides of it, BE, ED, are equal to one another, it is a square, and what was required is now done. But if they are not equal, produce one of them, BE to F, and make EF equal to ED; bisect BF in G, (1. 10) and from the centre G, at the distance GB, or GF, describe the semicircle BHF, produce DE to meet the circumference in H. Then the square described upon EH shall be equal to the given rectilineal figure A. Join GH. DEMONSTRATION Because the straight line BF is divided into two equal parts in the point G, and into two unequal parts in E, the rectangle BE, EF, together with the square of EG, is equal to the square of GF; (11. 5) but GF is equal to GH; (def. 15) 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; (1. 47) 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 Aviz., the square described upon EH. Q. E. F. 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. II. A straight line is said to touch a circle when it meets the circle, and being produced does not cut it. III. Circles are said to touch each other which meet, but do not cut each other. 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 V. 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 a figure contained by a straight line, and the circumference which it cuts off. 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 straight lines drawn from any point in the circumference of the segment, to the extremities of the straight line which is the base of the segment. IX. An angle is said to insist or stand upon the circumference intercepted between the straight lines that contain the angle. X. A sector of a circle is the figure contained by two straight lines drawn from the centre, and the circumference between them. XI. Similar segments of circles are those in which the angles are equal, or which contain equal angles. Draw within the circle any straight line AB, and bisect AB in D; (1. 10) from the point D draw DC at right angles to AB (1. 11), produce it to E, and bisect CE in F. Then the point F shall be the centre of the circle ABC. DEMONSTRATION For, if it be not, let, if possible, G be the centre, and join GA, GD, GB. Then, because DA is equal to DB, (constr.) 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 assumed to be equal to the base GB, (1. def. 15) because they are drawn from the centre G; therefore the angle ADG must be equal to the angle GDB; (1. 8) but when a straight line standing upon another straight line makes the adjacent angles equal to one another, each of the angles is a right angle;' (1. def. 10) therefore the angle GDB must be a right angle; but FDB is a right angle; (constr.) wherefore the angle FDB must be equal to the angle GDB, the greater to the less, 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 ; |