to each of these equals add the square of BC, and 3. 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, (II. 3.) 4. The rectangle DB, BC, is equal to the rectangle BC, CD, and the square of BC; and the doubles of these are equal; therefore 5. 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 BC is the straight line between the perpendicular and the acute angle at B; and it is manifest that (I. 47.) 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. Let A be the given rectilineal figure; it is required to describe a square that shall be equal to A. Describe (I. 45.) the rectangular parallelogram BCDE equal to the rectilineal figure 8. 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, and from the centre G, at the distance GB, or GF, describe the semicircle BHF, and produce DE to H: the square described upon EH shall be equal to the given rectilineal figure A. H B Join GH; and because the straight line BF is divided into two equal parts in the point G, and into two unequal at E, (II. 5.) 1. The rectangle BE, EF, together with the square of EG, is equal to the square of GF; but GF is equal (Def. 15.) to GH; therefore 2. 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 (I. 47.) to the square of GH; therefore 3. 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 of EH. But the rectangle contained by BE, EF, is the parallelogram BD, because EF is equal to ED; therefore 5. BD is equal to the square of EH; but BD is equal to the rectilineal figure 4; therefore 6. 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. EUCLID'S ELEMENTS OF GEOMETRY. 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 produced 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. 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 contained by a straight line and the circumference 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. And 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. PROP. I. PROBLEM. To find the centre of a given circle. Draw within it any straight line AB, and bisect (I. 10.) AB in D; from the point D draw (I. 11.) DC at right angles to , 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 (Constr.) 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 (I. Def. 15.) to the base GB, because they are drawn from the centre G; therefore (I. 8.) 1. The angle ADG is equal to the angle GDB: 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 (I. Def. 10.); therefore 2. The angle GDB is a right angle: but FDB is likewise (Constr.) a right angle; wherefore 3. The angle FDB is equal to the angle GDB, the greater to the less, which is impossible; therefore 4. 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 5. 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. If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle. |