square of GE, is equal to the squares of HE, GE; take away the square of GE from both, then the remaining rectangle under BE, EF, is equal to the square of EH. But the rectangle under BE, EF, is the parallelogram BD, because EF is equal to ED: therefore the parallelogram BD is equal to the square of HE. But the parallelogram BD (by const.) is equal to the right lined figure A: therefore the right lined figure a will be equal to the square Which was to be done. of EH. Deduction. To find a line D, the square of which shall be equal to the sum of the rectangles AB, AC, BC: A, B, C, being three given lines. In the demonstration of this, Dr. Keil, in his edition, has the words, "but if it be not, let either BE or ED be the greater, suppose BE, which let be produced to F," as if it was of any consequence, as Dr. Simson observes, whether the greater or less be produced; instead of these words there ought to be read, "But if they are not equal, produce one of them to F," as in the Oxford edition of Commandine. EUCLID'S ELEMENTS. BOOK III. DEFINITIONS. 1. Equal circles are those of which the diameters are equal, or from the centres of which the right lines drawn to the circumferences are equal. 2. A right line is said to touch a circle, which, touching the circle, and produced, does not cut it. 3. Circles are said to touch one another, which touching do not cut one another. 4. Right lines are said to be equally distant from the centre of a circle, when the right lines drawn from the centre perpendicular to them are equal. 5. And that right line on which the greater perpendicular falls, is said to be further from the centre. 6. A segment of a circle is the figure contained both by the right line and the circumference of the circle it cuts off. 7. The angle of a segment is that which is contained by a right line and the circumference of the circle. 8. The angle in a segment is, when some point is taken in the circumference, and from it at the ends of a right line, which is the base of the segment, right lines are joined, the angle contained by the right lines being joined. 9. When the right lines containing an angle assume some circumference, the angle is said to stand upon the circumference. 10. The section of a circle is, when the angle is placed at the centre of a circle, the figure contained by the right lines containing the angle and the circumference between them. 11. Similar segments of a circle are those in which the angles are equal to one another, or which contain equal angles. k PROPOSITION I.* PROBLEM. To find the centre of a given circle. Let ABC be the given circle; it is required to find the centre of the circle ABC. Draw in it anyhow the right line AB; which bisecta in the point D. From the point D drawb DC at right angles to AB, and produce it to E; and bisecta CE in F. Then is F the centre of the circle ABC. For if it be not; let & be the centre, if it be possible, and join GA, GD, GB. Therefore since DA is equal to DB, and DG common; the two AD, DG, are equal to the two DB, GD, each to each; and the base GA is equal to the base GB; for they are from the centre G. Therefore the angle ADG is equal to the angle GDB. But when a right line standing on another right line makes the adjacent angles equal to one another, each of them is a right angle. Therefore the angle GDB is a right angle; the angle FDB is also a right angle; therefore the angle FDB is equal to the angle GDB, the greater to the less, which is impossible. Wherefore G is not the centre of the circle ABC. In like manner it may be demonstrated that none other than F is the centre. Wherefore Fis the centre of the circle ABC. Q. E. F. COROLLARY. A F E If in a circle a right line bisects another right line at right angles, the centre of the circle shall be in the cutting line. PROPOSITION II. THEOREM. If in the circumference of a circle, any two points are taken, the right line which joins them shall fall within the circle. Let ABC be a circle; in its circumference take any two points A, B. The right line AB, which is drawn from A to B, falls within the circle. For in the right line AB *Tacquet, and other authors, have proposed various methods for finding the centre of a circle; but none, I think, so simple in its operation as that of Euclid. a 10. 1. b 11. 1. d 8. 1. e 10. Def. 1. a 5. 1. b 16. 1. a 1. 1. b 8. 1. take any point E; join DA, DE, DB; and in DE pro- C D B E are equal, the point F will be at the circumference of the circle. Therefore the point E must fall within the circumference of the circle. In like manner it can be demonstrated that of every other point of the right line AB, between the points A, B, is within the circumference of the circle. If, therefore, in the circumference of a circle, &c. Q. E. D. PROPOSITION III. If in a circle a right line drawn from the centre bisects another right line not drawn through the centre, it will also cut it at right angles; but if it cuts it at right angles, it shall also bisect it. Let ABC be a circle, and in it a right line CD drawn through the centre, which bisects the right line AB, not drawn through the centre in the point F. It will cut it at right angles. Find the centre of the circle ABC, which let be E, and join EA, EB. Therefore because AF is equal to FB, and FE common, the two AF, FE, are equal to the two BF, FE, and the base EA is equal to the base EB. Therefore also the angle AFE will be equal to the angle BFE. But when a right line standing on a right line makes the adjacent angles b *The truth of this theorem is evident from a consideration of the first, for as the construction of that problem is effected by drawing a right line dividing the same into two equal parts, and from the point of bisection drawing another line perpendicular to the former; also, as it is clearly demonstrated that to assume any other point as the centre which is not in the perpendicular would be absurd, it follows conversely that the right line passing through the centre bisects another line not passing through the centre, must cut it at right angles, and, on the contrary, if it cut it at right angles, it must bisect it. |