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 (5. 2.) 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 (47.1.) 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. THE ELEMENTS OF EUCLID. BOOK IIJ Definitions 1. 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 their centres are equal.' 2. A straight line is said to touch a circle, when it meets the circle, and being produced does not cut it. 3. Circles are said to touch one another which meet, but do not cut one another. 4. 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. 5. And the straight line on which the greater perpendicular falls, is said to be farther from the centre. '6. A segment of a circle is the figure contained by a straight line and the circumference it cuts off. 7. "The angle of a segment is that which i contained by the straight line and the circumfe rence. D 8. 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. 9. And an angle is said to insist or stand upon the circumference intercepted between the straight lines that contain the angle. 10. The sector of a circle is the figure contained by two straight lines drawn from the centre, and the circumference between them. 11. Similar segments of a circle are those in which the angles are equal, or which contain equal angles. Proposition I. Problem. 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 (10. 1.) it in D; from the point D draw (11.1.) 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. F G 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 G:* Therefore the angle ADG is equal (8. 1.) 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: (10. Def. 1.) Therefore the angle GDB is a right angle. E But FDB is likewise a right angle; wherefore the angle FDB is 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 shewn, 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. *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 tho circumference. Proposition II. Theorem. If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle. AEB 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, (1. 3.) 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 (5. 1.) to the angle DBA; and because AE, a side of the triangle DAE, is produced to B, the angle DEB is greater (16. 1.) than the angle DAE; but DAE is to the angle DBE; therefore the angle DEB is greater than the angle DBE but to the greater angle the greater side is opposite; (19. 1.) DB is therefore greater than DE: but DB is equal to DF; wherefore DF is greater than DE, the less than the greater, which is impossible : Therefore the straight line drawn from A to B does not fall without the circle. In the same manner, it may be demonstrated that it does not fall upon the circumference; it falls, therefore, within it. Where fore if any two points, &c. Q. E. D. If a straight line drawn through the centre of a circle bisect a straight line in it which does not pass through the centre, it shall cut it at right angles and if it cut it at right angles, it shall bisect it. Let ABC be a circle, and let CD, a straight line drawn through the centre, bisect any straight line AB, which does not pass through the centre, in the point F: it cuts it also at right angles. Take (1.3.) E the centre of the circle, and join EA, EB. Then because AF is equal to FB, and FE common to the two triangles AFE, BFE, there are two sides in the one equal to two sides in the other, and the base EA is equal to the base EB; therefore the angle AFE is equal (8. 1.) to the angle BFE. But when a straight line standing upon another, makes the adjacent angles equal to one another, each of them is a right (10 Def. 1.) angle: Therefore each of the angles AEF, BFE is a right angle; wherefore the straight line CD, drawn through the centre bisecting another AB that does not pass through the centre cuts the same at right angles. E F But let CD cut AB at right angles; CD also bisects it, that is, AF is equal to FB. The same construction being made, because EA, EB from the centre are equal to one another, the angle EAF is equal (5.1.) to the angle EBF; and the right angle AFE is equal to the angle BFE: Therefore in the two triangles, EAF, EBF, there are two angles in one equal to two angles in the orner, and the side EF, which is opposite to one of the equal angles in each, is common to both; therefore the other sides are equal; (26.1.) AF therefore is equal to FB. Wherefore if a straight line, &c. Q. E. D. Proposition IV. Theorem. If in a circle two straight lines cut one another which do not both pass through the centre, they do not bisect each the other. Let ABCD be a circle, and AC, BD two straight lines in it which cut one another in the point E, and do not both pass through the centre: AC, BD do not bisect one another. For, if it is possible, let AE be equal to EC and BE to ED; if one of the lines pass through the centre, it is plain that it cannot be bisected by the other which does not pass through the centre: But if neither of them pass through the centre, take (1.3.) F the centre of the circle, and join EF: and because FE, a straight line through the centre, bisects another AC, which does not pass through the centre, it shall cut it at right (3. 3.) angles; wherefore FEA is a right angle. Again, because the straight line FE B bisects the straight line BD, which does not pass through the centre, it shall cut it at right (3. 3.) angles; wherefore FEB is a right angle. A C And FEA was shewn to be a right angle; therefore FEA is equa! to the angle FEB, the less to the greater, which is impossible: Therefore AC, BD do not bisect one another. Wherefore, if in a circle, &c. Q. E. D. Proposition V. Theorem. If two circles cut one another, they shall not have the same centre. Let the two circles ABC, CDG cut one another in the points B, C ; they have not the same centre. C G ADE B For, if it be possible, let E be their cenue: Join EC, and draw any straight line EFG meeting them in F and G; and because E is the centre of the circle ABC, CE is equal to EF: Again, because E is the centre of the circle CDG, CE is equal to EG: But CE was shewn to be equal to EF; therefore EF is equal to EG, the less to the greater, which is impossible: Therefore E is not the centre of the circles ABC, CDG. Wherefore, if two circles, &c. Q. E. D. Proposition VI. Theorem. If two circles touch one another internally, they shall not have the same centre. Let the two circles ABC, CDE touch one another internally in the point C: They have not the same centre. For, if they can, let it be F; join FC and draw any straight line |