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 A — viz., the square described upon EH. Q. E. F. BOOK III. DEFINITIONS. 1. EQUAL circles are those of which the diameters are equal, or from the centres of which the straight lines to the circumference are eoual. II. go 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. PROP. I. - PROBLEM. To find the centre of a given circle. (References — Prop. I. 8, 10, 11.) Let ABC be the given circle. It is required to find its centre, 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 ; therefore 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. PROP. 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. (References—Prop. 1. 5, 16, 19 ; 111. 1.) Let ABC be a circle, and A, B, any two points in the circumference. Then the straight line drawn from A to B shall fall within the circle. Find D the centre of the circle ABC, (II. 1) and join AD, DB; in the circumference AB take any point F; join DF, and let it cut the straight line AB in E. DEMONSTRATION Because DA is equal to DB, (1. def. 15) the angle DAB is equal to the angle DBA; (1. 5) and because AE, a side of the triangle DAE, is produced to B, the exterior angle DEB is greater than the interior and opposite angle DAE; (1. 16) but the angle DAE was proved to be equal to the angle DBE; therefore the angle DEB is also greater than the angle DBE ; but to the greater angle the greater side is opposite ; (I. 19) therefore DB is greater than DE; but DB is equal to DF; (1. def. 15) |