Book II. a 15. 1. b 29. 1. c 34. 1. d 26. 1. e A, 2. PROP. B. THEOR. The sum of the squares of the diameters of any parallelogram is equal to the sum of the squares of the sides of the parallelogram. Let ABCD be a parallelogram, of which the diameters are AC and BD; the sum of the squares of AC and BD is equal to the sum of the squares of AB, BC, CD, DA. Let AC and BD intersect one another in E: and because the vertical angles AED, CEB are equal 2, a and also the alternate angles EAD, ECB, the triangles ADE, CEB have two angles in the one equal to two angles in the other, each to each: but the sides AD and BC, which are opposite to equal angles in these triangles, are also equale; therefore the other sides which are op posite to the equal angles are also e-B A E qual, viz. AE to EC, and ED to EB. D Since, therefore, BD is bisected in E, AB2 + AD2 = e 2BE2 + 2AE2; and for the same reason, CD2 + BC2 = 2BE2 + 2EC2 = 2BE2 + 2AE2, because EC = AE. Therefore AB2 + AD2 + DC2 + BC2= 4BE2+4AE2. f2.Cor.8.2. But 4BE2 = BD2, and 4AE2 = AC2f, because BD and AC are both bisected in E; therefore AB2 + AD2 + CD2 + BC2 BD2 + AC2. Therefore the sum of the squares, &c. Q. E. D. COR. From this demonstration, it is manifest that the diameters of every parallelogram bisect one another. २८ हूँ OF GEOMETRY. BOOK III. DEFINITIONS. A THE HE radius of a circle is the straight line drawn from Book III. the centre to the circumference. I. A straight line is said to touch a circle, when it meets the circle, and being produced, does not cut it. II. Circles are said to touch one another, which meet, but do not cut one another. III. 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. IV. And the straight line on which the greater perpendicular falls, is said to be farther from the centre. An arch of a circle is any part of the circumference. 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 bisecta it in Book III. D; from the point Ď drawb BC 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. 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; but the base GA is also equal to the base GB, because they are radii of the same circle: therefore 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 angled. Therefore the angle c F a 10. 1. b 11. 1. c 8. 1. D B GDB is a right angle: 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 shown, 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. d 7. def. 1. PROP. II. THEOR. If any two points be taken in the circumference of a circle, the straight line which joins them will fall within the circle. Book III. a 5. 1. b 16. 1. c 19. 1. a 1. 3. C Let ABC be a circle, and A, B any two points in the circumference: the straight line drawn from A to B will fall within the circle. a A b D B F Take any point in AB as E; find D the centre of the circle ABC; join AD, DB and DE, and let DE meet the circumference in F. Then because DA is equal to DB, the angle DAB is equal to the angle DBA; and because AE, a side of the triangle DAE, is produced to B, the angle DEB is greater than the angle DAE; but DAE is equal to the angle DBE; therefore the angle DEB is greater than the angle DBE: Now to the greater angle the greater side is opposite; DB is therefore greater than DE: but DB is equal to DF; wherefore DF is greater than DE, and the point E is therefore within the circle. The same may be demonstrated of any other point between A and B, therefore AB is within the circle. Wherefore, if any two points, &c. Q. E. D. PROP. III. THEOR. If a straight line drawn through the centre of a circle bisect a straight line in the circle, which does not pass through the centre, it will cut that line at right angles; and, "if it cut it at right angles, it will 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 also cuts it at right angles. Take a 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 |