43. If ABC be an isosceles triangle, and DE be drawn parallel to the base BC, and EB be joined; prove that BE2 = BC× DE + CE2. 44. If ABC be an isosceles triangle of which the angles at B and Care each double of A; then the square on AC is equal to the square on BC together with the rectangle contained by AC and BC. VI. 45. Shew that in a parallelogram the squares on the diagonals are equal to the sum of the squares on all the sides. 46. If ABCD be any rectangle, A and C being opposite angles, and O any point either within or without the rectangle: OA3 + OC2 = OB2 + OD3. 47. In any quadrilateral figure, the sum of the squares on the diagonals together with four times the square on the line joining their middle points, is equal to the sum of the squares on all the sides. 48. In any trapezium, if the opposite sides be bisected, the sum of the squares on the other two sides, together with the squares on the diagonals, is equal to the sum of the squares on the bisected sides, together with four times the square on the line joining the points of bisection. 49. The squares on the diagonals of a trapezium are together double the squares on the two lines joining the bisections of the opposite sides. 50. In any trapezium two of whose sides are parallel, the squares on the diagonals are together equal to the squares on its two sides which are not parallel, and twice the rectangle contained by the sides which are parallel. 51. If the two sides of a trapezium be parallel, shew that its area is equal to that of a rectangle contained by its altitude and half the sum of the parallel sides. 52. If a trapezium have two sides parallel, and the other two equal, shew that the rectangle contained by the two parallel sides, together with the square on one of the other sides, will be equal to the square on the straight line joining two opposite angles of the trapezium. 53. If squares be described on the sides of any triangle and the angular points of the squares be joined; the sum of the squares on the sides of the hexagonal figure thus formed is equal to four times the sum of the squares on the sides of the triangle. VII. 54. Find the side of a square equal to a given equilateral triangle. 55. Find a square which shall be equal to the sum of two given rectilineal figures. 56. To divide a given straight line so that the rectangle under its segments may be equal to a given rectangle. 57. Construct a rectangle equal to a given square and having the difference of its sides equal to a given straight line. 58. Shew how to describe a rectangle equal to a given square, and having one of its sides equal to a given straight line. DEFINITIONS. I. EQUAL circles are those of which the diameters are equal, or from the centers 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 centers coincide, the circles must likewise coincide, since the straight lines from the centers 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 center of a circle, when the perpendiculars drawn to them from the center are equal. V. And the straight line on which the greater perpendicular falls, is said to be further from the center. VI. A segment of a circle is the figure contained by a straight line, and the arc or the part of the circumference which it cuts off. VII. The angle of a segment is that which is contained by a straight line and a part of the circumference. VIII. An angle in a segment is any angle contained by two straight lines drawn from any point in the arc 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 part of 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 center and the arc between them. XI. Similar segments of circles are those in which the angles are equal, or which contain equal angles. Ꮐ To find the center of a given circle. Let ABC be the given circle: it is required to find its center. C FG D B Draw within it any straight line AB to meet the circumference in A, B; and bisect AB in D; (1. 10.) from the point D draw DC at right angles to AB, (I. 11.) meeting the circumference in C, produce CD to E to meet the circumference again in E, and bisect CẺ in F. Then the point F shall be the center of the circle ABC. For, if it be not, if possible, let G be the center, 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 equal to the base GB, (I. def. 15.) because they are drawn from the center G: therefore the angle ADG is 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 is a right angle: but FDB is likewise a right angle; (constr.) wherefore the angle FDB is equal to the angle GDB, (ax. 1.) the greater angle equal to the less, which is impossible; therefore G is not the center of the circle ABC. In the same manner it can be shewn that no other point out of the line CE is the center; and since CE is bisected in F, any other point in CE divides CE into unequal parts, and cannot be the center. Therefore no point but F is the center of the circle ABC. Which was to be found. COR. From this it is manifest, that if in a circle a straight line bisects another at right angles, the center of the circle is in the line which bisects the other. 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. Let ABC be a circle, and A, R any two points in the circumference. Then the straight line drawn from A to B shall fall within the circle. For if AB do not fall within the circle, let it fall, if possible, without the circle as AEB; find D the center of the circle ABC, (III. 1.) and join DA, DB; in the circumference AB take any point F, join DF, and produce it to meet AB in E. Then, because DA is equal to DB, (1. def. 15.) therefore the angle DBA is equal to the angle DAB; (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; (I. 16.) but DAE was proved to be equal to the angle DBE; but DB is equal to DF; (1. def. 15.) 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; therefore it falls within it. Wherefore, if any two points, &c. PROPOSITION III. Q. E.D. THEOREM. If a straight line drawn through the center of a circle bisect a straight line in it which does not pass through the center, it shall cut it at right angles: and conversely, 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 center, bisect any straight line AB, which does not pass through the center, in the point F. Then CD shall cut AB at right angles. Take E the center of the circle, (III. 1.) and join EA, EB. and FE common to the two triangles AFE, BFE, |