IV.

30. Shew that the square on the hypotenuse of a right-angled triangle, is equal to four times the area of the triangle together with the square on the difference of the sides.

31. In the triangle ABC, if AD be the perpendicular let fall upon the side BC; then the square on AC together with the rectangle contained by BC, BD is equal to the square on AB together with the rectangle CB, CD.

32. ABC is a triangle, right angled at C, and CD is the perpendicular let fall from Cupon AB; if HK is equal to the sum of the sides AC, CB, and LM to the sum of AB, CD, shew that the square an HK together with the square on CD is equal to the square on LM.

33. ABC is a triangle having the angle at B a right angle: it is required to find in AB a point P such that the square on AC may exceed the squares on AP and PC by half the square on AB.

34. In a right-angled triangle, the square on that side which is the greater of the two sides containing the right angle, is equal to the rectangle by the sum and difference of the other sides.

35. The hypotenuse AB of a right-angled triangle ABC is trisected in the points D, E; prove that if CD, CE be joined, the sum of the squares on the sides of the triangle CDE is equal to two-thirds of the square on AB.

36. From the hypotenuse of a right-angled triangle portions are cut off equal to the adjacent sides: shew that the square on the middle segment is equivalent to twice the rectangle under the extreme segments.

V.

37. Prove that the square on any straight line drawn from the vertex of an isosceles triangle to the base, is less than the square on a side of the triangle by the rectangle contained by the segments of the base: and conversely.

38. If from one of the equal angles of an isosceles triangle a perpendicular be drawn to the opposite side, the rectangle contained by that side and the segment of it intercepted between the perpendicular and base, is equal to the half of the square described upon the base.

39. If in an isosceles triangle a perpendicular be let fall from one of the equal angles to the opposite side, the square on the perpendicular is equal to the square on the line intercepted between the other equal angle and the perpendicular, together with twice the rectangle contained by the segments of that side.

40. The square on the base of an isosceles triangle whose vertical angle is a right angle, is equal to four times the area of the triangle. 41. Describe an isosceles obtuse-angled triangle, such that the square on the side subtending the obtuse angle may be three times the square on either of the sides containing the obtuse angle.

42. If AB, one of the sides of an isosceles triangle ABC be produced beyond the base to D, so that BD=AB, shew that

CD AB 2. BC2.

43. If ABC be an isosceles triangle, and DE be drawn parallel to the base BC, and EB be joined; prove that BE' = 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:

ОA + 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 triangle 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.

о

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.

G

PROPOSITION I. PROBLEM.

To find the center of a given circle.

Let ABC be the given circle: it is required to find its center.

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, (1. 11.) meeting the circumference in C, produce CD to E to meet the circumference again in E, and bisect CE in F.

Then the point Fshall 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 4D, DG, are equal to the two BD, DG, each to each; and the base GA is equal to the base GB, (1. def. 15.)

because they are drawn from the center G:

therefore the angle ÅDG 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; (I. 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 Fis 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.

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, B any two points in the circumference. Then the straight line drawn from A to B shall fall within the circle.