produced to E, and shew that the rectangle BC, CD is equal to the rectangle AC, CE.

41. Include the first two cases of Euc. II. 13, in one proof.

42. In the second case of Euc. II. 13, draw a perpendicular CE from the obtuse angle C upon the side AB, and prove that the square on AB is equal to the rectangle AB, AE together with the rectangle BC, BD.

43. Enunciate Euc. 11. 13, and give an Algebraical or Arithmetical proof of it. 44. The sides of a triangle are as 3, 4, 5. Determine whether the angles between 3, 4; 4, 5; and 3, 5; respectively are greater than, equal to, or less than, a right angle.

45. Two sides of a triangle are 4 and 5 inches in length, if the third side be 6 inches, the triangle is acute-angled, but if it be 6 inches, the triangle is obtuse-angled.

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46. A triangle has its sides 7, 8, 9 units respectively; a strip of breadth 2 units being taken off all round from the triangle; find the area of the remainder.

47. If the original figure, Euc. II. 14, were a right-angled triangle, whose sides were represented by 8 and 9, what number would represent the side of a square of the same area? Shew that the perimeter of the square is less than the perimeter of the triangle.

48. If the sides of a rectangle are 8 feet and 2 feet, what is the side of the equivalent square?

49. “All plane rectilineal figures admit of quadrature." Point out the succession of steps by which Euclid establishes the truth of this proposition.

50. Explain the construction (without proof) for making a square equal to a plane polygon.

51. Shew from Euc. II. 14, that any algebraical surd as a can be represented by a line, if the unit be a line.

52.

How may a rectangle be dissected so as to form an equivalent rectangle of any proposed length?

53. If the sides of the rectangle in Prop. 14, Book 11, be given numerically, what is the condition that the side of the square may be expressed by a rational number?

54. Shew that the area of every rectangular parallelogram, is equal to the rectangle contained by the diagonal and the perpendicular drawn from one of the angles upon the diagonal.

55. Classify all the properties of triangles demonstrated in the First and Second Books of Euclid.

56. Could any of the propositions of the Second Book be made corollaries to other propositions, with advantage? Point out any such propositions, and give your reasons for the alterations you would make.

BOOK IIL·

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.

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.

XI.

Similar segments of circles are those in which the angles are equal, or which contain equal angles.

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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; (I. 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 F' shall be the center of the circle ABC.

For, if it be not, if possible, let & 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. 11.) 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.

COR.

Therefore no point but Fis the center of the circle ABC.

Which was to be found.

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.

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 jein 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 exterier angle DEB is greater than the interior and opposite angle DAE; (1. 16.)

but DAE was proved to be equal to the angle DBE;
therefore the angle DEB is greater than the angle DBE;
but to the greater angle the greater side is opposite, (1. 19.)
therefore DB is greater than DE:

but DB is equal to DF; (1. def. 15.)
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;

therefore it falls within it. Wherefore, if any two points, &c.

Q. E. D.

PROPOSITION III. 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.
Then, because AF is equal to FB, (hyp.)

and FE common to the two triangles AFE, BFE,

there are two sides in the one equal to two sides in the other, each to each;

and the base EA is equal to the base EB; (1. def. 15.) therefore the angle AFE is equal to the angle BFE; (1. 8.) but when a straight line standing upon another straight line makes the adjacent angles equal to one another,

each of them is a right angle; (1. def. 10.)

therefore each of the angles AFE, BFÈ, is a right angle: wherefore the straight line CD, drawn through the center, bisecting another AB that does not pass through the center, cuts the same at right angles.