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I. 29.

and FH is equal to EG, therefore BD is equal to FH.

I. 29. 1. 29.

I. 19.

But if BH is not parallel to AG, draw BK and FL parallel to AG, meeting CD and FH in K and L respectively, then BK is equal to AC, and FL is equal to EG, therefore BK is equal to FL. Hence in the triangles BKD, FLH the angle KBD is equal to the angle LFH, 1. 23, Cor. 2. the angle BDK is equal to the angle FHL, 1. 23, Cor. 2. and the side BK is equal to the side FL, therefore the side BD is equal to the side FH.

Q.E.D. COR. 1. If there are three parallel straight lines, and the intercepts

made by them on any straight line that cuts them are equal, then the intercepts on any other straight line that

cuts them are also equal. COR. 2. The straight line drawn through the middle point of one of

the sides of a triangle parallel to the base passes through

the middle point of the other side. COR. 3. The straight line joining the middle points of two sides of

a triangle is parallel to the base. Ex. 58. The straight lines joining the middle points of the sides

of a triangle divide it into four identically equal tri

angles. *Ex. 59. The straight line joining the middle points of two sides

of a triangle is equal to half the base. Ex. 60. If a quadrilateral be formed by joining the middle

points of the sides of a given quadrilateral, it is a parallelogram.

EXERCISES.

61. Prove that the angle between the bisectors of two adjoining

angles of a quadrilateral is half the sum of the two remaining

angles. 62. If the sides of a regular pentagon be produced to meet, the

angles formed by these lines are together equal to two right

angles. 63. If a straight line parallel to BC, the base of an isosceles

triangle ABC, meet the sides AB, AC at D and E: shew that the triangles CDE, DCB have two sides and one angle of the one equal to two sides and one angle of the other.

Are they equal in all respects ? 64. Straight lines AD, BE, CF are drawn within the triangle

ABC, making the angles DAB, EBC, FCA all equal to one another. If the lines AD, BE, CF do not meet at a point : prove that the angles of the triangle formed by them are

equal to those of the triangle ABC, each to each. 65. The exterior angles at B and C of the triangle ABC are

bisected by lines meeting at D. Shew that the angle BDC

is equal to half the exterior angle at A. 66. If a quadrilateral has two sides parallel, and the other two

sides equal but not parallel, shew that the diagonals of the

quadrilateral are equal. 67. Two triangles ABC, DBC are upon the same base BC, and

AD is parallel to BC. If ABC is isosceles, shew that its

perimeter is less than that of DBC. 68. The sum of the distances of any point in the base of an

isosceles triangle from the two sides is constant.

69. If the base of an isosceles triangle be produced to any

point, the difference of the distances of the point from the

two sides is constant. 70. Shew that the sum of the distances of any point within an

equilateral triangle from the three sides is constant. 71. ABCD is a parallelogram, AD is bisected at E, and BC at

F: shew that BE and DF trisect the diagonal AC. *72. In the triangle ABC, D and E are the middle points of BC

and CA: shew that BE cuts off a third part from DA. *73. Shew that the three straight lines drawn from the vertices

of a triangle to the middle points of the opposite sides meet in a point.

SECTION IV.

PROBLEMS.

A Geometrical Problem is a proposition, of which the object is to effect some Geometrical construction.

The solution of a Problem depends on the instruments, the use of which is allowed ; and it will be readily understood that the more restricted the choice of instruments, the more limited will be the Problems which can be solved by their use, and the more difficult will the solution of many that are possible be found.

Owing to the existence of this arbitrary element in the treatment of Problems, they are grouped together in a separate section. Though important as applications of Geometrical truths, it should be clearly understood that Problems form no part of the chain of connected truths embodied in the Theorems of Geometry, so that, though they may advantageously be studied in connection with the Theorems on which they directly depend, they are not a necessary part of the pure Science of Geometry.

It is the recognised convention of Elementary Geometry that the only instruments to be employed are—the ruler, for drawing and producing straight lines, and the compasses for describing circles and for the transference* of distances.

This Convention is embodied in the following

POSTULATES OF CONSTRUCTION.

Let it be granted that 1. A straight line may be drawn from any one point to any other

point. 2. A terminated straight line may be produced to any length in a

straight line. 3. A circle may be drawn with any centre, with a radius equal to

any finite straight line.

DEF. 41. A circle is a plane figure contained by one line, which is

called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference are equal to one another. This point is

called the centre of the circle. DEF. 42. A radius of a circle is a straight line drawn from the

centre to the circumference. DEF. 43. A diameter of a circle is a straight line drawn through

the centre and terminated both ways by the circumference. It follows from the definition of a circle that a point is within a circle, when its distance from the centre is less than the radius of the circle.

* NOTE.—Euclid restricts the use of the compasses to describing circles, and shews in his Props. 2 and 3 how with this restriction to draw a line of given length and to cut off a given length from a given line.

It is evident that

(1) If a point in a straight line is within a closed figure the straight line if produced in either sense from the point will meet the boundary of the figure, and thus intersect it in two points at

least;

(2) If a point in the boundary of one closed figure lie within another closed figure, and also a point in the boundary of the latter lie within the former, the two boundaries intersect in two points at least;

For they cannot lie wholly outside each of the other : and if one were wholly inside the other, no point in the boundary of the second would lie within the first. Hence they must lie partly inside and partly outside each of the other, and their boundaries (circumferences) must meet in two points at least.

By the help of the above, it may be shewn that the straight lines and circles drawn in the Problems of this section intersect ; or the conditions that must be satisfied in order that they may intersect may be determined.

PROB. 1. To bisect a given angle.

Let BAC be the given angle: it is required to bisect it.

B

A

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