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EXAMPLE 1. To construct a triangle, having given the base, the altitude, and the length of the median which bisects the base.

Let AB be the given base, and P and Q the lengths of the altitude and median respectively :

then the triangle is known if its vertex is known. (i) Draw a straight line CD parallel to AB, and at a distance from it equal to P:

then the required vertex must lie on CD. (ii) Again, from the middle point of AB as centre, with radius equal to Q, describe a circle :

then the required vertex must lie on this circle. Hence any points which are common to CD and the circle, satisfy both the given conditions : that is to say, if CD intersect the circle in E, F each of the points of intersection might be the vertex of the required triangle. This supposes the length of the median Q to be greater than the altitude.

EXAMPLE 2. To find a point equidistant from three given points A, B, C, which are not in the same straight line.

(i) The locus of points equidistant from A and B is the straight line PQ, which bisects AB at right angles.

Ex. 1, p. 123. (ii) Similarly the locus of points equidistant from B and C is the straight line RS which bisects BC at right angles.

Hence the point common to PQ and RS must satisfy both con. ditions : that is to say, the point of intersection of PQ and RS will be equidistant from A, B, and C.

Obs. These principles may also be used to prove the theorems relating to concurrency already given on page 111.

EXAMPLE. To prove that the bisectors of the angles of a triangle are concurrent.

Let ABC be a triangle.
Bisect the _s ABC, BCA by straight

lines BO, CO: these must meet at
some point O.

Aix, 12.
Join OA.
Then shall OA bisect the L BAC.
Now BO is the locus of points equi.
distant from BC, BA; Ex. 3, p. 55.
:: OP=OR.

B

P
Similarly CO is the locus of points equi-
distant from BC, CA.

:. OP=OQ; hence OR=OQ.
:: O is on the locus of points equidistant from AB and AC:

that is, OA is the bisector of the L BAC.
Hence the bisectors of the three 28 meet at the point 0.

It may happen that the data of the problem are so related to one another that the resulting loci do not intersect. In this case the problem is impossible.

For example, if in Ex. 1, page 126, the length of the given median is less than the given altitude, the straight line CD will not be intersected by the circle, and no triangle can fulfil the conditions of the problem. If the length of the median is equal to the given altitude, one point is common to the two loci ; and consequently only one solution of the problem exists : and we have seen that there are two solutions, if the median is greater than the altitude.

In examples of this kind the student should make a point of investigating the relations which must exist among the data, in order that the problem may be possible ; and he must observe that if under certain relations two solutions are possible, and under other relations no solution exists, there will always be some intermediate relation under which one and only one solution is possible.

EXAMPLES.

1. Find a point in a given straight line which is equidistant from two given points.

2. Find a point which is at given distances from each of two given straight lines. How many solutions are possible?

3. On a given base construct a triangle, having given one angle at the base and the length of the opposite side. Examine the relations which must exist among the data in order that there may be two solutions, one solution, or that the problem may be impossible.

4. . On the base of a given triangle construct a second triangle equal in area to the first, and having its vertex in a given straight

line. V 5. Construct an isosceles triangle equal in area to a given triangle, and standing on the same base.

6. Find a point which is at a given distance from a given point, and is equidistant from two given parallel straight lines.

When does this problem admit of two solutions, when of one only, and when is it impossible?

BOOK II.

Book II. deals with the areas of rectangles and squares.

A Rectangle has been defined (Book I., Def. 37) as a parallelogram which has one of its angles a right angle.

It should be remembered that if a parallelogram has one right angle, all its angles are right angles.

[1. 46, Cor.]

DEFINITIONS.

1. A rectangle is said to be contained by any two of its sides which form a right angle: for it is clear that both the form and magnitude of a rectangle are fully determined when the lengths of two such sides are given. Thus the rectangle ACDB is said

A

B to be contained by AB, AC; or by CD, DB : and if X and Y are two straight lines equal respectively to AB and AC, then the rectangle contained by X and Y is equal to the rectangle contained by -AB, AC.

[See Ex. 12, p. 70.] X

After Proposition 3, we shall use the abbreviation rect. AB, AC to denote the rectangle contained by AB and AC.

2. In any parallelogram the figure formed by either of the parallelograms about a diagonal together with the two complements is called a gnomon.

A

E B Thus the shaded portion of the annexed diagram, consisting of the parallelogram EH together with the complements AK, G

АН KC is the gnomon AHF.

The other gnomon in the diagram is that which is made up of the figures AK, GF and FH, namely the gnomon AFH. D

F

INTRODUCTORY.

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Before entering upon Book II. the student is reminded of the following arithmetical rule :

RULE. To find the area of a rectangle, multiply the number of units in the length by the number of units in the breadth; the product will be the number of square units in the area. For example, if the two sides AB, AD A

B of the rectangle ABCD are respectively four and three inches long, and if through the points of division parallels are drawn as in the annexed figure, it is seen that the rectangle is divided into three rows, each containing four square inches, or into four columns, each containing three square inches.

D Hence the whole rectangle contains 3 x 4, or 12, square inches.

Similarly if AB and AD contain m and n units of length respectively, it follows that the rectangle ABCD will contain m xn units of area : further, if AB and AĎ are equal, each containing m units of length, the rectangle becomes a square, and contains m2 units of area.

From this we conclude that the rectangle contained by two straight lines in Geometry corresponds to the product of two numbers in Arithmetic or Algebra ; and that the square described on a straight line corresponds to the square of a number. Accordingly it will be found in the course of Book II. that several theorems relating to the areas of rectangles and squares are analogous to well-known algebraical formulæ.

In view of these principles the rectangle contained by two straight lines AB, BC is sometimes expressed in the form of a product, as AB . BC, and the square described on AB as ABP. This notation, together with the signs + and -, will be employed in the additional matter appended to this book; but it is not admitted into Euclid's text because it is desirable in the first instance to emphasize the distinction between geometrical magnitudes themselves and the numerical equivalents by which they may be expressed arithmetically.

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PROPOSITION 1. THEOREM. If there are two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the sum of the rectangles contained by the undivided straight line and the several parts of the divided line.

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1. 11.

Let P and AB be two straight lines, and let AB be divided into any number of parts AC, CD, DB.

Then shall the rectangle contained by P, AB be equal to the sum of the rectangles contained by P, AC, by P, CD, and by P, DB. Construction. From A draw AF perp. to AB; and make AG equal to P.

1. 3. Through G draw GH par to AB; 1. 31. and through C, D, B draw CK, DL, BH par to AG.

Proof. Now the fig. AH is made up of the figs. AK, CL, DH, and is therefore equal to their sum ; and of these,

the fig. AH is the rectangle contained by P, AB;

for it is contained by AG, AB; and AG =P:
and the fig. AK is the rectangle contained by P, AC;

for it is contained by AG, AC; and AG=P:
also the fig. CL is the rectangle contained by P, CD;

for it is contained by CK, CD;
and CK=the opp. side AG, and AG=P.

I. 34. Similarly the fig. DH is the rectangle contained by P, DB.

.:. the rectangle contained by P, AB is equal to the sum of the rectangles contained by P, AC, by P, CD, and by P, DB.

Q.E.D.

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