equal to AB drawn from the point B. But to form a triangle these lines must meet at a point. How shall we find this point? Looking at the figure, you will see that there are two points at which the circles cut one another, which we will call C and D, and at each of these, two radii, one drawn from the centre of each circle (i.e. from A and B), would meet. Take the point C and draw lines AC and BC from points A and B (post. 1). An equilateral triangle is thus described; for

A

B

since BC and AC are each equal to AB (def. 15), they are equal to each other (ax. 1).

Thus the two conditions of the enunciation are satisfied, for ABC is an equilateral triangle, and it is described upon the given

line AB.

It remains to translate the problem into formal mathematical language, and it will be identical with Euclid's first proposition.

PROPOSITION I., PROBLEM.

[The propositions are collected at the end of the book for convenient reference. It is intended that each should be read immediately after its analysis.]

SUMMARY OF PROPOSITION I., PROBLEM 1.

To describe an equilateral triangle on a given straight line AB.

.. BC= AC (ax. 1), and ▲ ABC is equilateral, &c.

Q.E.D.

Prop. II. We pass on to the second proposition, also a problem, 'From a given point to draw a straight line equal to a given straight line.' Mark any point A and draw a straight line BC to represent the given line. From our work in the last proposition you will remember that we can obtain lines equal to any given line by describing a circle of which it forms a radius, so that by taking BC as radius of a circle we can draw any number of lines equal to it, from the centre of the circle. Draw this circle, taking B as its centre. Now supposing the point A should fall anywhere on its circumference, the problem will be solved by drawing a line AB to the centre, which will form a radius and hence be equal to BC.

A

But supposing that A does not fall so, but at any point within or without the circumference of the circle, how shall we obtain a measure for any line drawn from the point A ? Here lies the difficulty of the problem, which Euclid overcomes by a very ingenious use of Prop. 1. He tells us to draw a straight line from A to B and on it describe an equilateral triangle (I. 1),

B

and to produce its sides DA and DB through the circle already drawn to any points E and F. This gives the key to the solution, which we ought to be able to complete.

D

The lines DF and DE are each divided into three parts; what do we know about them? Of the line passing through A we know that DA= DB, since they are sides of the equilateral triangle; we know nothing as to the length of the other parts. Looking at DF we have DB = DA, as was said, and also BG the given line (def. 15), or, adding these parts together, we may say that DG: = a side of the triangle+ the given line. Now drawing a

=

BC,

A

B

E

circle of which DG is radius and D centre, we cut off on line

DE a part DL = DG, i.e. a side of the triangle + the given line. Take away DA the side of the triangle, the remainder AL must be equal to the given line, and it is therefore the line required.

There are eight different ways in which the figure of this proposition can be drawn, according to the relative position of the point A and BC.

SUMMARY OF PROPOSITION II., PROBLEM 2.

From a given point to draw a straight line equal to a given straight line.

Proposition III. is also a problem, 'From the greater of two straight lines to cut off a part equal to the less.' The в given lines may, or may not, be drawn from one point. First, supposing that they are, take any point A, and from it draw two straight lines AB and AD, of which let AD be the shorter. Here we can at once apply our line-measure the circle, for, taking A as centre and

the shorter line as radius, a circle can be drawn which will cut AB in E, making a part AE equal to AD (def. 15). Next, supposing that the extremities of the lines do not meet, and that a line C is the shorter. From

the last proposition we have learnt 'to draw a line from any point equal to any given line;' hence from point A a line can be drawn equal to C, and then we can proceed as in the first case.

[NOTE.-In the formal statement of this proposition it is not necessary to divide it into two cases, as the first is included in the second. This has been done here only to make the solution more obvious.]

SUMMARY OF PROPOSITION III., PROBLEM 3.

From the greater of two given straight lines, AB and C,
to cut off a part equal to the less.

This proposition concludes the first group of problems, by which we have learnt to do three things in addition to those assumed in the postulates. We can

Describe an equilateral triangle (I. 1).

Draw a straight line equal to any other (I. 2).

Cut off from any line a part equal to any lesser line (I. 3).

Throughout these propositions the circle has been made use of as a line-measure.

LECTURE IV.

THE METHOD OF SUPERPOSITION.- -THE INDIRECT PROOF.-
PROPOSITIONS 4-6.

of theorems which you will remember,

PROPOSITION IV. is the first of a group continues till the ninth. A theorem, as demonstrates some geometrical truth. The truth here demonstrated is that: If one triangle has two sides and the angle included by these sides, severally equal to two sides, and the angle included by them of another, these triangles are equal in all respects.

Let us consider in what respects triangles can be said to be equal to one another :

(1) In their sides; each side of the one must be equal to the corresponding side of the other.

(2) In their angles; each angle of the one must be equal to the corresponding angle of the other.

(3) In their area, or the space enclosed by the sides. And these three equalities you will find very rigidly defined in Euclid's enunciation of this theorem, which is given below.

In the triangles which we are considering we are given(1) Two sides in one equal to two sides in the other, each to each.

(2) One angle (the angle included by the equal sides) in one equal to the corresponding angle in the other.

We are required to show that—

(1) The third sides, or bases, are equal.

(2) That the two remaining angles in each triangle are equal each to each.

(3) That the areas are equal.