It is assumed in this proposition that the straight line DB intersects the circle CEF. It is easily seen that it must intersect it in two points.

It will be noticed that in the construction of this proposition there are several steps at which a choice of two alternatives is afforded: (1) we can draw either AB or AC as the straight line on which to construct an equilateral triangle: (2) we can construct an equilateral triangle on either side of AB: (3) if DB cut the circle in E and I, we can choose either DE or DI as the radius of the circle which we describe with D as centre.

There are therefore three steps in the construction, at each of which there is a choice of two alternatives: the total number of solutions of the problem is therefore 2 × 2 × 2 or eight.

On the opposite page two diagrams are drawn, to represent two out of these eight possible solutions. It will be a useful exercise for the student to draw diagrams corresponding to some of the remaining six.

EXERCISES.

1. Draw a diagram for the case in which the given point is the middle point of the given straight line.

2. Draw a diagram for the case in which the given point is in the given straight line produced.

3. Draw from a given point a straight line (a) twice, (b) three times the length of a given straight line.

4. Draw from D in any one of the diagrams of Proposition 2 a straight line, so that the part of it intercepted between the two circles may be equal to the given straight line. Is a solution always possible?

PROPOSITION 3.

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

Let AB and CD be the two given straight lines, of which AB is the greater :

it is required to cut off from AB a part equal to CD. CONSTRUCTION. From A draw a straight line AE equal to CD;

with A as centre and AE as radius,

describe the circle EFG.

(Prop. 2.)

(Post. 6.)

The circle must intersect AB between A and B,
for AB is greater than AE.

Let F be the point of intersection :
then AF is the part required.

E

D

G

PROOF. Because A is the centre of the circle EFG,

AE is equal to AF.

(Def. 22.)

But AE was made equal to CD; (Construction.) therefore AF is equal to CD.

Wherefore, from AB the greater of two given straight

lines a part AF has been cut off equal to CD the less.

The demand made in Postulate 6, that "a circle may be described with any point as centre and with any straight line drawn from that point as radius," is equivalent, in practical geometry, to saying that a pair of compasses may be used in the following manner: the extremity of one leg of a pair of compasses may be put down on any point A, the compasses may then be opened so that the extremity of the other leg comes to any other point and then a circle may be swept out by the extremity of the second leg of the compasses, the extremity of the first leg remaining throughout the motion on the point A.

Compasses are also used practically for carrying a given length from any one position to any other: for instance, they would generally be used to solve the problem of Proposition 3 by opening the compasses out till the extremities of the legs came to the points C, D: they would then be shifted, without any change in the opening of the legs, until the extremity of one leg was on A and the extremity of the other in the straight line AB.

Euclid restricted himself much in the same way as a draughtsman would, if he allowed himself only the first mentioned use of the compasses: the first three propositions shew how Euclid with this selfimposed restriction solved the problem, which without such a restriction could have been solved more readily.

After the problems in the first three propositions have been solved, we may assume that we can draw a circle, as a practical draughtsman would, with any point as centre and with a length equal to any given straight line as radius.

EXERCISES.

1. On a given straight line describe an isosceles triangle having each of the equal sides equal to a second given straight line.

2. Construct upon a given straight line an isosceles triangle having each of the equal sides double of a second given straight line.

3. Construct a rhombus having a given angle for one of its angles, and having its sides each equal to a given straight line.

PROPOSITION 4.

If two triangles have two sides of the one equal to two sides of the other, and also the angles contained by those sides equal, the two triangles are equal in all respects.

(See Def. 21.)

Let ABC, DEF be two triangles, in which AB is equal to DE, and AC to DF, and the angle BAC is equal to the angle EDF:

it is required to prove that the triangles ABC, DEF are equal in all respects.

E

F

E'

F

PROOF.

Because the angles BAC, EDF are equal,
it is possible to shift the triangle ABC
so that A coincides with D,

and AB coincides in direction with DE,

and AC with DF. (Test of Equality, page 8.)

If this be done,

because AB is equal to DE,

B must coincide with E; and because AC is equal to DF,

C must coincide with F.

Again because B coincides with E and C with F,

BC coincides with EF;

(Post. 2.)

therefore the triangle ABC coincides with the triangle DEF,

and is equal to it in all respects.

Wherefore, if two triangles &c.

The proof of this proposition holds good not only for a pair of triangles such as ABC, DEF in the diagram: it holds good equally for a pair such as ABC, D'E'F', one of which must be reversed or turned over before the triangles can be made to coincide or fit exactly.

In this proposition Euclid assumed Postulate 2, that two straight lines cannot have a common part. When the triangle ABC is shifted, so that A is on D and AB is on DE, there would be no justification for the conclusion that B must coincide with E, because AB is equal to DE, if it were possible for two straight lines to have a common part. In fact, two curved lines might be drawn from the point D starting in the same direction DE but leading to two totally distinct points E and F although the lines were of the same length. It is tacitly assumed, that if the lines be straight lines, this is impossible.

EXERCISES.

1. If the straight line joining the middle points of two opposite sides of a quadrilateral be at right angles to each of these sides, the other two sides are equal.

2. If in a quadrilateral ABCD the sides AB, CD be equal and the angles ABC, BCD be equal, the diagonals AC, BD are equal.

3. If in a quadrilateral two opposite sides be equal, and the angles which a third side makes with the equal sides be equal, the other angles are equal.

4. Prove by the method of superposition that, if in two quadrilaterals ABCD, A'B'C'D', the sides AB, BC, CD be equal to the sides A'B', B'C', C'D' respectively, and the angles ABC, BCD equal to the angles A'B'C', B'C'D' respectively, the quadrilaterals are equal in all respects.