PROPOSITION III. PROBLEM.

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

Let AB be the greater of the two given st. lines AB, CD. It is required to cut off from AB a part=CD.

From A draw the st. line AE= CD.

With centre A and distance AE describe EFH,

I. 2.

Thus from AB a part AF has been cut off- CD.

EXERCISES.

Ax. 1.

Q. E. F.

1. Shew that if straight lines be drawn from A and B in the diagram of Prop. I. to the other point in which the circles intersect, another equilateral triangle will be described on AB.

2. By a construction similar to that in Prop. III. produce the less of two given straight lines that it may be equal to the greater.

3. Draw a figure for the case in Prop. II., in which the given point coincides with B.

4. By a similar construction to that in Prop. 1. describe on a given straight line an isosceles triangle, whose equal sides shall be each equal to another given straight line.

PROPOSITION IV. THEOREM.

If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal to one another, they must have their third sides equal; and the two triangles must be equal, and the other angles must be equal, each to each, viz. those to which the equal sides are opposite.

AA

In the As ABC, DEF,

let AB=DE, and AC=DF, and ▲ BAC=▲ EDF. Then must BC=EF and ▲ ABC = A DEF, and the other Ls, to which the equal sides are opposite, must be equal, that is, ▲ ABC= ▲ DEF and ▲ ACB= 2 DFE.

For, if ▲ ABC be applied to ▲ DEF,

so that A coincides with D, and AB falls on DE,
AB=DE, .. B will coincide with E.

then

And

AB coincides with DE, and ▲ BAC=▲ EDF, Hyp .. AC will fall on DF.

Then

AC=DF, .. C will coincide with F. And. B will coincide with E, and C with F, .. BC will coincide with EF;

for if not, let it fall otherwise as EOF: then the two st. lines BC, EF will enclose a space, which is impossible. Post. 5. .. BC will coincide with and.. is equal to EF,

Ax. 8.

À DEF,

L DEF,

LDFE.

Q. E. D.

NOTE 1. On the Method of Superposition.

Two geometrical magnitudes are said, in accordance with Ax. VIII. to be equal, when they can be so placed that the boundaries of the one coincide with the boundaries of the other.

Thus, two straight lines are equal, if they can be so placed that the points at their extremities coincide: and two angles are equal, if they can be so placed that their vertices coincide in position and their arms in direction: and two triangles are equal, if they can be so placed that their sides coincide in direction and magnitude.

In the application of the test of equality by this Method of Superposition, we assume that an angle or a triangle may be moved from one place, turned over, and put down in another place, without altering the relative positions of its boundaries.

We also assume that if one part of a straight line coincide with one part of another straight line, the other parts of the lines also coincide in direction; or, that straight lines, which coincide in two points, coincide when produced.

The method of Superposition enables us also to compare magnitudes of the same kind that are unequal. For example, suppose ABC and DEF to be two given angles.

Suppose the arm BC to be placed on the arm EF, and the vertex B on the vertex E.

Then, if the arm BA coincide in direction with the arm ED, the angle ABC is equal to DEF.

If BA fall between ED and EF in the direction EP, ABC is less than DEF.

If BA fall in the direction EQ so that ED is between EQ and EF, ABC is greater than DEF.

NOTE 2. On the Conditions of Equality of two Triangles.

A Triangle is composed of six parts, three sides and three angles.

When the six parts of one triangle are equal to the six parts of another triangle, each to each, the Triangles are said to be equal in all respects.

There are four cases in which Euclid proves that two triangles are equal in all respects; viz., when the following parts are equal in the two triangles.

1. Two sides and the angle between them.

I. 4.

2. Two angles and the side between them.

I. 26.

3. The three sides of each.

I. 8.

I. 26.

4. Two angles and the side opposite one of them.

The Propositions, in which these cases are proved, are the most important in our First Section.

The first case we have proved in Prop. IV.

Availing ourselves of the method of superposition, we can prove Cases 2 and 3 by a process more simple than that employed by Euclid, and with the further advantage of bringing them into closer connexion with Case 1. We shall therefore give three Propositions, which we designate A, B, and C, in the Place of Euclid's Props. V. VI. VII. VIIÍ.

The displaced Propositions will be found on pp. 108-112. Proposition A corresponds with Euclid I. 5.

B

с

I. 26, first part.

I. 8.

PROPOSITION A. THEOREM.

If two sides of a triangle be equal, the angles opposite those sides must also be equal.

In the isosceles triangle ABC, let AC=AB. (Fig. 1.)

Then must 4 ABC= 2 ACB.

Imagine the ABC to be taken up, turned round, and set down again in a reversed position as in Fig. 2, and designate the angular points A', B', C'.

Then in As ABC, A'C'B',

· AB=A'C', and AC=A'B', and ▲ BAC CA'B',

L

I. 4.

Ax. 1.

Q.E.D.

COR. Hence every equilateral triangle is also equiangular.

NOTE. When one side of a triangle is distinguished from the other sides by being called the Base, the angular point opposite to that side is called the Vertex of the triangle.