5. Euclid's reasoning is said to be Deductive, because by a connected chain of argument it deduces new truths from truths already proved or admitted. Thus each proposition, though in one sense complete in itself, is derived from the Postulates, Axioms, or former propositions, and itself leads up to subsequent propositions.

6. The initial letters Q.E. F., placed at the end of a problem, stand for Quod erat Faciendum, which was to be done.

The letters Q.E.D. are appended to a theorem, and stand for Quod erat Demonstrandum, which was to be proved.

7. A Corollary is a statement the truth of which follows readily from an established proposition; it is therefore appended to the proposition as an inference or deduction, which usually requires no further proof.

8. The attention of the beginner is drawn to the special use of the future tense in the Particular Enunciations of Euclid's propositions.

The future is only used in a statement of which the truth is about to be proved. Thus: "The triangle ABC SHALL BE equilateral" means that the triangle has yet to be proved equilateral. While, "The triangle ABC is equilateral" means that the triangle has already been proved (or given) equilateral.

9. The following symbols and abbreviations may be employed in writing out the propositions of Book I., though their use is not recommended to beginners.

and all obvious contractions of words, such as opp., adj., diag., etc.,

for opposite, adjacent, diagonal, etc.

Let AB be the given straight line.

It is required to describe an equilateral triangle on AB.

Construction. With centre A, and radius AB, describe the circle BCD.

Post. 3.

With centre B, and radius BA, describe the circle ACE.

Post. 3. From the point C at which the circles cut one another, draw the straight lines CA and CB to the points A and B.

Then shall the triangle ABC be equilateral.

Post. 1.

Proof. Because A is the centre of the circle BCD,

therefore AC is equal to AB.

Def. 15.

And because B is the centre of the circle ACE,

therefore BC is equal to AB.

Def. 15.

Therefore AC and BC are each equal to AB.

But things which are equal to the same thing are equal

to one another.

Therefore AC is equal to BC.

Therefore AC, AB, BC are equal to one another. Therefore the triangle ABC is equilateral; and it is described on the given straight line AB.

Ax. 1.

Q.E.F.

PROPOSITION 2. PROBLEM.

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

K

H

Let A be the given point, and BC the given straight line. It is required to draw from ▲ a straight line equal to BC.

Construction.

Join AB;

Post. 1. and on AB describe an equilateral triangle DAB. I. 1. With centre B, and radius BC, describe the circle CGH.

Produce DB to meet the circle CGH at G. With centre D, and radius DG, describe the circle Produce DA to meet the circle GKF at F. Then AF shall be equal to BC.

Post. 3.

Post. 2.

GKF.

Post. 2.

Proof.

Because B is the centre of the circle CGH,
therefore BC is equal to BG.

Def. 15.

therefore DF is equal to DG.

Def. 15.

And because D is the centre of the circle GKF,

And DA, a part of DF, is equal to DB, a part of DG; Def. 24. therefore the remainder AF is equal to the remainder BG.

Ax. 3.

But BC has been proved equal to BG; therefore AF and BC are each equal to BG. And things which are equal to the same thing are equal to one another.

Therefore AF is equal to BC;

and it has been drawn from the given point A.

Ax. 1.

Q.E.F.

PROPOSITION 3. PROBLEM.

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

E

Let AB and C be the two given straight lines, of which AB is the greater.

It is required to cut off from AB a part equal to C.

Construction. From the point A draw the straight line
AD equal to C;

I. 2.

and with centre A and radius AD, describe the circle DEF, cutting AB at E.

Post 3.

Then AE shall be equal to C.

Proof.

Because A is the centre of the circle DEF,
therefore AE is equal to AD.

Def. 15.

But C is equal to AD.

Therefore AE and C are each equal to AD.

Therefore AE is equal to C;

and it has been cut off from the given straight line AB.

Constr.

Ax. 1.

Q.E.F.

EXERCISES ON PROPOSITIONS 1 TO 3.

1. If the two circles in Proposition 1 cut one another again at F, prove that AFB is an equilateral triangle.

2. If the two circles in Proposition I cut one another at C and F, prove that the figure ACBF is a rhombus.

3. AB is a straight line of given length: shew how to draw from A a line double the length of AB.

4. Two circles are drawn with the same centre O, and two radii OA, OB are drawn in the smaller circle. If OA, OB are produced to cut the outer circle at D and E, prove that AD = BE.

5. AB is a straight line, and P, Q are two points, one on each side of AB. Shew how to find points in AB, whose distance from P is equal to PQ. How many such points will there be?

6. In the figure of Proposition 2, if AB is equal to BC, shew that D, the vertex of the equilateral triangle, will fall on the circumference of the circle CGH.

7. In Proposition 2 the point A may be joined to either extremity of BC. Draw the figure, and prove the proposition in the case when A is joined to C.

8. On a given straight line AB describe an isosceles triangle having each of its equal sides equal to a given straight line PQ.

9. On a given base describe an isosceles triangle having each of its equal sides double of the base.

10. In a given straight line the points A, M, N, B are taken in order. On AB describe a triangle ABC, such that the side AC may be equal to AN, and the side BC to BM.

NOTE ON PROPOSITIONS 2 AND 3.

Propositions 2 and 3 are rendered necessary by the restriction tacitly imposed by Euclid, that compasses shall not be used to transfer distances. [See Notes on the Postulates.]

In carrying out the construction of Prop. 2 the point A may be joined to either extremity of the line BC; the equilateral triangle may be described on either side of the line so drawn ; and the sides of the equilateral triangle may be produced in either direction. Thus there are in general 2×2×2, or eight, possible constructions. The student should exercise himself in drawing the various figures that may arise.