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8. Magnitudes which coincide with one another--that is, which exactly fill the same space-are equal to one another.

9. The whole is greater than its part.

10. Two straight lines cannot inclose a space.

11. All right angles are equal to one another.

12. If a straight line meets two straight lines, so as to make the two interior angles on the same side of it, taken together, less than two right angles, these straight lines being continually produced, shall at length meet upon that side on which are the angles which are less than two right angles.

EXPLANATION OF TERMS.

1. A Proposition in geometry, as its name implies, is something proposed either to be done or demonstrated.

2. Propositions fall into two classes, problems and theorems.

3. A Problem proposes some geometrical construction to be done -.g., the construction of a figure.

4. A Theorem proposes some geometrical property to be demonstrated.

5. A Postulate is a problem so simple that it is unnecessary to point out the method of doing it.

6. An Axiom is a theorem, the truth of which is self-evident.

7. A Corollary is an inference made immediately from the discussion of the proposition to which it is subjoined.

PROPOSITION 1.- Problem.

To describe an equilateral triangle upon a given finite straight line. Let AB be the given straight line.

required to describe an equilateral triangle upon AB.

B

Construction. From the centre A, at the distance AB, describe the circle BCD (Postulate 3); from the centre B, at the distance BA, describe the circle ACE; and from C, one of the points in which the circles cut one another, draw the straight lines CA, CB to the points A, B (Post. 1).

Then ABC shall be an equilateral triangle.
Proof. Because the point A is the centre of the circle BCD,

1. AC is equal to AB (Definition 15);
and because the point B is the centre of the circle ACE,

2. BC is equal to BA. But it has been proved that AC is equal to AB; therefore AC, BC are each of them equal to AB; but things which are equal to the same thing are equal to one another (Axiom 1); therefore

3. AC is equal to BC. Wherefore AB, BC, CA are equal to one another; and therefore

4. The triangle ABC is equilateral ; and it is described upon the given straight line AB.

Which was required to be done.

PROPOSITION 2.- Problem. From a given point, to draw a straight line equal to a given straight Let A be the given point, and BC the given straight line. It is required to draw, from the point A, a straight line equal to BC.

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Construction. From the point A to B draw the straight line AB (Post. 1); upon AB describe the equilateral triangle ABD (I. 1), and produce the straight lines DA, DB to E and F (Post. 2); from the centre B, at the distance BC, describe the circle CGH (Post. 3), cutting DF in the point G; and from the centre D, at the distance DG, describe the circle GKL, cutting AE in the point L.

Then the straight line AL shall be equal to BC.
Proof. Because the point B is the centre of the circle CGH,

1. BC is equal to BG (Def. 15);
and because D is the centre of the circle GKL,

2. DL is equal to DG, and DA, DB, parts of them are equal (I. 1); therefore

3. The remainder A L, is equal to the remainder BG (Ax. 3). But it has been shown that BC is equal to BG; wherefore, AL and BC are each of them equal to BG; and things which are equal to the same thing are equal to one another (Ax. 1); therefore

4. The straight line AL is equal to BC. Wherefore, from the given point A, a straight line AL has been drawn equal to the given straight line BC.

Which was to be done.

PROPOSITION 3.- Problem.

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

Let AB and C be the two given straight lines, of which AB is the greater. It is required to cut off from AB the greater, a part equal to C the less.

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F

Construction. From the point A draw the straight line AD equal to C (I. 2); and from the centre A, at the distance AD, describe the circle DEF (Post. 3), cutting AB in the point E.

Then AE shall be equal to C.
Proof. Because A is the centre of the circle DEF,

1. AE is equal to AD (Def. 15). But the straight line C is equal to AD (constr.); whence AE and C are each of them equal to AD; wherefore

2. The straight line AE is equal to C (Ax. 1). And therefore from AB, the greater of two straight lines, a part AE has been cut off equal to C the less. Q.E.F.*

PROPOSITION 4.-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 each other ; (1) they shall have their bases, or third sides, equal ; (2) and the two triangles shall be equal ; (3) and their other angles shall be equal, each to eachviz., those to which the equal sides are opposite.

Let ABC, DEF be two triangles, which have the two sides AB, AC equal to the two sides DE, DF, each to each-viz., AB to DE, and AC to DF, and the included angle BAC equal to the included angle EDF.

Then (1) shall the base BC be equal to the base EF; and (2) the triangle ABC to the triangle DEF; and (3) the other angles to which the equal sides are opposite shall be equal, each to eachviz., the angle ABC to the angle DEF, and the angle ACB to DFE.

* An abbreviation for quod erat faciendumthat is, " which was to be done." Demonstration. For, if the triangle ABC be applied to the triangle DEF, so that the point A may be on D, and the straight line on DE; then

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1. The point B shall coincide with the point E, because AB is equal to DE; and AB coinciding with DE,

2. The straight line AC shall fall on DF, because the angle BAC is equal to the angle EDF; therefore also

3. The point C shall coincide with the point F, because AC is equal to DF. But the point B was shown to coincide with the point Ē; wherefore, the base BC shall coincide with the base EF; because the point B coinciding with E, and C with F, if the base BC do not coincide with the base

EF, the two straight lines BC and EF would inclose a space which is impossible (Ax. 10); therefore

4. The base BC does coincide with EF, and is equal to it, and

5. The whole triangle ABC coincides with the whole triangle

DEF, and is equal to it; also, the remaining angles of one triangle coincide with the remaining angles of the other, and are equal to them, viz.,

6. The angle ABC is equal to the angle DEF, and the angle

ACB to the angle DFE. Therefore, if two triangles have two sides of the one equal to two sides, &c.

Which was to be demonstrated.

PROPOSITION 5.-Theorem. The angles at the base of an isosceles triangle are equal to each other ; and if the equal sides be produced, the angles on the other side of the base shall be equal.

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