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1. Things which are equal to the same thing are equal to one another.
2. If equals be added to equals the wholes are equal.
3. If equals be taken from equals the remainders are equal. 4. If equals be added to unequals the wholes are unequal. 5. If equals be taken from unequals the remainders are unequal.
6. Things which are double of the same are equal to one another.
7. Things which are halves of the same are equal to one another.
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 meet 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 on that side on which are the angles which are less than two right angles.
EXPLANATION OF TERMS.
A proposition in geometry is something proposed to be done, as a problem; or to be proved, as a theorem.
A problem proposes something to be done, such as the construction of a figure.
A theorem makes an assertion, the truth of which it proposes to demonstrate.
A postulate is a problem so simple that it is unnecessary to point out the method of performing it.
An axiom is a theorem the truth of which is so apparent as to be instantly admitted.
A corollary is an inference which arises easily and immediately from the discussion of the proposition to which it is affixed.
A lemma is a proposition of no importance in itself, merely introduced for the purpose of demonstrating some other propo
A scholium is a note or explanatory observation.,
To describe an equilateral triangle on a given finite straight line.
* (References-Def. 15; ax. 1; post. 1, 3.)
Given.—Let AB be the given straight line.
Sought. It is required to describe an equilateral triangle on AB.
Construction.-1. From the
centre A, at the distance AB, describe the circle BCD. (post. 3.)
2. From the centre B, at the distance BA, describe the circle ACE. (post. 3.)
3. From the point C, in which the circles cut one another, draw the straight lines CA, CB, to the points A and B. (post. 1.)
Then ABC shall be an equilateral triangle.
Proof.-1. Because the point A is the centre of the circle BCD, AC is equal to AB.
2. Because the point B is the centre of the circle ACE, BC is equal to BA. (def. 15.)
3. Therefore AC and BC are each of them equal to AB. 4. But things which are equal to the same thing are equal to one another. Therefore AC is equal to BC. (ax. 1.)
5. Therefore AB, BC, and CA are equal to one another. Conclusion. Therefore the triangle ABC is equilateral, and it is described on the given straight line AB. Which was to be done.
From a given point to draw a straight line equal to a given straight line.
(References-Prop. I. 1; post. 1, 2, 3; ax. 1, 3; def. 15.)
Given. Let A be the given point, and BC the given straight line.
Sought. It is required to draw from the point A a straight line equal to BC.
Construction.-1. From the point A to B draw the straight line AB. (post. 1.)
*To be repeated by the pupil before commencing to discuss the Troposition.
2. Upon AB describe the equilateral triangle DAB. (Book
I. prop. 1.)
3. Produce the straight lines DA, DB, to E and F. (post. 2.) 4. From the centre B, at the distance BC, describe the circle CGH, meeting DF in G. (post. 3.)
5. From the centre D, at the distance DG, describe the circle GKL, meeting DE in L. (post. 3.)
Then AL shall be equal to BC. Proof.-1. Because the point B is the centre of the circle CGH, BC is equal to BG. (def. 15.)
2. Because the point D is the centre of the circle GKL, DL is equal to DG. (def. 15.)
3. But DA, DB, parts of them, are equal. (construction.) 4. Therefore the remainder AL is equal to the remainder BG. (ax. 3.)
5. But it has been shown that BC is equal to BG.
6. Therefore AL and BC are each of them equal to BG.
7. But things which are equal to the same thing are equal to one another, therefore AL is equal to BC. (ax. 1.)
Conclusion. Therefore from the given point A a straight line AL has been drawn equal to the given straight line BC. Which was to be done.
From the greater of two given straight lines to cut off a part equal to the less.
(References-Prop. I. 2; post. 3; ax. 1; def. 15.)
Given.-Let AB and C be the two given straight lines, of which AB is the greater.
Sought. It is required to cut off from AB, the greater, a part equal to C, the less.
Construction.-1. From the point
A draw the straight line AD equal to
2. From the centre A, at the distance AD, describe the circle DEF,
cutting AB in E. (post. 3.)
Then AE shall be equal to C.
Proof.-1. Because the point A is the centre of the circle
DEF, AE is equal to AD.~ (def. 15.)
2. But C is also equal to AD. (construction.)
3. Therefore AE and C are each of them equal to AD. 4. Therefore AE is equal to C. (ax. 1.)
Conclusion.-Therefore, from AB, the greater of two given straight lines, a part AE has been cut off, equal to C, the less. Q. E. F*
If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one another: (1.) they shall have their bases, or third sides, equal; (2) the two triangles shall be equal; and (3.) their other angles shall be equal, each to each, viz. those to which the equal sides are opposite. Or,
If two sides and the contained angle of one triangle be respectively equal to those of another, the triangles are equal in every respect.
(References—Ax. 8, 10.)
Hypothesis.-Let ABC, DEF, be two triangles which have, 1. The two sides AB, AC, equal to the two sides DE, DF, each to each, viz. AB equal
to DE, and AC equal to DF. 2. And the angle BAC equal to the angle EDF:-thenSequence.-1. The base BC shall be equal to the base EF.
2. The triangle ABC shall be equal to the triangle DEF.
3. And the other angles to which the equal sides are opposite, shall be equal, each to each, viz. the angle ABC to the angle DEF, and the angle ACB to the angle DFE.
Demonstration.-1. For if the triangle ABC be applied to (or placed upon) the triangle DEF,
2. So that the point A may be on the point D, and the straight line AB on the straight line DE,
3. The point B shall coincide with the point E, because AB is equal to DE. (hypothesis.)
4. And AB coinciding with DE, AC shall coincide with DF, because the angle BAC is equal to the angle EDF. (hyp.) 5. Therefore also the point C shall coincide with the point F, because the straight line AC is equal to DF. (hypothesis.)
*Q. E. F. is an abbreviation for quod erat faciendum, that is, "which was to be done."
6. But the point B was proved to coincide with the point E. 7. Therefore the base BC shall coincide with the base EF.
8. Because the point B coinciding with E, and C with F, if the base BC do not coincide with the base EF, two straight lines would inclose a space, which is impossible. (ax. 10.)
9. Therefore the base BC coincides with the base EF, and is therefore equal to it. (ax. 8.)
10. Therefore the whole triangle ABC coincides with the whole triangle DEF, and is equal to it. (ax. 8.)
11. And the other angles of the one coincide with the remaining angles of the other, and are equal to them, viz. the angle ABC to DEF, and the angle ACB to DFE.
Conclusion. Therefore, if two triangles have, &c. (see Enunciation). Which was to be shown.
The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced, the angles upon other side of the base shall also be equal.
(References-Prop. I. 3, 4; ax. 3.)
Hypothesis.-1. Let ABC be an isosceles triangle, of which the side AB is equal to the side AC.
2. Let the straight lines AB, AC (the equal sides of the triangle), be produced to D and E.
Sequence.-1. The angle ABC shall be equal to the angle ACB (angles at the base).
2. And the angle CBD shall be equal to the angle BCE (angles upon the other side of the base).
Construction.-1. In BD take any point F.
2. From AE, the greater, cut off AG, equal to AF, the less. (I. 3.)
3. Join FC, GB.
Demonstration.-1. Because AF is equal to AG. (constr.) And AB is equal to AC. (hypoth.)
Therefore the two sides FA, AC, are equal to the two sides GA, AB, each to each.
2. And they contain the angle FAG, common to the two triangles AFC, AGB.
3. Therefore the base FC is equal to the base GB, (I. 4.)
4. And the triangle AFC to the triangle AGB. (I. 4.)