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DOOK I. ---POSTULATES AXIOMS.

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XXV. An isosceles triangle is that which has only two sides equal.

XXVI.
A scalene triangle is that which has three un qual
sides.

XXVII.
A right-angled triangle is that which has a right angle.

XXVIII.
An obtuse-angled triangle is that which has an obtuse
angle.

XXIX. An acute-angled triangle is that which has three acute angles.

XXX. Of four-sided figures, a square is that which has all its sides equal, and all its angles right angles.

XXXI. An oblong is that which has all its angles right angles, but has not all its sides equal.

XXXII. A rhombus is that which has all its sides equal, but its angles are not right angles.

XXXIII. A rhomboid is that which has its opposite sides equal to one another, but all its sides are not equal, nor its angles right angles.

XXXIV. All other four-sided figures, besides these, are called trapeziums.

XXXV. Parallel straight lines are such as are in the same plane, and which being produced ever so far both ways do not meet.

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POSTULATES. 1. Let it be granted that a straight line may be drawn from any one point to any other point.

2. That a terminated straight line may be produced to any length in a straight line.

3. And that a circle may be described from any centre, at any distance from that centre.

AXIOMS. 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 which is exactly fill the same space, are equal to one another.

9. The whole is greater than its part.
10. Two straight lines cannot enclose 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.

DEFINITIONS OF TERMS. 1. A PROPOSITION is something proposed either to be done or to be demonstrated, and is either a problem or a theorem. The words in which a proposition is delivered are called the enunciation of that proposition.

2. A PROBLEM is something which is proposed to be done; and when what is required to be done is so manifestly possible as to require no argument to prove it, such a problem is called a postulate.

3. A THEOREM is something which is proposed to be demonstrated. When the truth affirmed in a theoren is so obvious as to require no argument to prove it, the theorem is then called an axiom.

4. DEMONSTRATION is that process by which the truth affirmed is clearly made out. Here all the arguments must be drawn from first principles, such as the definitions which are always strictly to be taken in the same sense, the postulates, and the axioms; or, from propositions already demonstrated.

5. A DIRECT DEMONSTRATION is that which concludes with the certain and direct proof of the proposition in hand. See the 5th Prop. of Book I.

6. An INDIRECT, or NEGATIVE DEMONSTRATION, is that which shows a proposition to be true by some absurdity, which would necessarily follow if the affirmation in the theorem should be false. This is calied reductio ad absurdum, and shows the absurdity and falsehood of all suppositions but that which is contained in the proposition. See the 6th Prop. of Book I.

7. HYPOTHESIS is a supposition assumed to be true, and upon which we are to reason as if it were so. It is not absolutely necessary that an hypothesis should be true, but it must be possible, and should always be probable. Thus in the 6th Prop. of Book I. Euclid supposes AB to be greater than AC; but had he assumed AC greater than AB, a conclusion equally correct might have been drawn.

8. A COROLLARY is a consequent truth gained from some preceding truth or demonstration.

9. A SCHOLIUM is a remark or observation made upon something going before it.

10. A LEMMA is a short preparatory proposition, previously demonstrated, in order to shorten the demonstration of the main proposition, and to render what follows more easy.

Note.—The citations in the different propositions are to be understood as follow :-When two numbers are met with (I. 3.) or (I. 4.) the first shows the book referred to, and the second the proposition; as the third Proposition of the first Book, &c. Also (Post.) denotes postulate, (Ax.) axiom, (Def.) definition, (Cor.) corollary, (Hyp.) hypothesis, Cons.) construction; and tho figure, or figures, after the words, show the number of the postulate, axiom, definition, &c.; as (Post. 3.) the third postulate, (Def. 15.) the 15th definition, &c.

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Prop. I. (PROBLEM.) – To describe an equilateral triangle upon a given

finite straight line (AB). From the centre A, at the distance AB, describe (Post. 3) the circle BCD. From the centre B, at the distance BA, describe the circle A CE. From the point C, in which the circles cut one another, draw the straight lines (Post. 1) CA, CB, to the points A, B. Then ABC is an equilateral triangle.

AB E Because the point A is the centre of the circle BCD, AC is equal (Def. 15) to AB. And because the point B is the centre of the circle ACE, BC is equal to BA. But it has been proved that CA is equal to AB. Therefore the two straight lines CA, CB, are each of them equal to AB. But things which are equal to the same thing are equal (Ax. 1) to one another. Therefore CA is equal to CB. Wherefore the three sides CA, AB, BC, are equal to one another. The triangle ABC is, therefore, equilateral; and it is described upon the given straight lipe AB. Q. E. F.

PROP. II. (PROBLEM.)From a given point (A) to draw a straight line

equal to a given straight line (BC). From the point A to B draw (Post. 1) the straight line AB. Upon A B describe (I. 1) the equilateral triangle DAB. Produce (Post. 2) the straight lines DA, DB, to the points E and F. From the centre B, at the distance BC, describe (Post. 3) the circle CGH. From the centre D, at the distance DG, describe the circle GKL. Then, the straight line AL is equal to BC.

Because the point B is the centre of the circle CGH, BC is equal (Def. 15) to BG. And because the point D is the centre of the circle ĠKL, DL is equa! to DG. But (Const.) DA, DB, parts of these equals, are equal. Therefore

K к the remainder AL is equal (AX. 3) to the remainder BG. But it has been shown that BC is equal to BG. Wherefore AL and BC are each of them equal to BG. And things that are equal to the same thing are (Ax. 1) equal to one another. - Therefore the straight line AL is equal to the straight line BC. Wherefore from the given point A, a straight line AL has been drawn equal to the given straight line BC. Q. E. F.

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PROP. III. (PROBLEM.)–From the greater of two given straight lines

(AB) to cut off a part equal to the less (C). From the point A draw (I. 2) the straight line AD equal to C. From the centre A, at the distance AD, describe (Post. 3) the circle DEF. Then the part AE shall be equal to C.

Because A is the centre of the circle DEF, AE is equal (Def. 15) to AD. But the straight line C is likewise equal (Const.) to AD. Therefore AE and C are each of them equal to AD. Wherefore the straight line A E is equal (Ax. 1) to C. Therefore from AB the greater of two given straight lines, a part A E has been cut off equal to C the less. Q. E. F.

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Prop. IV. (Theorem.)-- If two triangles (ABC, DEF) have two sides of

the one (ÀB, AC) equal to two sides of the other (DE, DF), each to each (viz., AB to DÊ, and AC to DF), and have likewise the angles (BAC, EDF) contained by these sides equal to one another, their bases

BC and EF), or third sides, are equal; the two triangles (ABC, DE F) are equal; and their other angles are equal, each to each, viz. those to which the equal sides are opposite; and the remaining angles of the one are equal to the remaining angles of the other, each to each, viz., ihose to which the equal sides are opposite.

For, if the triangle ABC be applied to the triangle DEF, so that the point A may be on the point D, and the straight line AB upon the straight line DE. The point B shall coincide (that is, fall upon, so as to agree) with the point E, because AB is equal (Hyp.) to DÈ; and A B coinciding with DE, AC shall coincide with DF, because the angle BAC is equal (Hyp.) to the angle EDF Also, the point C shall coincide with the point F, because AC (Hyp.) is equal to DF. But the point B was proved to coincide with the point E. Therefore the base BC shall coincide with the base EF. For, the point B coinciding with the point E, and the point C with the point F, if the base BC does not coincide with the base EF, the two straight lines BC, EF, would enclose a space, which (Ax. 10) is impossible. Wherefore the base BC coincides with the base EF, and is, therefore, equal (A.c. 8) to it. Wherefore, also, the whole triangle ABC coincides with the whole triangle DEF, and is, therefore, equal to it. And the remaining angles of the one coincide with the remaining angles of the other, and are therefore equal to them, each to each, viz., the angle A BC to the angle DEF, and the angle ACB to the angle DFE. Therefore, if two triangles have two sides, &c. Q. E. D.

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Prop. V. (THEOREM.)The angles (ABC, ACB) at the base of an isoscele

triangle (ABC) are equal to one another; and if the egual sides (AB, AC) be produced (to D and E), the angles (DBC, ECB) upon the other side of the base shall be equal.

In BD take any point F, and from AE the greater, cut off AG equal (I. 3) to AF the less. Join FC, GB.

Because in the two triangles AFC, AGB, AP is equal to (Cunst.)

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AG, and AB to (Hyp.) AC, the two sides FA, AC are equal to the trvo sides G A, AB, each to each ; and they contain the angle FAG common to the two triangles AFC, AĞB. Therefore the base FC is equal (I. 4) to the base GB, and the triangle AFC to the triangle AGB. Also, the remaining angles of the one are equal (I. 4) to the remaining angles of the other, each to each, viz., those to which the equal sides are opposite; that is, the angle ACF to the angle ABG, and the angle AFC to the angle AGB. Again, because the whole AF is equal to the whole AG, of which the parts AB, AC, are equal. Therefore the remainder BF is equal (A23) to the remainder CG. But, FC was proved to be equal to GB. Therefore, in the two triangles BFC, GCB, the two sides BF, FC are equal to the two sides CG, GB, each to each. And the angle BFC was proved to be equal to the angle CGB. Wherefore the two triangles BFC, CGB are equal (I. 4), and their remaining angles are equal, each to each, viz., those to which the equal sides are opposite. Therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. But it has been demonstrated, that the whole anglo ABG is equal to the whole angle ACF, and the part CBG of the one, equal to the part BCF of the other. Therefore the remaining angle ABC is equ

(Ax. to the remaining angle ACB, and these are the angles at the base of the triangle ABC. It has also been proved that the angle FBC is equal to the angle GCB, and these are the angles upon the other side of the base. Therefore the angles at the base, &c. Q. E. D.

PROP. VI. (THEOREM.)- If txo angles (ABC, ACB) of a triangle

(ABC) be equal to one another, the sides (A'B, AC) which subtend or are opposite to the equal angles, are equal to one another.

For, if AB be not equal to AC, one of them is greater than the other. Let AB be the greater ; and froin BA cut (I. 3) off BD equal to AC, the less. And join DC.

Because in the two triangles DBC, ACB, the side DB is equal to the side AC, and BC is common to both, the two sides DB, BC are equal to the two sides AC, CB, each to each; and the angle DBC is equal to the angle (Hyp.) ACB. Therefore the base DC is equal to the base AB; and the triangle DBC is equal to the triangle (I. 4) ACB, the less to the greater, which is absurd. Therefore AB is not unequal to AC; that is, AB is equal to AC. Wherefor, if two angles, &c. Q. E. D.

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