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and produce DA, DB to E and F;
(Post. 2) from the centre B, at the distance BC, describe the circle CGH;
(Post. 3) from centre D, at the distance DG, describe the circle GKL.
Then the straight line AL shall be equal to BC.
Because B is the centre of the circle CGH, therefore BC is equal to BG ;
(Def. 15) and because D is the centre of the circle GKL, therefore DL is equal to DG; and DA, DB, parts of these, are equal;
(Constr.) therefore the remainder AL is equal to the remainder BG.
(Ax. 3) But it was shown that BC is equal to BG; therefore AL and BC are each equal to BG; therefore AL is equal to BC.
(Ax. 1) Wherefore, from the given point A, a straight line AL has been drawn equal to the given straight line BC.
Q. E. F. EXERCISE V. 1. Construct the diagram for Proposition II., similarly to the one which has been given, but making the equilateral triangle DAB on the other side of AB.
2. Perform the construction, as directed in the proposition, when A occurs in the continuation of CB.
3. Perform the same construction, placing A in BC itself.
[Other changes in the position of A may be made to furnish the learner with instructive exercises.]
4. Let BC be a given straight line; show how to draw from B a straight line equal to BC.
5. In the figure of Proposition I., produce AB both ways to meet the circles in D and E. Then prove that a straight line has thus been constructed which is equal to the sum of the three sides of the triangle.
6. Given any triangle, produce its base both ways to form a straight line equal to the sum of its sides.
Note.—The reasoning of Prop. III. will be found to be so easy that we need not give any preliminary consideration to it. Its importance must not, on that account, be under-estimated, so far as Euclid's Geometry is concerned. It will be found to meet with frequent application.
From the greater of two given straight lines to cut off a part
equal to the less.
It is required to cut off from AB a part equal to C.
(I. 2) and from the centre A, at the distance AD, describe the circle DEF.
(Post. 3) Then AE shall be equal to C.
Because A is the centre of the circle DEF, therefore AE is equal to AD;
(Def. 15) but the straight line C is equal to AD;
(Constr.) therefore AE and C are each equal to AD;
wherefore the straight line AE is equal to C. (Ax. 1)
Therefore from AB, the greater of two given straight lines, a part AE has been cut off equal to the less.
Q. E. F.
EXAMINATION XVII. 1. Give the derivation of the word 'proposition,' and infer its literal meaning. What is its usual meaning in Geometry? 2. Give the derivation and use of Enunciation. 3. Name the two classes of propositions in Geometry, and define 'problem.' 4. What is meant by solving' a problem ? 5. When is a straight line said to be finite ? 6. What is the
construction of a proposition ? 7. Explain the letters Q. E. F. 8. In the enunciation of Proposition I. separate what is given from what is required. 9. Similarly for Prop. II. 10. Also for Prop. III.
27. WHAT ARE THEOREMS ? Our Propositions now change character, and become “Theorems.'
A Theorem is a Proposition which places before us some truth which it is proposed to demonstrate.
In the enunciation of a Theorem, there are always two parts, called the · Hypothesis,' and the ‘Thesis,' or “Conclusion.'
The Hypothesis of a theorem states one or more conditions which are to be granted
The Thesis, or Conclusion, consists of an assertion which we propose to prove if the hypothesis be granted.
Thesis is a Greek word, meaning something placed or set before us ; while the prefix hypo means under. We may, therefore, compare the hypothesis to a foundation upon which is built up the thesis.
In the course of the argument, it will be necessary from time to time to refer to the hypothesis, and this will be indicated by an abbreviation in parentheses, thus—(Hyp.).
As an example of separating hypothesis from thesis, let us take a simple theorem, namely
If two radii of a circle form two sides of a triangle, the triangle will be isosceles.
Hypothesis : Two of the sides of a given triangle are radii of the same circle.
Thesis or Conclusion : That triangle is isosceles. In the following exercise, the phrase "each to each' means the same as • respectively. Thus, if A, B, C are equal to X, Y, Z, each to each ; this means that A is equal to X, B equal to Y, and C equal to Z; or that the following are pairs of equal angles : A and X, B and Y, C and Z.
EXERCISE VI. 1. If two triangles ABC, DEF have two angles A and B of the one equal to the two D, E of the other, each to each ; give the same equal angles in pairs of equals.
2. If angles A, B, C of the one triangle be equal to angles E, F, D of the other, each to each ; write the equal angles in pairs.
3. If two sides AB, BC of the triangle ABC be equal to the two sides DE, EF of the triangle DEF, each to each ; write those sides in pairs of equals.
4. Two sides of a triangle have been mentioned ; what name may be then given to the third side ?
5. Make ABC, DEF any two triangles, then write with three letters the angles opposite to AB, DE, AC, DF, BC, EF, respectively.
6. Distinguish between the angle BAC and the triangle BAC.
7. If we fit or apply one surface upon another, in order to try whether the two surfaces are equal, what axiom do we employ ?
8. In the process of applying one straight line to another in order to see whether they are equal or not, distinguish three separate steps.
9. Referring to the figure of Proposition IV., suppose we have taken up the triangle ABC from its place and have put it down so as to have AB upon DE exactly, and we then find that AC will also just fall upon DF; what must be true respecting the angles A and D ?
10. On the other hand, suppose we have AB upon DE, and we know that angle A is equal to angle D, where must AC fall if the triangle ABC be pressed down?
11. Suppose the boundaries of one figure fit upon the boundaries of another, what is the inference respecting their areas or surfaces inclosed ?
12. Suppose we observe that, of two triangles ABC, DEF, we can make, at the same time, AB fall upon DE, AC fall upon DF, and BC fall upon
what may we say of the areas of these triangles ? 13. Having thus found that AB and BC fit respectively upon DE, EF, draw the inference respecting the angles ABC, DEF.
Which sides must fit upon one another in order that we may infer that angles ACB and DFE are equal ?
14. When B falls upon E, and C upon F, quote the axiom which shows the straight line BC must coincide with the straight line EF.
Note.—Although Proposition IV. may appear at first somewhat difficult, it is proved by very simple means. It is not that the proof is really difficult, but rather that its phraseology is apt to be only partially understood. On this account, if the student has not seen his way clearly in each step of the preceding exercise, it will be quite worth his while to work it through again, in whole or in part. He will then have practised every process which occurs in proving Proposition IV., and there only remains to build up the separate parts of the argument into one whole.
28. PROPOSITION IV.—THEOREM. 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; then they shall have their bases equal, and the two triangles themselves shall be equal, and of their other angles those shall be equal to which equal sides are opposite. Let ABC, DEF be two triangles which have the two sides
AB, AC, and the angle BAC which they contain, respectively equal to the two sides DE, DF, and the angle
EDF which these contain; namely, AB equal to DE, AC to DF, and the angle BAC to the angle EDF.
Then shall be equal also : the bases BC, EF; the two triangles ABC, DEF; the angles ABC, DEF, to which AC, DF are opposite : and the angles ACB, DFE, to which AB, DE are opposite.
For, if the triangle ABC be applied to the triangle DEF, placing the point A upon D, and the straight line AB along DE; then, because AB is equal to DE,
(Hyp.) the point B will coincide with the point E.
And because AB thus coincides with DE, and the angle BAC is equal to the angle EDF, (Hyp.) therefore AC will fall along DF; also, because AC is equal to DF,
(Hyp.) therefore the point C will coincide with the point F.
en, since B coincides with E, and C with F, the base BC will fall along EF; or, otherwise, BC and EF will enclose a space; which is impossible.