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9. Suppose now it is asserted that DB, BC, and the angle DBC of the triangle DCB are respectively equal to AC, CB, and the angle ACB of the triangle ACB ; then, by Prop. IV., draw an inference respecting the areas of the triangles DBC, ACB. Quote an axiom by which such a conclusion in this case would be contradicted. What is the conclusion you come to on the point ?
10. State the sides which respectively subtend the angles BCD, BCA, BDC, ADC. What angles are both subtended by CD?
34. PROPOSITION VI.-THEOREM. If two angles of a triangle be equal, the sides which subtend
them shall also be equal. Let ABC be a triangle, having the angles ABC, ACB equal. Then the sides AB, AC opposite them shall be equal also. For, if AB be not equal to AC, one of them must be greater
than the other. If possible, let AB be the greater. From BA cut off BD equal to CA,
(I. 3) and join DC.
Then, in the triangles DBC, ACB, DB is equal to AC,
(Constr.) and BC common to both triangles; therefore 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 ACB. (Hyp.)
Therefore the triangle DBC is equal to the triangle ACB, (I. 4) the less to the greater; which is absurd.
Hence, AB and AC are not unequal; that is, they are equal. Wherefore, if two angles, &c.
Q. E. D. Cor.—Every equiangular triangle is also equilateral.
EXERCISE IX. 1. Given, in the triangle ABC, that CBA, CAB are equal angles, name the sides which are equal by Prop. VI.
2. Given, that BCA, BAC are also equal angles, name another pair of equal sides.
3. In the triangle XYZ, angles X, Z are equal; which sides are equal ? 4. Prove the corollary to Prop. VI.
5. Let BG and CF in the figure of Prop. V. meet in H, then show that HBC is an isosceles triangle. (Assume what is proved in Prop. V. about the triangles BCF and CBG.)
35. INDIRECT PROOF.
In the early portions of Geometry, we make our steps so simple that it is easy to see that the way in which they are deduced one from another is right. Certainly, it would be an advantage, in regard to this point, to know something of Logic. But we may feel safe enough for the present, if we are careful, and keep Euclid's propositions well before us as examples.
Assuming that we keep our mode of reasoning correct, if we commence our argument with a statement of some fact or facts, then every one of its steps will also form a true statement. Hence, if we should come to a step in the reasoning which we know is not true, we infer at once that the statement we started with will be also untrue. In other words, the hypothesis of the argument is wrong.
It may happen that we wished to prove this particular hypothesis wrong, in order to prove that some other statement is right. If so, our proof of the latter is called an indirect proof.
Suppose we say: “Here is a piece of glistening white substance; it is sugar.' In saying this, we have adopted a thesis, namely, the substance is sugar. Suppose we proceed to test this thesis, by making it the basis or hypothesis of further reasoning, and say, - We will put it into hot tea, and the tea will then taste sweet.” Afterwards, however, we make a trial of the tea, and find it not sweet at all, but having a briny flavour. What must be our conclusion ? It must be that our present hypothesis is false; for we know now the substance is not sugar. Instead of calling it false, however, we might call it a trial hypothesis, or a trial thesis in regard to the original proposition.
This kind of reasoning is often called experimental. In Geometry we argue in a way which is very similar to it, when we call the reasoning 'indirect.'
The proof of Proposition VI. is indirect; and the learner may now, perhaps, make out its false thesis, the reasoning based upon it as a new
hypothesis, the false conclusion we are led to, and the final or true conclusion.
Def.-An indirect proof of a proposition is one which shows that any hypothesis contradicting it could be made to lead to some false conclusion.
The student may hereafter observe that the second of two converse theorems is usually proved indirectly.
(In 1-13, Diagram I. is referred to.) 1. If the angles COE, HOK are equal, find three parts of the triangle COE equal to three of HOK, each to each, so as to correspond with the hypothesis of Prop. IV.
2. Draw the conclusions established for these triangles by the same proposition.
3. Given that OL bisects the angle HOK, find three elements of the triangle HOL and three of the triangle KOL which are equal in pairs.
4. Deduce four other equalities in respect to the same triangles. 5. State the common part of the two lines AT, FP. 6. State what angle is common to the two angles AOC, BOD? 7. Also state the side which is common to the two triangles HOG, HOR. 8. Also the angle which is common to the two triangles EOF and EOT. 9. And the angle common to the three triangles CDO, CRO, and QOR.
10. Write a pair of equal angles for each of the triangles AOC, FOG, QOR.
(In 11—14, say which angle is greater, giving the reason fully.) 11. OEC and OCD. 12. OHX and OKH. 13. OST and OTE. 14. EAB and BCA in Diagram IV.
15. In Diagram IV. ABCDEF has six equal sides; find six pairs of equal angles in the figure.
16. O is the centre of the circle ABC in the same; prove the angles QAF and OEF are equal.
36. THE MUTUAL RELATIONS OF PROPS. 1.-VI. The leading proposition of this section is the fourth, in which, under certain circumstances, two triangles are shown to be equal in all respects ; and it is the first case of the kind. The next two are deductions from it ; and the first three in the book are required in order to assist in making these deductions.
If, then, superposition proofs were adopted for V. and VI., the problems I.-III. might be dispensed with so far as Section I. is concerned.
EXAMINATION XVIII. 1. Give the definitions of theorem,' 'hypothesis,' and 'thesis' or conclusion.' 2. Separate hypothesis from thesis in the following theorems :
:-(a) If ABCD is a square, its diagonal will bisect it. (6) If ABC is a circle, its diameters are all equal. (c) If two adjacent sides of one oblong be equal to two adjacent sides of another, the two oblongs shall be equal. 3. Separate thesis from hypothesis in the enunciation of Proposition IV. 4. Show that the hypothesis is threefold. 5. Show that the thesis is fourfold, enumerating the four portions of it. 6. State the relation in the hypothesis of Prop. IV. between the equal angles and the two pairs of equal sides. 7. If the triangles ABC, DEF have AB, AC equal to DE, DF, each to each, and also angles ABC, DEF equal, is this an example of the hypothesis of Prop. IV.? Give your reason. 8. Given, in the triangles ABC, DEF, that BC, EF are equal, and also CA, FD; what further is necessary to enable us to say that AB, DE are equal? 9. Separate hypothesis from thesis for Prop. V. 10. Count, in the same proposition, the lines of printing which form the construction. 11. Define · Converse' Theorems. 12. Write the theorems which would be respectively converse to—(a) If the opposite sides of a figure be equal, they shall be parallel. (6) If A is B, then X shall be Y. (c) Proposition V. of Euclid, first part. 13. If ABC be a triangle, which side subtends the angle A, which the angle B, and which C? 14. Analyse the particular enunciation of Proposition VI. into hypothesis and thesis. 15. State the false or trial hypothesis of the reasoning, the false conclusion. What part of the particular enunciation does the false hypothesis contradict ? 16. Show that Prop. VI. is the second of two converse theorems proved by Euclid. 17. Explain “problem”; and say, in regard to each of the following, whether it is a theorem or a problem :-(a) Make a square equal to a given oblong. (6) If one angle of a square be right, all are so. (c) The centre bisects all diameters of a circle. 18. Show that an application of Prop. III. in Euclid is, strictly speaking, an application of all the first three propositions.
SECTION II. PROPOSITIONS VII.-XII.
Second case of equality of Two Triangles, with applica
tions to Problems on Bisectors and Perpendiculars.
EXERCISE XI. (Note—The word “vertex' is used in Euclid for an angle opposite to the
base of a triangle. In 1–5, Diagram IV. is referred to.) 1. Mention four vertices of triangles which are upon the base AC. 2. Also of all the triangles upon the base CD.
3. Name three triangles which are upon the base AE, and one one side of it.
4. Name the sides of the triangles upon base CD which are terminated in the extremity C; also those terminated in D.
5. Name two triangles upon base CE which have the vertex of one within the other.
6. In the figure to I. 5, if BG, CF interest in H, name all the triangles upon the base BC.
7. In the same figure, name two triangles on one side of BC which have the vertex of each outside the other.
8. Also two, of which the vertex of one lies on a side of the other.
9. In the annexed figure, if ABC be an isosceles triangle, prove that the angle EBC is greater than Ε ACB.
10. Also that angle DBC is less than ACB.
12. Prove that ABC and DBC in the figure cannot both be isosceles triangles.
13. In the annexed figure, having A and Don opposite sides of BC, let ABC be isosceles ; then prove that the angle BCD is greater than EBC.
В. 14. Also that angle DBC' is less than BCF.
15. If ABC and DBC be two triangles, and D be a point in the side AC of the first of them, É show that these triangles cannot have those sides equal which areterminated in the extremity C of the base.