NOTE ON PROPOSITIONS 5 AND 6. The enunciation of a theorem consists of two clauses. The first clause tells us what we are to assume, and is called the hypothesis ; the second tells us what it is required to prove, and is called the conclusion. For example, the enunciation of Proposition 5 assumes that in a certain triangle ABC the side AB=the side AC: this is the hypothesis. From this it is required to prove that the angle ABC=the angle ACB : this is the conclusion. If we interchange the hypothesis and conclusion of a theorem, we enunciate a new theorem which is called the converse of the first. For example, in Prop. 5 AB=AC; Thus we see that Prop. 6 is the converse of Prop. 5; for the hypothesis of each is the conclusion of the other. In Proposition 6 Euclid employs for the first time an indirect method of proof frequently used in geometry. It consists in shewing that the theorem cannot be untrue ; since, if it were, we should be led to some impossible conclusion. This form of proof is known as Reductio ad Absurdum, and is most commonly used in demonstrating the converse of some foregoing theorem. The converse of all true theorems are not themselves necessarily true. [See Note on Prop 8.] EXERCISES ON PROPOSITION 5. V 1. ABCD is a rhombus, in which the diagonal BD is drawn : shew that (i) the angle ABD=the angle ADB; (iii) the angle ABC=the angle ADC. V 2. ABC, DBC are two isosceles triangles drawn on the same base BC, but on opposite sides of it: prove (by means of 1. 5) that the angle ABD=the angle ACD. V 3. ABC, DBC are two isosceles triangles drawn on the same base BC and on the same side of it: employ 1. 5 to prove that the angle ABD=the angle ACD. PROPOSITION 7. THEOREM. On the same base, and on the same side of it, there cannot be two triangles having their sides which are terminated at one extremity of the base equal to one another, and likewise those which are terminated at the other extremity equal to one another. If it be possible, on the same base AB, and on the same side of it, let there be two triangles ACB, ADB in which the side AC is equal to the side AD, and also the side BC is equal to the side BD. CASE I. When the vertex of each triangle is without the other triangle. Proof. Then in the triangle ACD, because AC is equal to AD, Hyp. therefore the angle ACD is equal to the angle ADC. 1.5. But the whole angle ACD is greater than its part, the angle BCD ; therefore also the angle ADC is greater than the angle BCD; still more then is the angle BDC greater than the angle BCD. Again, in the triangle BCD, Нур. therefore the angle BDC is equal to the angle BCD: 1. 5. but it was shewn to be greater; which is impossible. CASE II. When one of the vertices, as D, is within the other triangle ACB. Construction. As before, join CD; and produce AC, AD to E and F. Proof. Then in the triangle ACD, Нур. therefore the angle ECD is equal to the angle FDC, these being the angles on the other side of the base. 1. 5. But the angle ECD is greater than its part, the angle BCD; therefore the angle FDC is also greater than the angle BCD: still more then is the angle BDC greater than the angle BCD. Again, in the triangle BCD, Нур. therefore the angle BDC is equal to the angle BCD: 1. 5. but it has been shewn to be greater; which is impossible. The case in which the vertex of one triangle is on a side of the other needs no demonstration. Therefore AC cannot be equal to AD, and at the same time, BC equal to BD. Q.E.D. NOTE. The sides AC, AD are called conterminous sides ; similarly the sides BC, BD are conterminous. PROPOSITION 8. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal, then the angle which is contained by the two sides of the one shall be equal to the angle which is contained by the two sides of the other. Let ABC, DEF be two triangles, in which the side AB is equal to the side DE, the side AC is equal to the side DF, so that the point B falls on the point E, Hyp. therefore the point C must coincide with the point F. Then since BC coincides with EF, it follows that BA and AC must coincide with ED and DF: for if they did not, but took some other position, as EG, GF, then on the same base EF, and on the same side of it, there would be two triangles EDF, EGF, having their conterminous sides equal: namely ED equal to EG, and FD equal to FG. But this is impossible. Therefore the sides BA, AC coincide with the sides ED, DF. That is, the angle BAC coincides with the angle EDF, and is therefore equal to it. Ax. 8. Q.E.D. I. 7. Note 1. In this Proposition the three sides of one triangle are given equal respectively to the three sides of the other; and from this it is shewn that the two triangles may be made to coincide with one another. Hence we are led to the following important Corollary. COROLLARY. If in two triangles the three sides of the one are equal to the three sides of the other, each to each, then the triangles are equal in all respects. [An alternative proof, which is independent of Prop. 7, will be found on page 26.] NOTE 2. Proposition 8 furnishes an instance of a true theorem of which the converse is not necessarily true. It is proved above that if the sides of one triangle are severally equal to the sides of another, then the angles of the first triangle are severally equal to the angles of the second. The converse of this enunciation would be as follows: If the angles of one triangle are severally equal to the angles of another, then the sides of the first triangle are equal to the sides of the second. But this, as the diagram in the margin shews, is by no means necessarily true. EXERCISES ON PROPOSITION 8. V 1. Shew (by drawing a diagonal) that the opposite angles of a rhombus are equal. v 2. If ABCD is a quadrilateral, in which AB=CD and AD=CB, prove that the angle ADC=the angle ABC. V 3. If ABC and DBC are two isosceles triangles drawn on the same base BC, prove (by means of 1. 8) that the angle ABD=the angle ACD, taking (i) the case where the triangles are on the same side of BC, (ii) the case where they are on opposite sides of BC. 4. If ABC, DBC are two isosceles triangles drawn on opposite sides of the same base BC, and if AD be joined, prove that each of the angles BAC, BDC will be divided into two equal parts. V 5. If in the figure of Ex. 4 the line AD meets BC in E, prove that BE=EC. |