If two angles of a triangle be equal to one another, then the sides also which subtend, or are opposite to, the equal angles, shall be equal to one another. B Let ABC be a triangle, in which Construction. For if AC be not equal to AB, Proof. Because Then in the triangles DBC, ACB, DB is equal to AC, and BC is common to both, I. 3. Constr. also the contained angle DBC is equal to the contained angle ACB; Hyp. therefore the triangle DBC is equal to the triangle ACB I. 4. in area, Therefore AB is not unequal to AC ; Q.E.D. COROLLARY. Hence if a triangle is equiangular it is also equilateral. 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 it is assumed that AB=AC; it is required to prove that the angle ABC= the angle ACB. Now in Prop. 6 it is assumed that the angle ABC= the angle ACB it is required to prove that 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. 1. ABCD is a rhombus, in which the diagonal BD is drawn : shew that (i) the angle ABD = the angle ADB ; (ii) the angle CBD the angle CDB; (iii) the angle ABC= the angle ADC. 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. 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. 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. Construction. Join CD. Proof. Then in the triangle ACD, 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, Hyp. therefore the angle BDC is equal to the angle BCD: I. 5. but it was shewn to be greater; which is impossible. CASE II. When one of the vertices, as D, is within Then in the triangle ACD, because AC is equal to AD, Hyp. therefore the angle ECD is equal to the angle FDC, these being the angles on the other side of the base. I. 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. 24 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 A A B Let ABC, DEF be two triangles, in which Then shall the angle BAC be equal to the angle EDF. Proof. If the triangle ABC be applied to the triangle DEF, 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. Q.E.D. |