AC, CB, each to each; and the angle DBC is equal to the angle ACB; [Hyp.] therefore the base DC is equal to the base AB, and the triangle DBC is equal to the triangle ACB, [I. 4.] the less to the greater; which is absurd. [Ax. 9.] Therefore AB is not unequal to AC, that is, it is equal to it. Therefore, if two angles, &c. Q.E.D. Corollary. Hence every equiangular triangle is also equilateral. If the hypothesis and predicate of a proposition be made to change places, the resulting proposition is called the converse' of the one from which it is thus derived. Propositions 5 and 6 may be thus exhibited. Prop. 5. Equal sided triangles have equal base-angles, conversely, Prof. 6. Triangles with equal base-angles have equal sides. From this it will be seen that prop. 6 is the converse' of prop. 5. 6 When we deny the truth of a proposition we assert its contradictory.' Thus the contradictory of prop. 6 is “Triangles with equal base-angles have not equal sides." Now when we get two propositions, one of which contradicts the other, one must be true and the other false. This principle justifies the method of proof called the ‘Indirect Demonstration, in which we establish the truth of a proposition by showing that we cannot contradict it without being led to an absurd conclusion. Thus in this prop. 6 it is asserted that AB=AC. We begin by denying this ; and following out this denial to its consequences we are led to this result, viz. : that “the less is equal to the greater." This being absurd it follows that the proposition cannot be contradicted, and is therefore true. Ex. If OB, OC bisect the angles B and C of an equilateral tri angle, then 0A also bisects the angle Å and 0A, BO and CO are all equal to one another. PROPOSITION 7. THEOREM. On the same base, and on the same side of it, there cannot be two triangles which have their sides which are terminated in one extremity of the base, equal to one another, and likewise those which are terminated in the other extremity. . If it be possible, on the C D 1. same base AB, and on the same side of it, let there be two triangles ACB, ADB, which have their sides CA, DA, terminated in the extremity A of the base, equal to one another, and likewise their sides CB, DB, that are terminated in B. А B Join CD. Then in the case in which the vertex 2. of each triangle is without the other triangle; because AC is equal to AD (Hyp.], the angle 3. ACD is equal to the angle ADC. (1. 5.] But the angle ACD is greater than the angle BCD [Ax. 9], therefore the angle ADC is also greater than the angle BCD; much more then is the angle BDC greater than the angle BCD. Again, because BC is equal to BD [Hyp.], the angle BDC is equal to the angle BCD. [I. 5.] But it has been proved to be greater ; which is impossible. But if one of the vertices as ,E D, be within the other triangle ACB, produce AC, AD to E, F. Then because AC is equal to 3. AD, in the triangle ACD [Hyp.], the angles ECD, FDC, on the other side of the base CD, are equal to one another. [I. 5.] But the angle ECD is greater than the A angle BCD [Ax. 9), therefore the angle FDC is also greater than the angle BCD; much more then is the angle BDC greater than the angle BCD. Again, because BC is equal to BD [Hyp.], the angle BDC is equal to the angle BCD. [I. 5.) But it has been proved to be greater ; which is impossible. a The case in which the vertex of one triangle is on a side of the other needs no demonstration. Therefore, on the same base &c. Q.E.D. This proposition is introduced for the sole purpose of leading on to prop. 8, and as this latter can be proved by other and easier methods, prop. 7 might have been ommitted. It follows from this proposition that the triangle is a rigid figure; i.e., its sides cannot be forced into any other relative position. In first figure to the prop., for example, the triangle ACB cannot be made to assume the position ADB, and so long as AB is fixed, the point C and the sides CA, CB are fixed also. This explains why a carpenter turns all his frames into triangles. Hence the use of the cross-bar in a door or gate, and of the tie-beam in a roof is to make the structure rigid. 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, 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, equal to them, of the other. Let ABC, DEF be two triangles, having the two 1. sides AB, AC equal to the two sides DE, DF, each to each, namely AB to DE and AC to DF, and also the base BC equal to the base EF; the angle BAC shall be equal to the angle EDF. A B For if the triangle ABC be applied to the 2, 3. triangle DEF, so that the point B may be on E, and the straight line BC on the straight line EF, the point C will also coincide with the point F, because BC is equal to EF. [Hyp.] Therefore BC coinciding with EF, BA and AC will coincide with ED and DF. For if the base BC coincides with the base EF, but the sides BA, CA do not coincide with the sides ED, FD, but ħave a different situation as EG, FG; then on the same base and on the same side of it there will be two triangles having their sides which are terminated in one extremity of the base equal to one another, and likewise their sides which are terminated in the other extremity. But this is impossible. [I. 7.] Therefore if the base BC coincides with the base EF, the sides BA, AC must coincide with the sides ED, DF. Therefore also the angle BAC coincides with the angle EDF, and is equal to it. [Ax. 8.] Therefore, if two triangles &c. Q.E.D. Notice that Euclid does not prove the triangles equal in this prop. When triangles are to be proved equal , he always appeals to prop. 4. The demonstration of this prop. may be conducted thus: For let the triangle DEF be applied to ABC so that the bases may coincide, and the vertices fall on opposite A D sides of the base. Let GBC represent the position of DEF, so that BG is the position assumed by DE, and GC the position of DP. Join AG. Then the angle BAG=BGA because BA=BG, and the angle GAC=AGC because AC=GC. Therefore the whole anglo BAC=BGC, that is EDF. Q.E.D. Ex. 1. The diagonal of a rhombus bisects each of the angles through which it passes. Hence show that 2. The diagon ıls of a square bisect each other at right angles. [Prop. 4.] PROPOSITION 9. PROBLEM. 2. To bisect a given rectilineal angle, that is to divide it into two equal angles. Let BAC be the given recti1. A lineal angle. It is required to bisect it. Take any point D in AB, and from AC cut off AE equal to AD [I. 3]; join DE, and on DE on E the side remote from A describe the equilateral triangle DEF. [I. 1.) Join AF. The straight line AF shall bisect the angle BAC. 3. Because AD is equal to AE B [Con.], and AF is common to с the two riangles DAF, EAF, the two sides A, AF are equal to the two sides EA, AF, each to each ; and the base DF is equal to the base EF [Def. 24]; therefore the angle DAF is equal to the angle EAF. [I. 8.] Therefore the given rectilineal angle BAC is bisected by the straight line AF. Q.E.F. If the equilateral triangle were not described on the side of DE remote from A the point F might fall either, 1. Within the angle BAC, as in the prop. 2. On the point A; in which case the construction would fail; or, 3. Outside the angle BAC; when the line FA would have to be produced in order to bisect the angle, and the proof would require modification. The restriction introduced insures that F shall fall within BAC as in the figure. If BA and AC were in one straight line the construction and proof would exactly correspond to prop. 11, the reason of which is obvious. Ex. 1. If two isosceles triangles be described on opposite sides of the same base the straight line joining their vertices will bisect their common base at right angles. [See fig. to the prop.] 2. If the triangles were described on the same side of the base would the same result follow ? |