DF is equal to AC; therefore, the point F will fall on C, and the third side EF, will coincide with the third side BC (Ax. 11.): therefore, the triangle EDF is equal to the triangle BAC (Ax. 13.). Cor. When two triangles have these three things equal, namely, the side ED=BA, the side DF-AC, and the angle D=A, the remaining three are also respectively equal, namely, the side EF=BC, the angle E=B, and the angle F=C PROPOSITION VI. THEOREM. If two triangles have two angles and the included side of the one, equal to two angles and the included side of the other, each to each, the two triungles will be equal. Let the angle E be equal to the angle B, the angle F to the angle C, and the inIcluded side EF to the included side BC; then will the triangle EDF be equal to the triangle BAC. For to apply the one to the other, let the side EF be placed on its equal BC, the point E falling on B, and the point F on C; then, since the angle E is equal to the angle B, the side ED will take the direction BA; and hence the point D will be found somewhere in the line BA. In like manner, since the angle F is equal to the angle C, the line FD will take the direction CA, and the point D will be found somewhere in the line CA. Hence, the point D, falling at the same time in the two straight lines BA and CA, must fall at their intersection A: hence, the two triangles EDF, BAC, coincide with each other, and re therefore equal (Ax. 13.). Cor. Whenever, in two triangles, these three things are equal, namely, the angle E=B, the angle F-C, and the included side EF equal to the included side BC, it may be inferred that the remaining three are also respectively equal, namely, the angle D=A, the side ED=BA, and the side DF=AC. Scholium. Two triangles are said to be equal, when being applied to each other, they will exactly coincide (Ax. 13.). Hence, equal triangles have their like parts equal, each to each, since those parts must coincide with each other. The converse of this proposition is also true, namely, that two triangles which have all the parts of the one equal to the parts of the other, each B E to each, are equal; for they may be applied to each other, and the equal parts will mutually coincide. PROPOSITION VII. THEOREM. The sum of any two sides of a triangle, is greater than the third side. Let ABC be a triangle: then will the sum of two of its sides, as AC, CB, be greater than the third side AB. For the straight line AB is the shortest distance between the points A and B (Def. 3.); hence AC+CB is greater A. than AB. PROPOSITION VIII. THEOREM. B C If from any point within a triangle, two straight lines be drawn to the extremities of either side, their sum will be less than the sum of the two other sides of the triangle. Let any point, as O, be taken within the triangle BAC, and let the lines OB, OC, be drawn to the extremities of either side, as BC; then will OB+OC<BA+AC. Let BO be produced till it meets the side AC in D: then the line OC is shorter than OD+DCB (Prop. VII.): add BO to each, and we have BO+OC<BO+ OD+DC (Ax. 4.), or BO+OC<BD+DC. Again, BD<BA+AD: add DC to each, and we have BD+ DC BA+AC. But it has just been found that BO+OC< BD+DC; therefore, still more is BO+OC<BA+AC. PROPOSITION IX. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, and the included angles unequal, the third sides will be unequal; and the greater side will belong to the triangle which has the greater included angle. Let BAC and EDF be two triangles, having the side AB-DE, AC =DF, and the angle A>D; then will BC> EF. Make the angle CAGB =D; take AG=DE, and draw CG. The C F triangle GAC is equal to DEF, since, by construction, they have an equal angle in each, contained by equal sides, (Prop. V.); therefore CG is equal to EF. Now, there may be three cases in the proposition, according as the point G falls without the triangle ABC, or upon its base BC, or within it. First Case. The straight line GC<GI+IC, and the straight line AB<AI+IB; therefore, GC+AB<GI+AI+IC+ÏB, or, which is the same thing, GC+AB<AG+BC. Take away AB from the one side, and its equal AG from the other; and there remains GC<BC (Ax. 5.) ; but we have found GC=EF, therefore, BC>EF. Second Case. If the point G fall on the side BC, it is evident that GC, or its equal EF, will be shorter than BC (Ax. 8.). Да B Third Case. Lastly, if the point G fall within the triangle BAC, we shall have, by the preceding theorem, AG+ GC AB+BC; and, taking AG from the one, and its equal AB from the other, there will remain GC < BC or BC>EF. B Scholium. Conversely, if two sides BA, AC, of the triangle BAC, are equal to the two ED, DF, of the triangle EDF, each to each, while the third side BC of the first triangle is greater than the third side EF of the second; then will the angle BAC of the first triangle, be greater than the angle EDF of the second. CE E H For, if not, the angle BAC must be equal to EDF, or less than it. In the first case, the side BC would be equal to EF, (Prop. V. Cor.); in the second, CB would be less than EF; but either of these results contradicts the hypothesis: therefore, BAC is greater than EDF. PROPOSITION X. THEOREM. If two triangles have the three sides of the one equal to the three sides of the other, each to each, the three angles will also be equal, each to each, and the triangles themselves will be equal, Let the side ED=BA, the side EF-BC, and the side DF-AC; then will the angle D=A, the angle E=B, and the angle F =C. E D A FB For, if the angle D were greater than A, while the sides ED, DF, were equal to BA, AC, each to each, it would fol low, by the last proposition, that the side EF must be greater than BC; and if the angle D were less than A, it would follow, that the side EF must be less than BC: but EF is equal to BC, by hypothesis; therefore, the angle D can neither_be_greater nor less than A; therefore it must be equal to it. In the same manner it may be shown that the angle E is equal to B, and the angle F to C: hence the two triangles are equal (Prop. VI. Sch.). Scholium. It may be observed that the equal angles lie opposite the equal sides: thus, the equal angles D and A, lie opposite the equal sides EF and BC. PROPOSITION XI. THEOREM. In an isosceles triangle, the angles opposite the equal sides are equal. Let the side BA be equal to the side AC; then will the angle C be equal to the angle B. For, join the vertex A, and D the middle point of the base BC. Then, the triangles BAD, DAC, will have all the sides of the one equal to those of the other, each to each; for BA is equal to AC,B by hypothesis; AD is common, and BD is equal A to DC by construction: therefore, by the last proposition, the angle B is equal to the angle C. Cor. An equilateral triangle is likewise equiangular, that is to say, has all its angles equal. Scholium. The equality of the triangles BAD, DAC, proves also that the angle BAD, is equal to DAC, and BDA to ADC, hence the latter two are right angles; therefore, the line drawn from the vertex of an isosceles triangle to the middle point of its base, is perpendicular to the base, and divides the angle at the vertex into two equal parts. In a triangle which is not isosceles, any side may be assumed indifferently as the base; and the vertex is, in that case, the vertex of the opposite angle. In an isosceles triangle, however, that side is generally assumed as the base, which is not equal to either of the other two. PROPOSITION XII. THEOREM. Conversely, if two angles of a triangle are equal, the sides opposite them are also equal, and the triangle is isosceles. Let the angle ABC be equal to the angle ACB; then will the side AC be equal to the side AB. D For, if these sides are not equal, suppose AB to be the greater. Then, take BD equal to AC, and draw CD. Now, in the two triangles BDC, BAC, we have BD-AC, by construction; the angle B equal to the angle ACB, by hypothesis;B and the side BC common: therefore, the two triangles, BDC, BAC, have two sides and the included angle in the one, equal to two sides and the included angle in the other, each to each: hence they are equal (Prop. V.). But the part cannot be equal to the whole (Ax. 8.); hence, there is no inequality between the sides BA, AC; therefore, the triangle BAC is isosceles. PROPOSITION XIII. THEOREM. The greater side of every triangle is opposite to the greater an gle; and conversely, the greater angle is opposite to the greater side. First, Let the angle C be greater than the angle B; then will the side AB, opposite C, be greater than AC, opposite B. For, make the angle BCD=B. Then, in the triangle CDB, we shall have CD=BD (Prop. XII.). Now, the side AC<AD+CD; but AD+CDAD+DB AB: therefore AC<AB. A B Secondly, Suppose the side AB>AC; then will the angle C, opposite to AB, be greater than the angle B, opposite to AC. For, if the angle C<B, it follows, from what has just been proved, that AB<AC; which is contrary to the hypothesis. If the angle C-B, then the side AB=AC (Prop. XII.); which is also contrary to the supposition. Therefore, when AB>AC, the angle C must be greater than B. |