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+IB, 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.). AA 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 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 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 follow, 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. A D 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, by hypothesis; AD is common, and BD is equal to DC by construction: therefore, by the last proposition, the angle B is equal to the angle C. B 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;B4 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 angle; 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+CD= C AD+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 Č<B, it follows, from what has just been proved, that AB AC; which is contrary to the hypothesis. If the angle CB, 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. PROPOSITION XIV. THEOREM. From a given point, without a straight line, only one perpendicular can be drawn to that line. Let A be the point, and DE the given line. B E Let us suppose that we can draw two perpendiculars, AB, AC. Produce either of them, as AB, till BF is equal to AB, and De draw FC. Then, the two triangles CAB, CBF, will be equal: for, the angles CBA, and CBF are right angles, the side CB is common, and the side AB equal to BF, by construction; therefore, the triangles are equal, and the angle ACB=BCF (Prop. V. Cor.). But the angle ACB is a right angle, by hypothesis; therefore, BCF must likewise be a right angle. But if the adjacent angles BCA, BCF, are together equal to two right angles, ACF must be a straight line (Prop. III.): from whence it follows, that between the same two points, A and F, two straight lines can be drawn, which is impossible (Ax. 11.): hence, two perpendiculars cannot be drawn from the same point to the same straight line. Scholium. At a given point C, in the line AB, it is equally impossible to erect two perpendiculars to that line. For, if CD, CE, were those two perpendiculars, the angles BCD, BCE, would both be right angles:. hence they would be equal (Ax. 10.); and E B the line CD would coincide with CE; otherwise, a part would be equal to the whole, which is impossible (Ax. 8.). PROPOSITION XV. THEOREM. If from a point without a straight line, a perpendicular be let fall on the line, and oblique lines be drawn to different points : 1st, The perpendicular will be shorter than any oblique line. 2d, Any two oblique lines, drawn on different sides of the perpendicular, cutting off equal distances on the other line, will be equal. 3d, Of two oblique lines, drawn at pleasure, that which is farther from the perpendicular will be the longer. Let A be the given point, DE the given line, AB the perpendicular, and AD, AC, AE, the oblique lines. Produce the perpendicular AB till BF is equal to AB, and draw FC, FD. First. The triangle BCF, is equal to the triangle BCA, for they have the right angle CBF CBA, the side CB common, and the = D A BE F side BF=BA; hence the third sides, CF and CA are equal (Prop. V. Cor.). But ABF, being a straight line, is shorter than ACF, which is a broken line (Def. 3.); therefore, AB, the half of ABF, is shorter than AC, the half of ACF; hence, the perpendicular is shorter than any oblique line. Secondly. Let us suppose BC=BE; then will the triangle CAB be equal to the the triangle BAE; for BC=BE, the side AB is common, and the angle CBA=ABE; hence the sides AC and AE are equal (Prop. V. Cor.): therefore, two oblique, lines, equally distant from the perpendicular, are equal. Thirdly. In the triangle DFA, the sum of the lines AC, CF, is less than the sum of the sides AD, DF (Prop. VIII.); therefore, AC, the half of the line ACF, is shorter than AD, the half of the line ADF: therefore, the oblique line, which is farther from the perpendicular, is longer than the one which is nearer. Cor. 1. The perpendicular measures the shortest distance of a point from a line. Cor. 2. From the same point to the same straight line, only two equal straight lines can be drawn; for, if there could be more, we should have at least two equal oblique lines on the same side of the perpendicular, which is impossible. PROPOSITION XVI. THEOREM. If from the middle point of a straight line, a perpendicular be drawn to this line; 1st, Every point of the perpendicular will be equally distant from the extremities of the line. 2d, Every point, without the perpendicular, will be unequally distant from those extremities. |