DE, AC to DF and BC to EF, then will the three angles be equal, each to each, and the triangles will be every way equal. to DE, DF, it would follow, that the side BC must be greater than EF: (Prop. 9:) and, if the angle A were less than D, it would follow that the side BC must be less than EF. But BC is equal to EF; therefore the angle A can neither be greater nor less than D; consequently they must be equal. In the same manner it may be shown, that B is equal to E, and C to F. Hence, Any two triangles are every way equal, when the three sides of the one are respectively equal to the three sides of the other. Scholium. It may be observed, that the equal angles lie opposite to the equal sides: thus the equal angles A and D lie opposite to the equal sides BC and EF. PROPOSITION XI. THEOREM. In an isosceles triangle, the angles opposite the equal sides are equal. Let the side AB be equal to AC, the angle C will be equal to B. A B C D Join A the vertex, and D the middle point of the base BC. The triangles ADB, ADC, have all the sides of the one respectively equal to those of the other; for AD is common, AB AC by hypothesis and BD= DC by construction; therefore, the angle B must be equal to the angle C. (Prop. 10.) Hence, In any isosceles triangle, the angles opposite the equal sides are equal. Cor. An equilateral triangle is likewise equiangular, that is to say, has all its angles equal. For, in the equilateral triangle ABC, the side AB=AC, hence the angle B= C; and since the side AB=BC, therefore the angle A=C; hence the angle A=B; (Ax. 1;) therefore the three angles A, B, C, are equal to each other. B A C Scholium. The equality of the triangles ABD, ACD, proves also that the angle BAD DAC, and BDA=ADC; therefore the last two are right-angles. (Def. 12.) Hence, The line drawn from the vertex of an isosceles triangle to the middle point of its base, is perpendicular to that 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 commonly assumed as the base, which is not equal to either of the other two. Conversely, if two angles of a triangle are equal, the sides opposite them will be equal, and the triangle will be isosceles. Let the angle ABC be equal to ACB; then will the side AC be equal to the side AB. For, if these sides are not equal, let AB be the greater. Take BD=AC, and join DC. By hypothesis, the angle DBC is equal to ACB; and the two sides DB, B D A BC, are equal to the two AC, CB, by construction; therefore the triangle DBC must be equal to ACB. (Prop. 5.) But the part cannot be equal to the whole; (Ax. 9;) hence the side AB is not greater than AC. In like manner it may be proved that AB is not less than AC, hence there is no inequality between the sides AB, AC; therefore the triangle ABC is isosceles. (Def. 17.) Hence, If two angles of any triangle are equal, the sides opposite them will be equal, and the triangle will be isosceles. Of two sides of a triangle, ABC, the greater is that which lies opposite the greater angle; and conversely, of two angles of a triangle, the greater is that which lies opposite the greater side. First. Let the angle C be greater than B; then will the side AB, opposite to C, be greater than AC, opposite to B. Make the angle BCD=B. Then in the triangle BDC, we shall have BD=DC,(Prop. C A D B 12,) hence AB=AD+DC. But AD+DC > AC; (Prop. 7;) therefore AB>AC. Second. 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 would follow, from what has just been proved, that the side AB < AC, which is contrary to the hypothesis. If the angle CB, it would follow, (Prop. 12,) that the side AB AC, which is also contrary to the hypothesis. Therefore, the angle C must be greater than B. Hence, In every triangle, the greater side lies opposite the greater angle; and conversely, the greater angle lies opposite the greater side. From a given point ▲ without the straight line DE, only one perpendicular can be drawn to that line. For, suppose we can draw two, AB and AC. Produce one of them AB, till BF is equal to AB, and join FC. D A E C B F The triangle CBF ABC; for the angles CBF and CBA are right-angles, the side CB is common, and the side BF AB by construction: therefore these triangles are every way equal; hence the angle BCF=BCA. (Prop. 5.) But BCA is a right-angle by hypothesis; therefore BCF must also be a right-angle. Now since the adjacent angles BCA, BCF, are together equal to two right-angles, the line ACF must be straight; (Prop. 3;) from whence it follows, that between the same two points A and F, two straight lines can be drawn, which is impossible. (Ax. 12.) In like manner, if AE were joined, it may be proved that AB and AE cannot both be perpendicular to DE. Hence, From any given point without a straight line, only one perpendicular can be drawn to that 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 and CE were those two perpendiculars, the angles DCB, BCE would be right-angles, (Def. 12,) and consequent A E D B ly would be equal to each other; i. e., a part equal to the whole, which is impossible. (Ax. 9.) If from a point A situated without the straight line DE, a perpendicular AB be let fall on that straight line, and several oblique lines AE, AC, AD, be drawn to several points in the same line; then, 1. The perpendicular AB will be shorter than any oblique line. 2. The two oblique lines, AC, AE, drawn on different sides of the perpendicular, and equally distant from it, will be equal. 3. Of two oblique lines, AC, AD, or AE, AD, drawn at pleasure, the one which lies farther from the perpendicular, will be the longer. Produce the perpendicular AB till BF is equal to AB, and join FC, FD. First. The triangle BCF, is equal D to the triangle BCA, for they have the right-angle CBF CBA, the side CB common, and the side A E C B F BF BA by construction; hence the third sides, CF and AC are equal. (Prop. 5.) But ABF, being a straight line, is shorter than ACF, which is a broken line; (Def. 5;) hence AB, the half of ABF, is shorter than AC, the half of ACF: therefore the perpendicular is shorter than any oblique line. Second. If we suppose BE=BC; then, since the side AB is common, and the angle ABE=ABC, the triangle ABE must be equal to the triangle ABC; (Prop. 5;) hence the sides AE, AC are equal; therefore two oblique lines equally distant from the perpendicular are equal. Third. In the triangle DFA, the sum of the lines AC, CF, is less than the sum of the sides AD, DF; (Prop. 8;) hence AC, the half of the line ACF, is shorter than AD, the half |