they have the side C B common, the side A B equal to the side B F, and the angle A B C equal to the angle F B C, both being right angles (Prop. I. Cor. 1); hence the third sides, CF and A C, are equal (Prop. V. Cor.). But ABF, being a straight line, is shorter than ACF, which is a broken line (Art. 34, Ax. 10); therefore AB, the half of ABF, is shorter than A C, the half of ACF; hence the perpendicular is shorter than any oblique line. Secondly. If BE is equal to B C, then, since A B is common to the triangles, A B E, A B C, and the angles ABE, ABC are right angles, the two triangles are equal (Prop. V.), and the side AE is equal to the side A C (Prop. V. Cor.). Hence the two oblique lines, meeting the given line at equal distances from the perpendicular, are equal. Thirdly. The point C being in the triangle ADF, the sum of the lines A C, C F is less than the sum of the sides AD, DF (Prop. XII.) But AC has been shown to be equal to CF; and in like manner it may be shown that AD is equal to DF. Therefore A C, the half of the line ACF, is shorter than AD, the half of the line ADF; hence the oblique line which meets the given line the greater distance from the perpendicular, is the longer. 73. Cor. 1. The perpendicular measures the shortest distance of any point from a straight line. 74. Cor. 2. From the same point to a given straight line only two equal straight lines can be drawn. 75. Cor. 3. Of any two straight lines drawn from a point to a straight line, that which is not shorter than the other will be longer than any straight line that can be drawn between them, from the same point to the same line. PROPOSITION XV. - THEOREM. 76. If from the middle point of a straight line a perpendicular to this line be drawn, 1st. Any point in the perpendicular will be equally distant from the extremities of the line. 2d. Any point out of the perpendicular will be unequally distant from those extremities. Let D C be drawn perpendicular to the straight line A B, from its middle point C. First. Let D and E be points, taken at pleasure, in the perpendicular, and join DA, DB, and also AE, EB. A Then, since AC is equal to CB, the Ꭰ F E B two oblique lines D A, D B meet points which are at the same distance from the perpendicular, and are therefore equal (Prop. XIV.). So, likewise, the two oblique lines EA, E B are equal; therefore any point in the perpendicular is equally distant from the extremities A and B. Secondly. Let F be any point out of the perpendicular, and join F A, FB. Then one of those lines must cut the perpendicular, in some point, as E. Join EB; then we have E B equal to EA. But in the triangle FE B, the side FB is less than the sum of the sides EF, EB (Prop. IX.), and since the sum of FE, EB is equal to the sum of FE, EA, which is equal to FA, FB is less than FA. Hence any point out of the perpendicular is at unequal distances from the extremities A and B. 77. Cor. If a straight line have two points, of which each is equally distant from the extremities of another straight line, it will be perpendicular to that line at its middle point. PROPOSITION XVI. - THEOREM. 78. If two triangles have two sides of the one equal to two sides of the other, each to each, and the included angle of the one greater than the included angle of the other, the third side of that which has the greater angle will be greater than the third side of the other. Of the two sides D E, D F, let D F be the side which is not shorter than the other; make the angle EDG equal to BAC; and make D G equal to AC or D F, and join EG, GF. Since D F, or its equal D G, is not shorter than D E, it is longer than DH (Prop. XIV. Cor. 3); therefore its extremity, F, must fall below the line EG. The two triangles, ABC and DE G, have the two sides AB, AC equal to the two sides DE, DG, each to each, and the included angle BAC of the one equal to the included. angle EDG of the other; hence the side BC is equal to EG (Prop. V. Cor.). In the triangle DFG, since D G is equal to DF, the angle D F G is equal to the angle D G F (Prop. VII.); but the angle DGF is greater than the angle EGF; therefore the angle DFG is greater than EGF, and much more is the angle EFG greater than the angle EGF. Because the angle EFG in the triangle EFG is greater than EGF, and because the greater side is opposite the greater angle (Prop. X.), the side EG is greater than EF; and EG has been shown to be equal to BC; hence BC is greater than EF. 79. If two triangles have two sides of the one equal to two sides of the other, each to each, but the third side of the one greater than the third side of the other, the angle contained by the sides of that which has the greater third side will be greater than the angle contained by the sides of the other. Let ABC, DEF be two triangles, the side AB equal to D E, and AC equal to DF, and A D For, if it be not greater, it must either be equal to it or less. But the angle A cannot be equal to D, for then the side B C would be equal to E F (Prop. V. Cor.), which is contrary to the hypothesis; neither can it be less, for then the side BC would be less than EF (Prop. XVI.), which also is contrary to the hypothesis; therefore the angle A is not less than the angle D, and it has been shown that is not equal to it; hence the angle A must be greater than the angle D. PROPOSITION XVIII.-THEOREM. 80. If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles themselves will be equal. E, and the angle C to the angle F, and the two triangles will also be equal. For, if the angle A were greater than the angle D, since the sides A B, A C are equal to the sides DE, DF, each to each, the side BC would be greater than EF (Prop. XVI.); and if the angle A were less than D, it would follow that the side BC would be less than EF. But by hypothesis BC is equal to EF; hence the angle A can neither be greater nor less than D; therefore it must be equal to it. In the same manner, it may be shown that the angle B is equal to E, and the angle C to F; hence the two triangles must be equal. 81. Scholium. In two triangles equal to each other, the equal angles are opposite the equal sides; thus the equal angles A and D are opposite the equal sides B C and EF. PROPOSITION XIX.-THEOREM. 82. If two right-angled triangles have the hypothenuse and a side of the one equal to the hypothenuse and a side of the other, each to each, the triangles are equal. equal to the triangle DEF. B The two triangles are evidently equal, if the sides BC and EF are equal (Prop. XVIII.). If it be possible, let |