and the side BA the direc BU taking the direction EF, tion ED. Then, because BC is equal to EF, the vertex ✔ will coincide with the vertex F; and because the angle C is equal to the angle F, the side CA will take the Now, the vertex A being at the same time direction FD. and FD, it must be at their intersection hence, the triangles coincide throughout, and are therefore equal in all their parts (I., D. 14); which was to be proved. 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 any two sides, as AB, BC, be greater than the third side For, the distance from A AC. 2 to C, measured on any broken line AB, BC, A is greater than the distance measured on the straight line AC (A. 12): hence, the sum of AB and BC is greater than AC; which was to be proved. Cor. If from both members of the inequality, ACAB+ BC, we take away either of the sides AB, BC, example, there will remain (A. 5), AC – BC < AB`; as BC, for that is, the difference between any two sides of a triangle is Less than the third side. Scholium. In order that any three given lines may re present the sides of a triangle, the sum of any two must be greater than the third, and the difference of any two must be less than the third. PROPOSITION VIII. THEOREM. If from any point within a triangle two straight lines be drawn to the extremities of any side, their sum will be. less than that of the two remaining sides of the triangle. A Let be any point within the triangle BAC, and let the lines OB, OC, be drawn to the extremities of any side, as BC: then will the sum of BO and 00 be less than the sum of the sides BA and AC. B D C Prolong one of the lines, as BO, till it meets the side AC in D; then, from Prop. VII., we shall have, OC < OD + DC ; adding BO to both members of this inequality, recollecting that the sum of BO and OD is equal to BD, we have (A. 4), BO+OC < BD + DC. From the triangle BAD, we have (P. VII.), BD < BA + AD ; adding DC to both members of this inequality, recollecting that the sum of AD and DC is equal to AC, we have, BD + DC <BA + AC. But it was shown that BO+ OC is less than BD + DC; still more, then, is BO+ 00 less than BA+ AC; which was to be proved. 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. In the triangles BAC and DEF, let AB be equal to DE, AC to DF, and the angle A greater than the angle D: then will BC be greater than EF. Let the line AG be drawn, making the angle CAG equal to the angle D (Post. 7); make AG equal to -DE, and draw GC. Then will the triangles AGC and DEF have two sides and the included angle of the one equal to two sides and the included angle. of the other, each to each; consequently, GC is equal to EF (P. V.). Now, the point G may be without the triangle ABC, it may be on the side BC, or it may be within the triangle ABC. Each case will be considered separately. GI+IC > GC, and BI + IA > AB; whence, by addition, recollecting that the sum of BI and IC is equal to BC, and the sum of GI and IA, to GA, we have, AG + BC > AB + GC Or, since AG = AB, and GC = EF, we have, AB + BC > AB + EF. Taking away the common part AB, there remains (A. 5), 3°. When G is within the triangle ABC. From Proposition VIII., we have, Hence, in each case, BC is greater than EF; which was to be proved. Conversely: If in two triangles ABC and DEF, the side AB is equal to the side DE, the side AC to DF, and BC greater than EF, then will the angle BAC be greater than the angle EDF. For, if not, BAC must either be equal to, or less than, EDF. In the former case, BC would be equal to EF (P. V.), and in the latter case, BC would be less than EF; either of which would be contrary to the hypothesis hence, BAC must be 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 triangles will be equal in all their parts. then will the tri In the triangles ABC and DEF, let AB be equal to DE AC to DF, and BC to EF: angles be equal in all their parts. For, since the sides AB, AC, are equal to DE, DF, each to each, if the angle A were greater than D, it would follow, by the last Pro position, that the side AA B E F BC would be greater than EF; and if the angle A were less than D, the side BC would be less than EF. But BC is equal to EF, by hypothesis; therefore, the angle A can neither be greater nor less than D: hence, it must be equal to it. The two triangles have, therefore, two sides and the included angle of the one equal to two sides and the included angle of the other, each to each; and, consequently, they are cqual in all their parts (P. V.); which was to be proved. Scholium. In triangles, equal in all their parts, the equal sides lie opposite the equal angles; and conversely. PROPOSITION XI. THEOREM. In an isosceles triangle the angles opposite the equal sides are equal. Let BAC be an isosceles triangle, having the side AB equal to the side AC: then will the angle C be equal to the angle B. |