PROPOSITION IV. THEOREM. If a straight line meet two other straight lines at a comof the contiguous angles two lines met will form mon point, making the sum equal to two right angles, the one and the same straight line. Let DC meet AC and BC at C, making the sum of the angles DCA and DCB equal to two right angles: then will CB be the prolongation of AC. D A B C E For, if not, suppose CE to be the prolongation of AU, then will the sum of the angles of the angles DCA and DCE be equal to two right angles (P. I.): have (A. 1), We shall, consequently, DCA + DCB DCA + DCE; Taking from both the common angle DCA, there remains, DCB = DCE, which is impossible, since a part cannot be equal to the whole (A. 8). Hence, CB must be the prolongation of AC; which was to be proved. PROPOSITION V. THEOREM. If two triangles have two sides and the included angle of the one equal to two sides and the included angle of the other, each to each, the triangles will be equal in all their parts. In the triangles ABC and DEF, let AB be equal to DE, AC to DF, and the angle A to the angle D: then will the triangles be equal in all their parts. equal to DE, the vertex B will coincide with the vertex E; and because AC is equal to DF, the vertex C will coincide with the vertex F; consequently, the side BC will coincide with the side EF (A. 11). The two triangles, therefore, coincide throughout, and are consequently equal in all their parts (I., D. 14); which was to be proved. 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 triangles will be equal in all their parts. will the triangles be equal in all their parts. For, let ABC be applied to DEF in such a manner that the angle B shall coincide with the angle E, the side BC taking the direction EF, and the side BA the direc tion ED. Then, because BC is equal to EF, the vertex C will coincide with the vertex F; and because the angle C is equal to the angle F, the side CA will take the direction FD. Now, the vertex A being at the same time on the lines ED and FD, it must be at their intersection D (P. III., C.) 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 AC. For, the distance from A to C, measured on any broken line AB, BC, A B 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, AC < AB + BC, we take away either of the sides AB, BC, as BC, for example, there will remain (A. 5), 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. 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 B A D 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, recollectin; 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+OC 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 |