will CB be the prolongation of AC; or, AC and CB will form one and the same straight line. For, if CB is not the prolongation of AC, let CE be that prolon gation. Then the line ACE being straight, the sum of the angles ACD, DCE, will be equal to two right angles (P. 1). But by hypothesis, the sum of the angles ACD, DCB, is also equal to two right angles: therefore (A. 1), D A -B E ACD+DCE must be equal to ACD+DCB. Taking away the angle ACD from each, there remains the angle DCE equal to the angle DCB: that is, a whole equal to a part, which is impossible (A. 8): therefore, AC and CB form one and the same straight line. PROPOSITION IV. THEOREM. When two straight lines intersect each other, the opposite or vertical angles, which they form, are equal. Let AB and DE be two straight lines, intersecting each other at C; then will the angle ECB be equal to the angle ACD, and the angle ACE to the angle DCB. For, since the straight line DE A is met by the straight line AC, the sum of the angles ACE, ACD, E is equal to two right angles (P. I.); D B and since the straight line AB is met by the straight line EC, the sum of the angles ACE, and ECB, is equal to two right angles: hence (A. 1), ACE+ACD is equal to ACE+ECB. Take away from both, the common angle ACE, there remains (A. 3) the angle ACD, equal to its opposite or vertical angle ECB. In a similar manner it may be proved that ACE is equal to DCB. Scholium. The four angles formed about a point by two straight lines, which intersect each other, are together equal to four right angles. For, the sum of the two angles ACE, ECB, is equal to two right angles (P. I); and the sum of the other two, ACD, DCB, is also equal to two right angles therefore, the sum of the four, is equal to four right angles. In general, if any number of straight lines CA, CB, CD, &c., meet in a common point C, the sum of all the successive angles, ACB, BCD, DCE, ECF, FCA, will be equal to four right angles. For, if four right angles were formed about the point C, by two lines F B E perpendicular to each other, their sum would be equal to the sum of the successive angles ACB, BCD, DCE, ECF, FCA. 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 two triangles will be equal. In the two triangles EDF and BAC, let the side ED be equal to the side BA, the side DF to the side AC, and the angle D to the angle A; then will the triangle EDF be equal to the triangle BAC. For, if these trian gles be applied the one to the other, they will exactly coincide. Let the side ED be placed on the equal side BA; E FB A then, since the angle D is equal to the angle A, the side DF will take the direction AC. But DF is equal to AC; therefore the point F will fall on C, and the third side EF, will coincide with the third side BC (A. 11): consequently, the triangle EDF is equal to the triangle BAC (A. 14). Cor. When two triangles have these three things equal, viz., the side ED=BA, the side DF-AC, and the angle D=A, the remaining three are also respectively equal, viz., the side EF-BC, the angle E=B, and the angle F=C. 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 two triangles will be equal. Let EDF and BAC be two triangles, having the angle E equal to the angle B, the angle F to the angle C, and the included side EF to the included side BC; then will the triangle EDF be equal to the triangle BAC. For, let the side EF be placed on its equal E D FB A BC, the point E falling on B, and the point Fon C. Then, since the angle E is equal to the angle B, the side ED will take the direction BA; and hence, the point D will be found somewhere in the line BA. In like manner, since the angle F is equal to the angle C the line FD will take the direction CA, and the point D will be found somewhere in the line CA. Hence, the point D, falling at the same time in the two straight lines BA and CA, must fall at their intersection A: hence, the two triangles EDF, BAC, coincide with each other, and consequently, are equal (a. 14). Cor. Whenever, in two triangles, these three things are equal, viz. the angle E=B, the angle FC, and the included side EF equal to the included side BC, it may be inferred that the remaining three are also respectively equal, viz.: the angle D=A, the side_ED=BA, and the side DF-AC. Scholium. Two triangles which being applied to each other, coincide in all their parts, are equal (A. 14). The like parts are those which coincide with each other; hence, they are also equal each to each. The converse of this proposition is also true; viz., if two triangles have all the parts of the one equal to the parts of the other, each to each, the triangles will be equal: for, when applied to each other, they will mutually coincide. 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 two of its sides, as AB, BC, be greater than the third side AC For the straight line AC is the shortest distance between the points A and C (A. 12); hence, AB+BC is greater than AC. Cor. If from both members of the inequality AC AB+BC B C we take away either of the sides, as BC, we shall have (A. 5) AC-BC<AB: that is, the difference between any two sides of a triangle is less than the third side. PROPOSITION VIII. THEOREM. If from any point within a triangle, two straight lines be drawn to the extremities of either side, their sum will be less than that of the two remaining sides of the triangle. Let O be any point within the triangle BAC, and let the lines OR, OC, be drawn to the extremities of either side, as BC; then will OB+OC<BA+AC. Let BO be prolonged till it meets the side AC in D: then OC<OD+DC (P. 7): add BO to each, and we have RO+OC<BO+OD+DC (A. 4): add DC to each, and we have BD+DC<BA+AC. But it has been shown that BO+OC<BD+DC: therefore, still more is BO+OC <BA+AC. 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. Let BAC and EDF be two triangles, having the side AB=DE, AC=DF, and the angle A>D; then will the side BC be greater than EF. Make the angle CAG=D; take AG=DE, and draw CG. Then, the triangles GAC and EDF will be equal, since they have two sides and an included angle in each equal, each to each (P. 5); consequently, CG is equal to EF (P. 5, c). There may be three cases in this proposition. 1st. When the point G falls without the triangle BAC. 2d. When it falls on the side BC; and 3d. When it falls within the triangle. Case I. In the triangles AGC and ABC, we have, its equal AB from the other, and there will remain BC |