F 12 E C D Cor. III. And as all the angles that can be made on the other side of AB are also equal to two right angles; therefore all the angles that can be made around the point D, by any number of lines, are together equal to four right angles. Cor. IV. If two adjacent angles, ADC and BDC, are supplementary, their exterior sides. AD and BD will form one and the same line. Scholium. Since a straight line has two different directions, exactly opposite each other, it is sometimes considered as making with itself an angle equal to two right angles. A. D B THEOREM II. If through the vertex of any angle, lines are drawn perpendicular respectively to its sides, they will form a new angle, either equal to the first, or supplementary to it. Let BAC be the given angle, DE perpendicular to AB, and FG perpendicular to AC. F A D G C B Then we shall have the angle DAF equal to BAC, since each is the complement of CAD (D. XV.). The angle EAG being opposite DAF must also equal BAC (T. I). But the angle DAG, or its opposite angle FAE, is supplementary with DAF, and consequently supplementary with its equal angle BAC. E THEOREM III. Two angles having their corresponding sides parallel are either equal or supplementary. First. When the sides AB and AC, forming the angle at A, have respectively the same directions as the sides DE and DF forming the angle at D. C H Produce CA and DE until they cross each other at G. Now, since AB and DE are parallel, they have the same direction in reference to the line CG (D. X.), consequently the angle BAC is equal to KGC. And since AC and DF are parallel they have the same direc tion in reference to the line DG; consequently, the angle KGC is equal to EDF; therefore (A. I.), the angle BAC is equal to EDF. Secondly. When the sides AB and AC are respectively in opposite directions to DE and DF. As before, we have the angle BAC equal to KGC, and the angle EDF equal to EGH. But KGC is equal to its opposite angle EGH (T. I.), consequently, the angle BAC is equal to EDF. Thirdly. When one of the sides, as AB, has an opposite direction to DE, its corresponding side. As before, the angle BAC is equal to KGC, and EDF is equal to EGC. But KGC and E F H E D C DA B K Ꮐ H C F B K EGC are supplementary (T. I.), consequently BAC and EDF are supplementary. Cor. Two angles having their corresponding sides perpendicular, are either equal or supplementary. For, if we draw through the vertex of the first angle lines respectively parallel to the sides of the second angle, we should thus form a third angle, which will be equal to the second. But this third angle is either equal to the first or supplementary to it (T. II.). Consequently the second angle is either equal to the first or supplementary to it (A. I.). THEOREM IV. When all the sides of a polygonal figure are produced, in the same direction, the sum of all the exterior angles will be equal to four right angles. For, if from any point in the same plane, straight lines be drawn parallel respectively to the sides of the figure, the angles contained by the straight lines about that point will be equal to the exterior angles of the figure (T. III.), each to each. Thus the angles a, b, c, etc., are respectively equal to the exterior F A D C d angles A, B, C, etc.; but the former angles are together equal to four right angles (T. I., C. III.); therefore all the exterior angles of the figure are together equal to four right angles. F Scholium. This proposition must be restricted to the case in which the polygon is convex. A convex polygon may be defined to be one, such that no side, by being produced in either direction, can divide the polygon. The polygon ABCDFG is not convex, since it may be divided by producing either of the sides CD or FD. This polygon is said to have a re-entering angle at D. G C D A R THEOREM V. In any convex polygon, the sum of all the interior angles, taken together, is equal to twice as many right angles as the polygon has sides, wanting four right angles. G F g a A D d b B с Let ABCDFG be a convex polygon. Conceive the sides to be produced all in the same direction, forming exterior angles, which we will denote by the capital letters A, B, C, etc., while their corresponding interior angles are denoted by the small letters a, b, c, etc. Now any exterior angle, together with its adjacent interior angle, as A+a, is equal to two right angles (T. I.), therefore the sum of all the interior angles, together with all the exterior angles, is equal to twice as many right angles as the polygon has sides; but the sum of all the exterior angles is equal to four right angles (T. IV.); therefore the sum of all the interior angles is equal to twice as many right angles as the polygon has sides, wanting four right angles. Cor. I. In any triangle, the sum of the three angles is equal to two right angles. Cor. II. If one angle of a triangle is a right angle, as in the case of a right-angled triangle (D. XXII.), each of the other angles must be acute, and they are complementary, since their sum must equal a right angle. Each of the angles of a triangle may be acute, but only one of the angles can be obtuse. Cor. III. If two triangles have two angles of the one respectively equal to two angles of the other, their third angles will be equal, and the triangles will be mutually equiangular. Cor. IV. In any quadrilateral, the sum of the four interior angles is equal to four right angles. Cor. V. If two angles of a quadrilateral are right, the other two angles will be supplementary. Cor. VI. If from a point D within a triangle ABC, we draw two lines to the extremities of one of the sides AB, the angle ADB formed by these lines will be greater than the angle ACB, opposite this side. A D C B For, the sum of the two angles DAB, DBA is obviously less than the sum of the two angles CAB, CBA. And since the sum of the three angles of every triangle is equal to two right angles, it follows that the remaining angle ADB, of the triangle ADB, must be greater than the remaining angle ACB of the triangle ACB. THEOREM VI. The exterior angle, formed by producing one of the sides of a triangle, is equal to the sum of the two interior and opposite angles of the triangle. In the triangle ABC, if the side AB be produced to D, the exterior angle CBD will be equal to the sum of the two angles at A and at C. C A B D For, the angle CBD, together with its adjacent angle, CBA, is equal to two right angles (T. I.), and the sum of A and C, together with the same angle CBA, is also equal to two right angles (T. V., C. I.); hence, the exterior angle is equal to the sum of the two interior and opposite angles. Cor. The exterior angle is greater than either of the interior and opposite angles. OF THE SIDES OF TRIANGLES. THEOREM VII. Either side of any triangle is less than the sum of the other two sides, and greater than their difference. C First. The straight line AB is the shortest distance between A and B (A. XI.), and therefore shorter than the broken line AC+ CB. The same reasoning applies to each of the sides. Hence, we have AB < AC+ CB; AC <AB+CB; BC < AB+ AC. A B Secondly. Since AC < AB + CB, we have AC - CB < AB (A. V.). That is, AB > AC- CB. In a similar manner we deduce AC > AB- CB; CB > AB- AC. THEOREM VIII. If from any point within a triangle two lines be drawn to the extremities of either side, the sum of these two lines will be less than the sum of the other two sides. From the point D, suppose the lines DA and DB to be drawn to the extremities of the side AB; then will DA+DB < AC+ BC. For, producing AD until it meets BC at E, we have (T. VII.) DB < DE+EB, consequently (A. IV.), Again, we have AE < AC+ CE, consequently, AE+EB < AC+ CE+EB, that is, Comparing these conditions, we have AD+DB < AC+ CB. C E A B |