divided into as many triangles as it has sides. Now, the sum of the three angles in each of these triangles is equal to two right angles (Prop. XXVIII.); therefore the sum of the angles of all these triangles is equal to twice as many right angles as there are triangles, or sides, to the polygon. But the sum of all the angles about the point P is equal to four right angles (Prop. IV. Cor. 2), which sum forms no part of the interior angles of the polygon; therefore, deducting the sum of the angles about the point, there remain the angles of the polygon equal to twice as many right angles as the figure has sides, less four right angles. 103. Cor. 1. The sum of the angles in a quadrilateral is equal to four right angles; hence, if all the angles of a quadrilateral are equal, each of them is a right angle; also, if three of the angles are right angles, the fourth is likewise a right angle. 104. Cor. 2. The sum of the angles in a pentagon is equal to six right angles; in a hexagon, the sum is equal to eight right angles, &c. 105. Cor. 3. In every equiangular figure of more than four sides, each angle is greater than a right angle; thus, in a regular pentagon, each angle is equal to one and one fifth right angles; in a regular hexagon, to one and one third right angles, &c. E 106. Scholium. In applying this proposition to polygons which have re-entrant angles, or angles whose vertices are directed inward, as BP C, each of these angles must be considered greater than two right angles. But, in order to avoid ambiguity, we shall hereafter limit our reasoning to polygons with salient angles, or with angles directed outwards, and which may be called convex polygons. Every convex polygon is such that a A D P B с straight line, however drawn, cannot meet the perimeter of the polygon in more than two points. PROPOSITION XXX.-THEOREM. 107. The sum of all the exterior angles of any polygon, formed by producing each side in the same direction, is equal to four right angles. A Let each side of the polygon ABCDE be produced in the same direction; then the sum of the exterior angles A, B, C, D, E, will be equal to four right angles. For each interior angle, together with its adjacent exterior angle, is equal to two right angles (Prop. I.); hence the sum of all the angles, both interior and exterior, is equal to twice as many right angles as there are sides to the polygon. But the sum of the interior angles alone, less four right angles, is equal to the same sum (Prop. XXIX.); therefore the sum of the exterior angles is equal to four right angles. E Let ABCD be a parallelogram ; then the opposite sides and angles are equal to each other. D D PROPOSITION XXXI.-THEOREM. 108. The opposite sides and angles of every parallelogram are equal to each other. B C B Draw the diagonal BD, then, since the opposite sides AB, DC are paral- A lel, and BD meets them, the alternate angles ABD, BDC are equal (Prop. XXII.); and since AD, BC are parallel, and BD meets them, the alternate angles ADB, DBC are likewise equal. Hence, the two triangles ADB, DBC have two angles, A BD, A D B, in the one, equal to two angles, BDC, DBC, in the other, each to each; and since divided into as many triangles as it has sides. Now, the sum of the three angles in each of these triangles is equal to two right angles (Prop. XXVIII.); therefore the sum of the angles of all these triangles is equal to twice as many right angles as there are triangles, or sides, to the polygon. But the sum of all the angles about the point P is equal to four right angles (Prop. IV. Cor. 2), which sum forms no part of the interior angles of the polygon; therefore, deducting the sum of the angles about the point, there remain the angles of the polygon equal to twice as many right angles as the figure has sides, less four right angles. 103. Cor. 1. The sum of the angles in a quadrilateral is equal to four right angles; hence, if all the angles of a quadrilateral are equal, each of them is a right angle; also, if three of the angles are right angles, the fourth is likewise a right angle. 104. Cor. 2. The sum of the angles in a pentagon is equal to six right angles; in a hexagon, the sum is equal to eight right angles, &c. 105. Cor. 3. In every equiangular figure of more than four sides, each angle is greater than a right angle; thus, in a regular pentagon, each angle is equal to one and one fifth right angles; in a regular hexagon, to one and one third right angles, &c. E 106. Scholium. In applying this proposition to polygons which have re-entrant angles, or angles whose vertices. are directed inward, as BPC, each of these angles must be considered greater than two right angles. But, in order to avoid ambiguity, we shall hereafter limit our reasoning to polygons with salient angles, or with angles directed outwards, and which may be called convex polygons. Every convex polygon is such that a A F D B с straight line, however drawn, cannot meet the perimeter of the polygon in more than two points. PROPOSITION XXX.-THEOREM. 107. The sum of all the exterior angles of any polygon, formed by producing each side in the same direction, is equal to four right angles. Let each side of the polygon ABCDE be produced in the same direction; then the sum of the exterior angles A, B, C, D, E, will be equal to four right angles. A For each interior angle, together with its adjacent exterior angle, is equal to two right angles (Prop. I.); hence the sum of all the angles, both interior and exterior, is equal to twice as many right angles as there are sides to the polygon. But the sum of the interior angles alone, less four right angles, is equal to the same sum (Prop. XXIX.); therefore the sum of the exterior angles is equal to four right angles. Let ABCD be a parallelogram ; then the opposite sides and angles are equal to each other. D D C PROPOSITION XXXI.-THEOREM. 108. The opposite sides and angles of every parallelogram are equal to each other. B C B Draw the diagonal BD, then, since the opposite sides A B, DC are paral- A lel, and BD meets them, the alternate angles ABD, BDC are equal (Prop. XXII.); and since AD, BC are parallel, and BD meets them, the alternate angles ADB, DBC are likewise equal. Hence, the two triangles ADB, DBC have two angles, ABD, AD B, in the one, equal to two angles, BDC, DBC, in the other, each to each; and since the side BD included between these equal angles is common to the two triangles, they are equal (Prop. VI.); hence the side A B opposite the angle ADB is equal to the side DC opposite A the angle DBC (Prop. VI. Cor.); and, in like manner, the side A D is equal to the side BC; hence the opposite sides of a parallelogram are equal. B D с Again, since the triangles are equal, the angle A is equal to the angle C (Prop. VI. Cor.); and since the two angles DBC, ABD are respectively equal to the two angles AD B, BD C, the angle ABC is equal to the angle AD C. 109. Cor. 1. The diagonal divides a parallelogram into two equal triangles. 110. Cor. 2. The two parallels A D, B C, included between two other parallels, A B, CD, are equal. D PROPOSITION XXXII.-THEOREM. 111. If the opposite sides of a quadrilateral are equal, each to each, the equal sides are parallel, and the figure is a parallelogram. Let ABCD be a quadrilateral having its opposite sides equal; then will the equal sides be parallel, and the figure be a parallelogram. A B For, having drawn the diagonal BD, the triangles A BD, BDC have all the sides of the one equal to the corresponding sides of the other; therefore they are equal, and the angle A D B opposite the side. A B is equal to D B C opposite CD (Prop. XVIII. Sch.); hence the side A D is parallel to BC (Prop. XX.). For a like reason, AB is parallel to CD; therefore the quadrilateral A B C D is a parallelogram. C |