Elements of Geometry and Trigonometry

A CB and AC D, equal to the sum of A B C, BCA, and CAB. But the sum of A CB and A CD is equal to two right angles (Prop. I.); hence the sum of the three angles ABC, BCA, and C A B is equal to two right angles (Art. 34, Ax. 2).

96. Cor. 1. Two angles of a triangle being given, or merely their sum, the third will be found by subtracting that sum from two right angles.

97. Cor. 2. If two angles in one triangle be respectively equal to two angles in another, their third angles will also be equal.

98. Cor. 3. A triangle cannot have more than one angle as great as a right angle.

99. Cor. 4. And, therefore, every triangle must have at least two acute angles.

100. Cor. 5. In a right-angled triangle the right angle is equal to the sum of the other two angles.

101. Cor. 6. Since every equilateral triangle is also equiangular (Prop. VII. Cor. 3), each of its angles will be equal to two thirds of one right angle.

PROPOSITION XXIX.-THEOREM.

102. The sum of all the interior angles of any polygon is equal to twice as many right angles, less four, as the figure has sides.

Let ABCDE be any polygon; then the sum of all its interior angles, A, B, C, D, E, is equal to twice as many right angles as the figure has sides, less four right angles.

For, from any point P within the polygon, draw the straight lines PA, PB, PC, PD, PE, to

the vertices of all the angles, and the polygon will be

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.

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

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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

straight line, however drawn, cannot meet the perimeter of the polygon in more than two points.

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. 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.

108. The opposite sides and angles of every parallelogram are equal to each other.

Let ABCD be a parallelogram;

then the opposite sides and angles are equal to each other.

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, 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 AB 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.

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.

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.

For, having drawn the diagonal BD, the triangles A B D, BD C 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 (Prep. XX.). For a like reason, AB is parallel to CD; therefore the quadrilateral A B C D is a parallelogram.

112. If two opposite sides of a quadrilateral are equal and parallel, the other sides are also equal and parallel, and the figure is a parallelogram.

Let ABCD be a quadrilateral, having the sides A B, CD equal and parallel; then will the other sides also be equal and parallel.

Draw the diagonal BD; then, since A

A B is parallel to CD, and B D meets them, the alternate angles A BD, BDC are equal (Prop. XXII.); moreover, in the two triangles A B D, D B C, the side BD is common; therefore, two sides and the included angle in the one are equal to two sides and the included angle in the other, each to each; hence these triangles are equal (Prop. V.), and the side AD is equal to B C. Hence the angle A D B is equal to D B C, and consequently A D is parallel to BC (Prop. XX.); therefore the figure ABCD is a parallelogram.

PROPOSITION XXXIV. - THEOREM.

113. The diagonals of every parallelogram bisect each other.

Let ABCD be a parallelogram,

and A C, DB its diagonals, intersecting at E; then will A E equal E C,

and BE equal E D.

For, since A B, CD are parallel, and BD meets them, the alternate angles CDE, ABE are equal (Prop. XXII.); and since AC meets the same parallels, the alternate angles BAE, ECD are also equal; and the sides AB, CD are equal (Prop. XXXI.). Hence the triangles ABE, CDE have two angles and the in

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Elements of Geometry and Trigonometry: With Practical Applications