11. Let there be a six-sided figure as shown, having all its sides produced in one direction; and denote the exterior angles thus formed by a, b, c, d, e, f, and the interior angles by 1, 2, 3, 4, 5, 6. What is the value of each of the pairs of angles: a and 1, b and 2, c and 3, &c. ? 12. What is the sum of the six interior and the six exterior angles taken altogether? 13. Assuming the result of Question 4 of this exercise, deduce from 12 the sum of the outside angles a, b, c, d, e, f taken alone. e/5 2/6 14. After the manner indicated in Questions 11-13, calculate the sum of the exterior angles of a five-sided figure, such as that of Corollary 2, which follows. 15. Apply a similar process of reasoning to obtain the sum of the exterior angles of a decagon formed by producing its sides successively in one direction. 78. COROLLARY 2.-THEOREM. All the exterior angles of any rectilineal figure, made by producing the sides successively in the same direction, are together equal to four right angles. Since every interior angle ABC, and its adjacent exterior angle ABD, are together equal to two right angles, (I. 13) therefore all the interior angles, together with all the exterior angles of the figure, are equal to twice as many right angles as the figure has sides. But, by the foregoing Corollary, all the interior angles, together with four right angles, are equal to twice as many right angles as the figure has sides; therefore all the interior angles, together with all the exterior angles, are equal to all the interior angles and four right angles. Take from these equals all the interior angles; (Ax. 1) therefore all the exterior angles of the figure are equal to four right angles. (Ax. 3) Q. E. D. 79. THE DETERMINATION OF THE MAGNITUDES OF THE ANGLES OF EQUIANGULAR POLYGONS. In the preceding exercise we have practised calculations of the angles of equiangular polygons by means of Corollary 1. The procedure we followed renders clear to the mind the nature of that Corollary; and it also appears to be that which is usually preferred for the purpose to which it was applied. It will probably be found, however, that Corollary 2 affords a readier means of effecting the same calculations. This is certainly the case in such as require the angles to be expressed in degrees. The student may judge for himself while working the ensuing exercise. EXERCISE XXXIX. Note. In this exercise the exterior angles of a figure are supposed to be made by producing its sides successively in the same direction. 1. If the interior angles of a figure be all equal, prove that its exterior angles are also equal. 2. If an exterior angle be one-third of a right angle, find the adjacent interior angle. 3. If an exterior angle of an equiangular figure be right angle, calculate each interior angle. 4. Calculate each exterior angle of an equiangular pentagon. Thence deduce each interior angle. 5. By the procedure indicated in 4, calculate each interior angle of an equiangular hexagon. 6. Prove this rule : To find the interior angle of an equiangular polygon, divide four right angles by the number of sides, and subtract the result from two right angles. 7. Apply this rule in finding the angle of an equiangular octagon, decagon, duodecagon. 8. Find the same angles in the same way, calling four right angles 360°, and obtaining each result in degrees. figure are also equal, find two sides in the one triangle respectively equal to two sides in the other. 3. Show, by I. 4, that these triangles are equal in all respects. 4. Write out all the resulting equalities established by I. 4; and state which of them proves that AD, BC are parallel. 5. Which result proves that the same sides are also equal? 6. Prove that the opposite sides of a rhombus are parallel. Note.-If two parallel straight lines be placed so as to be horizontal, those ends which are both to the right would be said to be "towards the same parts,” and similarly those which are on the left. The expression would be applied to other positions of the lines in a similar manner. 80. PROPOSITION XXXIII.—THEOREM. The straight lines which join the extremities of two equal and parallel straight lines towards the same parts, are also themselves equal and parallel. Let AB, CD be equal and parallel straight lines, and joined towards the same parts by the straight lines AC, BD. Then AC, BD shall be equal and parallel. Join BC. A D Then, because AB is parallel to CD, and BC meets them, the angle ABC is equal to the alternate angle BCD. (I. 29) And because AB is equal to CD, and BC common to the triangles ABC, DCB; the sides AB, BC are equal to DC, CB, each to each; and the angle ABC was proved equal to DCB; therefore the base AC is equal to the base BD. (I. 4) Also the angles are equal to which the equal sides AB, CD are opposite; therefore the angle ACB is equal to CBD. (I. 4) And because BC meets the two straight lines AC, BD, and makes the alternate angles ACB, CBD equal; 1. What proposition may be, in part, enunciated thus ?— Any quadrilateral having two sides equal and parallel is a parallelogram. How much more than this is proved in the proposition referred to ? 2. Show that, if two opposite sides of a quadrilateral are equal and parallel, the diagonal will bisect it. 3. In the figure to XXXIII., given AB, CD parallel, and also AC, BD; find two angles in the triangle ABC equal to two in BCD, each to each. 4. Show that the side adjacent to these angles of the one triangle is equal to the side adjacent to those of the other. 5. Draw three inferences by I. 26. 6. From the two equalities of Question 3, deduce that the angles ABD, ACD are equal, quoting an axiom. 7. If on opposite sides of a base two equilateral triangles be made, the two together form a parallelogram. 8. If two opposite sides of a quadrilateral be equal and parallel, the straight line bisecting both will be parallel to the remaining sides of the figure. 81. PROPOSITION XXXIV.-THEOREM. The opposite sides and angles of a parallelogram are equal to one another; and either diameter bisects it, that is, divides it into two equal triangles. Let ABDC be a parallelogram, of which BC is a diameter. Then its opposite sides and angles shall be equal, and its diameter BC shall bisect it. (I. 29) Because AB is parallel to CD, and BC meets them, the angle ABC is equal to the alternate angle BCD. And because AC is parallel to BD, and BC meets them, the angle ACB is equal to the alternate angle CBD. (I. 29) Hence the triangles ABC, CBD have the two angles ABC, BCA in the one equal to the two BCD, CBD in the other, each to each; and one side BC, adjacent to these equal angles, common; therefore their other sides are equal, each to each, and also their third angles; (I. 26) namely, AB equal to CD, AC to BD, and the angle BAC to the angle BDC; and these are the opposite sides and one pair of opposite angles of the parallelogram. Also the angle ABC has been proved equal to BCD, and the angle CBD to ACB; therefore the whole ABD is equal to the whole ACD ; (Ax. 2) and these are the other opposite angles of the figure. Hence the opposite sides and angles of a parallelogram are equal to one another. Again, in the triangles ABC, BCD, since AB is equal to CD and BC common, |