From a point F within the polygon, draw a straight line to each of the angles, dividing the polygon into as many triangles as it has sides. The angles of each triangle =hm L? .. The sum of the angles of all the triangles 54. = hm L? 29.3. .. The angles at A, B, C, D, and E, = hm L? = twice .. The interior angles of the polygon are together as many right angles as the polygon has sides, wanting how many? § 62. II. Given, any pol ygon ABCDE. XXI. F K Required, the sum of its ex- B terior angles. H Each interior angle + its adjacent exterior = hm L? 59. C. 61. To the angles at what point in Fig. XX. (§ 61) are the exterior angles of a polygon equal? § 63. THEOR. VIII. The interior angles of any polygon are together equal to twice as many right angles, less four, as the figure has sides; and the exterior angles are equal to four right angles. [Proved by dividing the polygon into triangles, &c.] = hm L? == a.) The interior angles of a quadrilateral hm? of a pentagon? of a hexagon? of a heptagon? of an octagon? of a decagon? of a triangle (applying Theor. VIII.)? How many sides has a polygon, if its interior angles are together equal to 4L? if they are = 6L? 8L? 2L? In what kind of a polygon is the sum of the interior angles equal to the sum of the exterior? In what, is the sum of the interior angles half that of the exterior? In what, twice as great? In what, 3 times as great? b.) In an equiangular triangle, how many degrees does each interior angle contain? each exterior angle? What fraction of a right angle is each? Answer the same questions in respect to an equiangular quadrilateral, pentagon, hexagon, octagon, and decagon. In equiangular polygons, as the number of sides increases, does each interior angle increase or diminish? each exterior angle? In what kind of a figure is each interior angle twice as great as its adjacent exterior angle? In what, equal? In what, half as great? In what, 4 times as great? c.) Can you fill up the space about a point with equiangular triangles? how many? with equiangular quadrilaterals? how many? with equiangular pentagons? with equiangular hexagons? how many? Can you fill up the space about a point with equiangular polygons of any other kind? Why? Which of these figures has the bee, that admirable geometer, adopted in the construction of its cell? d.) If, in Fig. XXI. (§ 62), BAE BCD 110°, and CDE = 70°, ABC: 140°, = 130°, hm does each of the other angles of the pentagon, both interior and exterior, contain? 100°, BAF = 90°, CBG = If AEK = 70°, hm does each of the other angles of If BAF: = 100°, BCD = 120°, EDI 90°, hm does each of the other angles tain? § 64. e.) aACD + aACB hm L? And aBAC+ aABC + aACB hm L? = 60°, and DCH= the figure contain? 80°, and AED of the figure con = 59. C. = 54. B 20. a. 20. f. .. ACD + ACB ≈≈≈ BAC + ABC +ACB? ACD BAC+ ABC? COR. Any exterior angle of a triangle is equal to the sum of the two interior opposite angles. It is of course greater than either singly. Draw CE | BA, and prove the same from Theor. V. Join AD; and then (as the angles of tADC are together = the angles of tADB) show, by subtracting the same angles from both sums, that aACD = @BAC + aABC. B XXII. A Prove from this Corollary, that the sum of the exterior angles of a triangle is equal to twice the sum of the interior angles. 65. f.) Divide a polygon into triangles, by drawing diagonals from one of the angles; and show, by this means, what is the sum of the interior angles. A XXIII. G F E (h) Fr. rentrant, from Lat. re-, again, and intro, to enter. (i) L. saliens, springing forth, projecting. angle made by one side with the other side produced falls within the figure. Still there is no need of excepting this case in the doctrine of exterior angles, if one distinction is only borne in mind; viz. that, while the exterior angle must be added to a salient angle to obtain 180°, it must be subtracted from a re-entrant angle to obtain the same. The exterior angle of a re-entrant angle is therefore a negative quantity, and must be subtracted in obtaining the sum of the exterior angles of a polygon. § 68. i.) That the exterior angles of every polygon must be equal to 4 right angles, is also evident from the consideration that, if we depart from the direction of CD (Fig. XXI. or XXIII.), we must pass through the whole circuit of divergence (§ 10), before we can come round to it again; and that it is indifferent how many bends we make in completing the circuit. Thus, in Fig. XXIII., the bend at D carries us from the direction of DE to that of DA; the bend at A, from that of DA to that of DI (|| AG); the bend at B, from that of DI to that of DK (|| BH); and the bend at C, from that of DK back to that of DE. k.) In the case of a re-entrant angle, there is a turning back in the journey round, and a passing over the same ground twice. Thus (Fig. XXV.), after passing from the direction of DF to that of DE, we turn back at E to the direction of DH (|| EK), and then proceed again at A to the direction of DI (|| AL). But in every journey, the distance travelled back must, of course, be subtracted from the whole amount, to ascertain the onward progress. = = 90°, EDF = If, in Fig. XXV., CBL = 100°, DCN 120°, and AEG 50°, hm does BAK contain? hm° does each interior angle of the figure contain? 7.) A polygon which has no re-entrant angles is termed a convex polygon. PROPOSITION IX. § 69. I. Given, in tABC, AC = АВ. Required, aB aC. Draw AD bisecting aA. Then, as XXVI. B C 40. 40. R. XXVII. A D aABD aADB? 69. 64. ABD ACB? 20.b. ABC ABD? ADB DCB (or ACB)? ABC ACB? § 71. III. Given, in tABC, aB= aC. Required, AC≈≈ AB. Must AC be either, >, or < AB? If AC > AB, then aB≈≈≈ aC? 20. C. XXVIII. B 70. 23. a. (k) Lat. bis, in two, and seco, to cut, to divide into two equal parts. |