But with this exception, two polygons may be mutually equilateral without being mutually equiangular, or mutually equiangular without being mutually equilateral. If two polygons are both mutually equilateral and mutually equiangular, they are equal. For they can evidently be applied one to the other so as to coincide throughout. 125. Two polygons are equal when they are composed of the same number of triangles, equal each to each, and similarly placed. For they can evidently be applied one to the other so as to coincide throughout. PROP. XLV. THEOREM. 126. The sum of the angles of any polygon is equal to two right angles taken as many times, less two, as the polygon has sides. Given a polygon of n sides. To Prove the sum of its equal to n 2 times two rt. 4. Proof. The polygon may be divided into n-2 A by drawing diagonals from one of its vertices. The sum of the of the polygon is equal to the sum of the of the A. But the sum of the s of each ▲ is two rt. 4. [The sum of the of any ▲ is equal to two rt. 4.] Hence, the sum of the s of the polygon is n two rt. s. (§ 84) 2 times 127. Cor. I. The sum of the angles of any polygon is equal to twice as many right angles as the polygon has sides, less four right angles. For if R represents a rt. 2, and n the number of sides of a polygon, the sum of its is (n − 2) × 2 R, or 2 nR – 4 R. 128. Cor. II. The sum of the angles of a quadrilateral is equal to four right angles; of a pentagon, six right angles; of a hexagon, eight right angles; etc. 129. If the sides of any polygon be produced so as to make an exterior angle at each vertex, the sum of these exterior angles is equal to four right angles. Given a polygon of n sides with its sides produced so as to make an ext. at each vertex. To Prove the sum of these ext. 4 equal to 4 rt. Æ. Proof. The sum of the ext. and int. at any one vertex is two rt. s. [If two adj. is equal to two rt. have their ext. sides in the same str. line, their sum Hence, the sum of all the ext. and int. is 2n rt. 4. (§ 32) - 4 rt. s. [The sum of the of any polygon is equal to twice as many rt. as the polygon has sides, less 4 rt, Æ.] Whence, the sum of the ext. 4 is 4 rt. 4. (§ 127) EXERCISES. 34. How many degrees are there in each angle of an equiangular hexagon? of an equiangular octagon? of an equiangular decagon ? of an equiangular dodecagon? 35. How many degrees are there in the exterior angle at each vertex of an equiangular pentagon ? 36. If two angles of a quadrilateral are supplementary, the other two angles are supplementary. 37. If, in a triangle ABC, ZA = ZB, a line parallel to AB makes equal angles with sides AC and BC. (To prove CDE = / CED.) 38. If the equal sides of an isosceles triangle be produced, the exterior angles made with the base are equal. (§ 31, 2.) D B 39. If the perpendicular from the vertex to the base of a triangle bisects the base, the triangle is isosceles. (Fig. of Prop. XXX. ▲ ACD and BCD are equal by § 63.) 40. The bisectors of the equal angles of an isosceles triangle form, with the base, another isosceles triangle. 41. If from any point in the base of an isosceles triangle perpendiculars to the equal sides be drawn, they make equal angles with the base. (<ADE = / BDF, by § 31, 1.) 42. If the angles adjacent to one base of a trapezoid are equal, those adjacent to the other base are also equal. (Given ZA = ≤D; to prove ▲ B = ≤ C.) 43. Either exterior angle at the base of an isosceles triangle is equal to the sum of a right angle and one-half the vertical angle. B (≤ DAE is an ext. 2 of ▲ ACD.) A B D 44. The straight lines bisecting the equal angles of an isosceles triangle, and terminating in the opposite sides, are equal. (^ ABD = ▲ ABE.) E X45 B 45. Two isosceles triangles are equal when the base and vertical angle of one are equal respectively to the base and vertical angle of the other. (Each of the remaining of one ▲ is equal to each of the remaining of the other.) 46. If two parallels are cut by a transversal, the bisectors of the four interior angles form a rectangle. (EH|| FG, by § 73; in like manner, EF || GH; then use Exs. 12 and 33.) 47, Prove Prop. XXVI. by drawing through B a line parallel to AC. (Sum of 4 at B = 2 rt. 4.) A -B F B A MISCELLANEOUS THEOREMS. X PROP. XLVII. THEOREM. 130. The line joining the middle points of two sides of a triangle is parallel to the third side, and equal to one-half of it. Given line DE joining middle points of sides AB and AC, respectively, of ▲ ABC. Proof. Draw line BF AC, meeting ED produced at F. In A ADE and BDF, ZADEL BDF. [If two str. lines intersect, the vertical ▲ are equal.] Also, since Is AC and BF are cut by AB, ZA=2DBF. (§ 40) [If two Is are cut by a transversal, the alt. int. are equal.] (§ 72) [Two are equal when a side and two adj. of one are equal respectively to a side and two adj. of the other.] [In equal figures, the homologous parts are equal.] (§ 68) (§ 66) Then since, by hyp., AE EC, BF is equal and I to CE. Whence, BCEF is a . = [If two sides of a quadrilateral are equal and ||, the figure is a .] 131. Cor. The line which bisects one side of a triangle, and is parallel to another side, bisects also the third side. Given, in ▲ ABC, D the middle point of side AB, and line DE || BC. To Prove that DE bisects AC. D Proof. A line joining D to the middle B point of AC will be | BC. [The line joining the middle points of two sides of a ▲ is || to the third side.] Then this line will coincide with DE. (§ 130) [But one str. line can be drawn through a given point || to a given str. line.] Therefore, DE bisects AC. (§ 53) |