THEOREM XXX. 215. In a convex polygon, the sum of the exterior angles, one at each vertex, made by producing each side in order, is a perigon. HYPOTHESIS. A convex polygon of n sides, each in order produced at one end, CONCLUSION. Sum of exterior X's perigon. PROOF. Every interior angle, as 4 A, and its adjacent exterior angle, as X. *, together = st. 2. .. all interior X's + all exterior X's = n st. *'s. But, by 211, all interior *'s (n − 2) st. {'s. sum of all exterior X's = a perigon. ... EXERCISES. 30. How many diagonals can be drawn in a polygon of n sides? 31. The exterior angle of a regular polygon is one-third of a right angle: find the number of sides in the polygon. 32. The four bisectors of the angles of a quadrilateral form a second quadrilateral whose opposite angles are supplemental. 33. Divide a right-angled triangle into two isosceles triangles. 34. In a right-angled triangle, the sect from the mid-point of the hypothenuse to the right angle is half the hypothenuse. III. Parallelograms. THEOREM XXXI. 216. If two opposite sides of a quadrilateral are equal and parallel, it is a parallelogram. HYPOTHESIS. ABCD a quadrilateral, with AB = and 11 CD. By hypothesis, AB = CD, and AC is common, * CAD, (124. Triangles are congruent if two sides and the included angle are equal in each.) .. i. BC || AD. EXERCISES. 35. Find the number of elements required to determine a parallelogram. 36. The four sects which connect the mid-points of the consecutive sides of any quadrilateral, form a parallelogram. 37. The perpendiculars let fall from the extremities of the base of an isosceles triangle on the opposite sides will include an angle supplemental to the vertical angle of the triangle. 38. If BE bisects the angle B of a triangle ABC, and CE bisects the exterior angle ACD, the angle E is equal to one-half the angle A. THEOREM XXXII. 217. The opposite sides and angles of a parallelogram are equal to one another, and each diagonal bisects it. HYPOTHESIS. Let AC be a diagonal of ABCD (using the sign for the word "parallelogram "). CONCLUSIONS. AB = CD. BC= DA. * DAB = XBCD. *B= D. A ABC A CDA. X CAD = % BCA; therefore, adding, X BAC + * CAD = X BAD = % ACD + 4 ACB = = 4 BCD. Again, the side AC included between the equal angles is common, (128. Triangles are congruent if two angles and the included side are equal in each.) EXERCISES. 39. If one angle of a parallelogram be right, all the angles are right. 40. If two parallelograms have one angle of the one equal to one angle of the other, the parallelograms are mutually equiangular. 218. COROLLARY I. Any pair of parallels intercept equal sects of parallel transversals. 219. COROLLARY II. If two lines be respectively parallel to two other lines, any angle made by the first pair is equal or supplemental to any angle made by the second pair. 220. COROLLARY III. If two angles have their arms respectively perpendicular, they are equal or supplemental. For, revolving one of the angles through a right angle around its vertex, its arms become perpendicular to their traces, and therefore parallel to the arms of the other given angle. 221. COROLLARY IV. If one of the angles of a parallelogram is a right angle, all its angles are right angles, and it is called a Rectangle. 222. COROLLARY V. If two consecutive sides of a parallelogram are equal, all its sides are equal, and it is called a Rhombus. 223. A Square is a rectangle having consecutive sides equal. c 224. Both diagonals, AC and BD, being drawn, it may with a few exceptions be proved that a quadrilateral which has any two of the following properties will also have the others : 1. AB || CD. B B 3. AB = CD. 4. BC= DA. 5. * DAB = * BCD. 6. * ABC = * CDA. 7. The bisection of AC by BD. 8. The bisection of BD by AC. 9. The bisection of the by AC. 10. The bisection of the by BD. N D |