SUGGESTIONS ON THE TREATMENT OF EXERCISES 219 In Arithmetic and Algebra definite rules are given for the solution of problems. No such specific directions can be given for the treatment of original exercises in Geometry. The following general directions, however, will greatly assist the beginner: 1 Draw general figures. Thus, if you are dealing with a triangle in general, draw a scalene triangle; if with a parallelogram, draw a rhomboid. Observe that this is always done in the text. 2 Draw figures accurately. An accurate figure frequently suggests a clue to the proof; an inaccurate figure leads into error. 3 Fix clearly in mind the things given about the figure, and the precise thing in the figure to be proved. This direction is fundamental; its disregard is the prime cause of the beginner's frequent failure. Observe under "Hypothesis" and "Conclusion" how carefully and clearly this injunction is observed in every proposition of the text. 4 In the proof make use of every condition given in the hypothesis. Work along lines involving only a part of the things given will end in certain failure. No relation would be given were it not a necessary part upon which to build the proof. Examine the proof of any proposition in the text, and observe that in such proof every condition given in the hypothesis is always used. 5 Begin by supposing the theorem true; then, step by step, trace out the relations which follow this supposition, until some known theorem is reached. Now begin with this known theorem and arrange the steps in reverse order, thus leading up to the required proof. In illustration consider the following THEOREM. The bisectors of two adjacent angles of a parallelogram are perpendicular to each other. In the DEFG let DH and EK bisect the 4D and E respectively. We are to prove that Ze is a rt. angle. Suppose that c is a rt. Z. Then ≤a + <b = a rt. (§ 155). Then D+ZE = D G a K H E F 2 rt. 4 (Ax. 7). But this last conclusion is a known theorem (§ 124) with which we begin the direct This is the usual method of attacking most exercises in Geometry, and is called Analysis. The reverse process, the orderly building up from known facts until a new truth is obtained, is called Synthesis. It should be observed that the synthetic method of proof is used in most text propositions. 6 When other means fail, especially in converse theorems, try the indirect method. See §§ 115, 116. 7 Draw such auxiliary lines as may be necessary to give a clue to the line of argument. See the dotted lines in the figures of many of the text propositions; also, Exs. 138, 150, 151. The ability to attack exercises successfully and rapidly implies a comprehensive knowledge of the chief propositions of the text, and a keen recognition of their application at sight. To this end we here remind the student of some of the cardinal truths which should ever be vividly in his mind. Two lines are equal: 1 If they are homologous sides of equal triangles. § 166 § 176 § 196 $ 200 Two angles are equal: 1 If they are complements or supplements of equal angles. 2 If they are vertical angles. §§ 104, 105 § 112 3 If they are alternate-interior or corresponding angles of parallel lines. §§ 121, 122 § 166 § 173 4 If they are homologous angles of equal triangles. 5 If they are opposite equal sides in a triangle. 6 If they are the opposite angles of a parallelogram. § 203 Two triangles are equal: 1 If two sides and the included angle of the one are equal respectively to two sides and the included angle of the other. § 162 2 If two angles and the included side of the one are equal respectively to two angles and the included side of the other. § 167 3 If the three sides of the one are equal respectively to the three sides of the other. Two right triangles are equal: 1 If their legs are equal, each to each. § 172 § 163 2 If the hypotenuse and an acute angle of the one are equal respectively to the hypotenuse and an acute angle of the other. § 168 3 If a leg and an acute angle of the one are equal respectively to a leg and the homologous acute angle of the other. § 169 4 If the hypotenuse and a leg of the one are equal respectively to the hypotenuse and a leg of the other. Two lines are parallel : § 178 1 If they are perpendicular to the same straight line. § 115 2 If they are parallel to a third straight line. § 118 (d) the interior angles on the same side of the transversal supplementary. § 128 4 If they are the opposite sides of a parallelogram. $ 200 5 If they are the bases of a trapezoid. § 197 A quadrilateral is a parallelogram: 1 If the opposite sides are parallel. 2 If the opposite sides are equal. 3 If two sides are equal and parallel. MISCELLANEOUS EXERCISES 168 If from any point within a triangle lines are drawn to the extremities of any side, the angle included by them is greater than the angle included by the other two sides. [<x> <y> <z. § 159.] § 195 § 204 $ 205 169 If from any point within a triangle lines are drawn to the extremities of any side, their sum is less than the sum of the other two sides. [PC < PD + DC. .. PB+ PC < BD + DC. For BD substitute the greater value AB + AD, and PB + PC < AB+ AC.] A B C [By Ex. 169, < a + b (3). x 170 The sum of the lines drawn from any point within a triangle to the three vertices is less than the perimeter. x + y ≤ b + c (1), y + z < a + c (2), x + z Add (1), (2), (3), and reduce.] 2 b 171 The sum of the lines drawn from any point within a triangle to the three vertices is greater than half the perimeter. [x + y > a (1), y + z > b (2), x + z > c (3). and reduce.] Add (1), (2), (3), 172 Perpendiculars drawn from any point in the base of an isosceles triangle to the legs make equal angles with the base. [Ex. 40.] 173 Perpendiculars dropped from any two points in the legs of an isosceles triangle to the base make equal angles with the legs. [Ex. 40.] 174 BC is the base of the isosceles triangle ABC. BA is produced its own length to D. Prove that DC is perpendicular to BC. [4 B = ≤ x. <D = 2 BCD = a rt. 2.] 2D = 2y. .. < B + 175 A triangle is divided into four equal triangles by the lines joining the mid-points of its sides. [By § 201, show that ▲ 1 is equal to each of the ▲ 2, 3, and 4.] 4 3 176 The triangle formed by joining the mid-points of the sides of an equilateral triangle is equilateral. 177 In the triangle ABC, BE and CD are altitudes intersecting at H. Prove that the DHE is the supplement of A. [In the quadl. ADHE, ≤ D = < E = art. 2. ...H is sup. of ▲ A (§ 215).] 178 In the equilateral triangle ABC, AX Prove XYZ an equilateral triangle. [§ 162.] = BY = CZ. 179 If the base angles of two isosceles triangles are complementary, the vertical angles are supplementary. B B D HE C 180 If the vertical angles of two isosceles triangles are supplementary, the base angles are complementary. 181 The median to the hypotenuse of a right triangle is equal to half the hypotenuse. [CD = EF = DA = AB.] E [In the Fig. 182 If one acute angle of a right triangle is double the other acute angle, the hypotenuse is double the shorter leg. of Ex. 181, let A = 2 ZB. Then ▲ A = 60°, and ▲ ADC is equilateral.] 183 The feet of two altitudes of a triangle are equidistant from the mid-point of the third side. [In rt. ▲ ADC, DM is the median to the hypotenuse AC. ... DM = AC by Ex. 181. Likewise in rt. ▲ AEC, EM = AC. ... DM 184 The altitudes of an equilateral triangle are equal. M E |