3. The following drawing shows the method of making a map cdefg of the river bank CDEFG by use of the plane table. Study the drawing, and explain how the points c, d, e, f, and g of the map are obtained as the intersections of corresponding lines drawn to the points C, D, E, F, and G from two points in the base line. 640 ft., ab = 16 in., If AB = and ad = 20 in., how far is it from A to D? If cf is measured and found to be 10 in., how far is it from C to F? NOTE. It will be found a very interesting and valuable experience, if the school will provide, or the students make, a plane table and use it out of doors in doing such work as is suggested in the exercises above. CHAPTER VII INEQUALITIES: METHODS OF ATTACK 142. Inequalities. An inequality is a statement that two numbers or magnitudes are unequal. The symbols of inequality are >, meaning "is greater than," and <, meaning "is less than." Thus "a is greater than b" is written a > b. And " α is less than b" is written a < b. The inequalities a > b and x > y are of the same order. The inequalities a > b and x < y are of the reverse order. 143. Theorem. If two sides of a triangle are unequal, the angles opposite these sides are unequal, the angle opposite the greater side being the greater. 2. Then let CD be a part of AC equal to BC. Draw DB. 144. Theorem. If two angles of a triangle are unequal, the sides opposite these angles are unequal, the side opposite the greater angle being the greater. C A Hypothesis. In ▲ ABC, <B>LA. Conclusion. AC>BC. Suggestions. Use indirect proof. than, equal to, or less than BC. cannot equal BC. Show by § 143 than BC. Write the proof in full. B AC is either greater Show by § 70 that AC that AC cannot be less 145. Corollary. - The perpendicular from a point to a line is the shortest line-segment that can be drawn from the point to the line. The proof is left to the student. 146. Theorem. The sum of any two sides of a triangle is greater than the third side. Suggestions. Produce AC through C to D, making CD = BC. Compare ≤1 and ≤ D, Z≤ DBA and ≤ 1, then ▲ DBA and D. Why, then, is AC + CD > AB? Why, then, is AC+ BC AB? 147. Theorem. If two triangles have two sides of one equal respectively to two sides of the other, but the included angle of the first greater than the included angle of the second, the sides opposite these angles are unequal, the greater side being opposite the greater angle. Hypothesis. A ABC and ▲ DEF have AC = DF, BC = EF, and ACB > < DFE. Conclusion. AB> DE. = = Proof. 1. AC DF, BC EF, ACB > < DFE. Hyp. 2... A DEF may be placed upon ▲ ABC so that DF coincides with AC, taking the position of AAMC. 3. ACB > < ACM and MC BC. 4. .. CM lies within ACB. = § 10 Ax. XII § 13 and Ax. X 5. Draw CN bisecting / MCB and intersecting AB at N, and draw MN. 6. Then ▲ MNCA BNC. 7. ... MN = NB. 8. AN+MN> AM. 9. .. AN + NB > AM, or AB > DE. § 63 Def. congruence § 146 Ax. XII In the special case when point M falls on AB, how may the proof be shortened? Is it possible to draw a figure in which M falls within ▲ ABC? Draw a figure in which the included angles ACB and DFE are both acute, and prove the theorem. 148. Theorem. If two triangles have two sides of one equal respectively to two sides of the other, but the third side of the first greater than the third side of the second, the angles opposite these sides are unequal, the greater angle being opposite the greater side. A B D E F Hypothesis. In ▲ ABC and DEF, AC = DF, BC = EF, AB> DE. Suggestions. Use indirect proof. Either <C><F, < C= ZF, or C<ZF. Show by § 63 that C is not equal to ZF. Show by § 147 that Cis not less than F. Write the complete proof. EXERCISES State in which cases it is possible to draw triangles with sides of the following lengths: 1. 4 in., 5 in., 6 in. 2. 2 in., 4 in., 6 in. 3. 6 in., 12 in., 20 in. 4. 6 in., 7 in., 13 in. 5. The hypotenuse of a right triangle is greater than either leg. 6. If an isosceles triangle is obtuse, the base is the longest side. 7. In any triangle, any side is greater than the difference between the other two sides. 8. Any side of a polygon is less than the sum of all of the other sides. SUGGESTION. - Draw all diagonals possible from one end of the given side. |