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. A C B Hypothesis. In △ ABC, ∠ B > ∠ A. Conclusion. AC>BC. Suggestions. Use indirect proof. AC is either greater than, equal to, or less than BC. cannot equal BC. Show by § 143 than BC. Write the proof in full. 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 21 and ∠ D, ∠ 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. △ABC and A DEF have AC = DF, BC = EF, and ∠ACB > ∠ DFE. Conclusion. AB>DE. Proof. 1. AC = DF, BC = EF, ∠ ACB > < DFE. Hyp. 2... △ DEF may be placed upon & ABC so that DF co incides with AC, taking the position of A AMC. 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, 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 C B D E F Hypothesis. In AABC and DEF, AC = DF, BC = EF, Suggestions. Use indirect proof. Either / C > < F, ∠ C= ∠F, or < C <∠F. Show by § 63 that ∠C is not equal to ∠F. Show by § 147 that ∠C is 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. 61 in., 71 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 9. The perimeter of any quadrilateral is greater than the sum of the diagonals. side. 10. The sum of the diagonals of any quadrilateral is greater than the sum of a pair of opposite sides. 11. In ∆ABC, AB>AC and DB = EC. Prove BE > CD. 12. If P is any point between B and C in the base BC of an isosceles triangle ABC, then AP <AB. 13. The median to one side of a triangle is less than one half of the sum of the other two sides. SUGGESTIONS. — Produce CD through D to E so that DE = CD. Now it may be proved that CD < (AC + BC), if it is first proved that 2 CD or CE <AC + BC. 14. In any triangle the sum of two sides is greater than the sum of two line-segments drawn from a point within the triangle to the extremities of the third side. SUGGESTIONS. - Produce AD to meet BC at A E. Prove AC + CB > AE + EB. prove AE + EB > AD + DB. Ax. XI. Similarly, 15. D is a point within ∆ABC such that BD = BC. Prove AC > AD. 16. If two right triangles have their hypotenuses equal and an acute angle of one greater than the corresponding acute angle of the other, the legs opposite these angles are unequal, the longer leg being oppo site the larger angle. SUGGESTIONS. - If BC=EF and ∠C>∠F, produce AB through A M A to M, making AM = AB, and draw CM; also produce DE through D to N, making DN = DE, and draw FN. Now show that MB > NE, and hence AB > DE. 17. If two sides of a triangle are unequal, the median to the third side makes the larger angle with the shorter of the unequal sides. SUGGESTIONS. - Draw a triangle ABC with AC>CB; let CD be the median. Prove that ∠DCB > ∠ACD. Produce CD to E, making CD = DE ; draw AE. Compare ∠ DCB with ∠ E, and ∠ ACD with ∠E. 18. Using the figure drawn for Ex. 17, prove that CD < (AC + CB). 19. Give a direct proof that any point not on the perpendicular bisector of a line-segment is not equidistant from the ends of the segment. SUGGESTION. - If CD is the perpendicular bisector of AB and P any point not on CD, let P be on the same side of CD as B, and let AP intersect CD at M. Draw MB. Prove CP A D B PB < PA. PB is less than the sum of what two line-segments? 20. Give a direct proof that a point within an angle but not on the bisector of the angle is not equidistant from the sides. SUGGESTION. - If BD bisects ∠ABC, and P is within ∠ABC but not on BD, let P be between AB and BD. Prove PE<PF. Draw MNBA and draw PN. Show that PE < PN, PN < PM + MN, etc. B C F D M P NEA 149. Method of attack. - The student's experience in proving the unproved theorems and corollaries and the exercises in the preceding pages should have revealed the fact that there is a definite procedure or method of attack in such proofs. Success in demonstrations requires, first of all, thorough familiarity of the student with all preceding theorems, corollaries, and definitions upon which the proofs may depend. Most proofs involve a comparison of line-segments, angles, or ratios. It would be well for the student at this point to review all of the theorems, corollaries, and definitions, and to fix them thoroughly in mind as providing means of proving one or other of the following fundamental relations : |