EXERCISES 33 perpendicular). One could also use the properties of isosceles triangles (23, Re- mark). However, one must observe that our way of proceeding has the advantage of showing which of the two distances is the greater when they are unequal. Remark II. The proposition which we just stated: Any point which is equidis- tant from A and B is on the perpendicular bisector of AB, is the converse of the first part of the preceding theorem. We have here two different ways of proving a converse. The first consists in following the original reasoning in reverse. This is what we did in the preceding remark. The original reasoning (1◦ in the previous theorem) started from the hypothesis that point M is on the perpendicular bisector in other words, that the feet of MA, MB were equally distant from the perpendic- ular from M , and deduced from this that MA and MB are equal. This time, we started from the hypothesis that MA and MB are equal, and we concluded that their feet are equally distant. The second method of proving the converse consists in proving what we call the inverse statement. This name is given to the statement whose hypothesis is the negation of the original hypothesis, and whose conclusion is the negation of the original conclusion. Thus the second part of the preceding theorem is the inverse of the first, and is equivalent to its converse. We will find later on (see, for example, 41) a third method of proving a converse. 33. Let us now make use of the definition of 1b. Using this definition, the preceding theorem can be restated as follows: Theorem. The locus of points equidistant from two given points is the perpen- dicular bisector of the segment joining these two points. This is true because the figure formed by the points equidistant from A and B is in fact the perpendicular bisector of AB. We remark that, to establish this fact, one must prove, as we have, that: 1◦. every point on the perpendicular bisector satisfies the given condition 2◦. every point satisfying this condition is on the perpendicular bisector or, equivalently, that no point outside the bisector satisfies the condition. This kind of double argument is necessary in all problems about geometric loci. Exercises Exercise 16. If the legs of a first right triangle are respectively smaller than the legs of a second, then the hypotenuse of the first is smaller than the hypotenuse of the second. Exercise 17. If the angles B and C of a triangle ABC are acute, and the sides AB, AC unequal, then the lines starting from A are, in decreasing order of length, as follows: larger side, median (see Exercise 7 in Chapter II), angle bisector, smaller side, altitude. Exercise 18. The median of a non-isosceles triangle is greater than the bisector from the same vertex, bounded by the third side.
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