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:
every point on the perpendicular bisector satisﬁes the given condition;
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.
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.