EXERCISES 37 Suppose the angle is ABC, and M is a point on its bisector. If we drop perpen- diculars MD, ME to the sides of the angle, then right triangles AMD and AME are congruent, because they have a common hypotenuse, and an equal acute angle (at A) by hypothesis. Therefore the perpendiculars MD, ME are equal. 2◦. The distances from any point inside the angle, but not on its bisector, to the two sides are unequal. Suppose point M lies, for instance, between the bisector and side AC. Then angle BAM will be greater than M AC. If we drop perpendiculars M D , M E onto sides AB, AC, then right triangles AM D , AM E will have a common hy- potenuse and an unequal angle at A. Therefore M D will (35) be greater than M E . As in the case of the theorem of 32, we could have proved, in place of the inverse proposition in 2◦ above, the converse proposition: Any point inside an angle, and equidistant from its two sides, is on its bisector. To do this, following the original reasoning in reverse, we would have considered a point M equidistant from the two sides, and would have applied the second case of congruence (34) to the two right triangles AMD, AME, which have a common hypotenuse and in which MD = ME. We would have concluded that the angles at A are equal, so that AM is the angle bisector. However, we would not have established which is the larger distance when they are unequal. Corollaries. I. This theorem allows us to give a second definition of the bisector of an angle, namely: The bisector of an angle is the locus of the points inside the angle which are equidistant from the sides. We observe that this second definition is exactly equivalent to the one given in 11. II. The locus of all points equidistant from two intersecting lines is composed of the two bisectors (16) of the angles formed by these two lines. Exercises Exercise 19. Show that a triangle is isosceles if it has two equal altitudes. Exercise 20. More generally, in any triangle, the smaller altitude corresponds to the larger side.
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