II. ON TRIANGLES 23 Figure 21 Theorem. In an isosceles triangle, the bisector of the angle at the vertex is perpendicular to the base, and divides it into two equal parts. In isosceles triangle ABC (Fig. 21), let AD be the bisector of A. In turning angle BAC around on itself, this bisector does not move, and therefore neither does the point D where this bisector cuts the base. Segment DB falls on DC and angle ADB on ADC. Therefore DB = DC and ADB = ADC. QED Remark. In triangle ABC we can consider: 1◦. The bisector of A 2◦. The altitude from A 3◦. The median from the same point 4◦. The perpendicular to the midpoint of BC. In general, these four lines are distinct from each other (see Exercise 17). The preceding theorem shows that, in an isosceles triangle, all of these are the same line, which is a line of symmetry of the triangle (19b). This theorem can thus be restated: the altitude of an isosceles triangle is also an angle bisector and a median or the median of an isosceles triangle is at the same time an angle bisector and an altitude the perpendicular bisector of the base passes through the vertex and bisects the angle at the vertex. Corollary. In an isosceles triangle, the altitudes dropped from the endpoints of the base are equal the same is true about the medians from the endpoints of the base, about the bisectors of the angles at these points, etc., because these segments are symmetric to each other. 24. The following propositions, known under the name of cases of congru- ence of triangles, give necessary and suﬃcient conditions for two triangles to be congruent. Figure 22

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