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
Figure 22
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