Figure 50
In triangle ABC (Fig.51), we draw the bisectors of angles B and C; they
intersect at a point O inside the triangle. This point, being on the bisector of angle
B, is equidistant from the sides AB and BC. Likewise, being on the bisector of
angle C, the point O is equidistant from AC and BC. It is therefore equidistant
from AB and AC and, being interior to angle A, is on the bisector of this angle.
Since the exterior angles CBx, BCy have a sum less than four right angles,
halves of each have a sum less than two right angles. The bisectors of these angles
will therefore meet (41, Corollary) at a point O inside angle A. This point O , like
point O, will be equidistant from the three sides of the triangle. Therefore it will
belong to the bisector of A.
55. Theorem. The segment joining the midpoints of two sides of a triangle
is parallel to the third side, and equal to half of it.
In triangle ABC (Fig. 52), let D be the midpoint of AB and let E be the
midpoint of AC. We extend line DE past E by its own length, to a point F .
Quadrilateral ADCF will be a parallelogram (47), and therefore CF will be equal
and parallel to DA, or, equivalently, to BD. Thus quadrilateral BDCF is also a
parallelogram. Therefore:
1◦. DE is parallel to BC;
2◦. DE, which is equal to half of DF , is also half of BC.
56. Theorem. The three medians of a triangle are concurrent at a point
situated on each of them one third of its length from the corresponding side.
First, let BE, CF be two medians of triangle ABC (Fig. 53). We claim that
their intersection G lies at one third the length of each of them. To see this, let M
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