Perpendiculars and Oblique Line Segments
29. Theorem. If, from a given point outside a line, we draw a perpendicular
and several oblique line segments:
1◦. The perpendicular is shorter than any oblique segment;
2◦. Two oblique segments whose feet are equally distant from the foot of the
perpendicular are equal;
3◦. Of two oblique segments, the longer is the one whose foot is further from
the foot of the perpendicular.
Consider perpendicular OH and oblique segment OA from point O to line
xy (Fig. 31). If we extend OH by a length HO equal to itself, then O is symmetric
to O with respect to line xy. Therefore O A is the symmetric image of OA, and
the two are equal. Now in triangle OO A we have
OO OA + O A,
and we can replace OO with 2OH and OA + OA with 2OA. Thus we ﬁnd 2OH
2OA, or OH OA.
Consider next the oblique segments OA, OB such that HA = HB. These
two oblique segments will be equal by symmetry with respect to line OH.
3◦. Finally, let OA and OC be oblique segments such that HC HA (Fig.
31). Suppose ﬁrst that points A and C are on the same side of H. Then point A
is inside triangle OO C. By (27) we have
OA + O A OC + O C
and, as we saw before, OA = O A and OC = O C. Dividing by two, we have, as