CHAPTER III

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.

x y

Figure 31

1◦.

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.

2◦.

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

before,

OA OC.

31