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

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