The Parallel Projection Theorem. We assume the Parallel Postulate. Let j j be a length function for segments. Summary: The Parallel Projection Theorem. We assume the Parallel Postulate. Let j  j be a length function for segments. Theorem. Suppose L; L 0 are distinct lines, o 2 L, o 0 2 L 0 . Let P : L ! L 0 be such that for each p 2 L either p 2 L 0 and P (p) = p or p 62 L 0 and P (p) is that point p 0 in L 0 such that l(p; p 0 ) k l(o; o 0 ): Then (1) js(P (c); P (d))j js(P (a); P (b))j = js(c; d)j js(a; b)j whenever a; b; c; d 2 L, a 6= b and c 6= d. Proof. Let us rst suppose that s(a; b) ' s(c; d). Let M be the line passing through a which is parallel to L 0 and let e be the point of intersection of M and l(b; P (b)). Let N be the line passing through c which is parallel to L 0 and let f be the point of intersection of M and l(d; P (d)). By ASA we nd that s(a; e) ' s(c; f ). Because opposite sides of parallelograms are congruent we nd that s(a; e) ' s(P (a); P (b)) and that s(c; f) ' s(P (c); P (d)). Thus, in this case, both sides of (1) are 1 and are therefore equal. Next let us suppose p; q are positive integers and p < s(a; b) >= q < s(c; d) > : Let e be that point in r(a; b) such that < s(a; e) >= p < s(a; b) > and let f be that point in r(c; d) such that < s(c; f) >= q < s(c; d) >. Note that < s(P (a); P (e)) >= p < s(P (a); P (b)) > and < s(P (c); P (f)) >= q < Collections: Mathematics