Question P.413: We are given two parallel lines, and a common perpendicular ......

One key to this problem is the observation that because the area and the height of the trapezoid are both constant, the sum AC + BD of its bases is also constant (252b).

{{1}}^{\circ}. Suppose segments AC, BD are on the same side of the common perpendic-ular, as in figure t413a. Then if we join the midpoints E, F of AB, CD respectively, then E F={\textstyle{\frac{1}{2}}}(A C+B D), which is also constant. Thus point F is fixed as C and D vary. If we drop perpendicular EH to line CD, then as CD moves, H lies on a circle with (fixed) diameter EF (78f). Thus the locus of point H is the part of this circle lying outside triangle AFB.
Note that this solution depends on the length of EF being fixed.

2^{\circ}. Suppose segments AC, BD are on opposite sides of the common perpendic-ular, as in figure t413b. Then we can still find a segment of fixed length, if w edrop a perpendicular C C_{0} from C to line BD. Then A C=B C_{0}, so C0B+BD = AC+BD, which is again constant. Thus as C and D vary, the legs C C_{0},\,C_{0}D of right triangle C C_{0}D do not vary, so the triangle retains its shape (24, 2^{\circ}). This means that angle \widehat{{C D B}} has the same measure, and diagonal CD is merely translated parallel to itself. If we drop perpendicular EH onto any position of CD, then the locus of H is asegment of the perpendicular line (EK in figure t413b). Because A is the limiting position of point C, and B is the limiting position of point D, the endpoints of this segment are the intersections M, N of line EK with the parallels to CD through points A and B.