We are given two parallel lines, and a common perpendicular which intersects them in A, B. Points C, D are taken on these lines so that trape-zoid ABCD has an area equal to that of a given square. Let H be the projection on CD of the midpoint of AB. Find the locus of H. (One must distinguish two

Question

We are given two parallel lines, and a common perpendicular which intersects them in A, B. Points C, D are taken on these lines so that trape-zoid ABCD has an area equal to that of a given square. Let H be the projection on CD of the midpoint of AB. Find the locus of H. (One must distinguish two cases, according as the points C, D are on the same or opposite sides of the common perpendicular.)

Accepted Answer

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).

[latex]{{1}}^{\circ}.[/latex] 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 [latex]E F={ extstyle{\frac{1}{2}}}(A C+B D),[/latex] 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.

[latex]2^{\circ}.[/latex] 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 [latex]C C_{0}[/latex] from C to line BD. Then [latex]A C=B C_{0},[/latex] so C0B+BD = AC+BD, which is again constant. Thus as C and D vary, the legs [latex]C C_{0},\,C_{0}D[/latex] of right triangle [latex]C C_{0}D[/latex] do not vary, so the triangle retains its shape (24, [latex]2^{\circ}[/latex]). This means that angle [latex]\widehat{{C D B}}[/latex] 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.

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

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