A trapezoid $ABCD$ and a line $\ell$ perpendicular to its bases $AD$ and $BC$ are given. A point $X$ moves along  $\ell$. The perpendiculars from $A$ to $BX$ and from $D$ to $CX$ meet at point $Y$ . Find the locus of $Y$ .

by D.Prokopenko
1. **Identify the given elements and their properties:**
   - Trapezoid \(ABCD\) with bases \(AD\) and \(BC\).
   - Line \(\ell\) perpendicular to the bases \(AD\) and \(BC\).
   - Point \(X\) moves along \(\ell\).
   - Perpendiculars from \(A\) to \(BX\) and from \(D\) to \(CX\) meet at point \(Y\).

2. **Define the intersection point \(K\):**
   - Let \(K = BA \cap CD\).

3. **Identify the circumcenters:**
   - Let \(O\) be the circumcenter of \(\triangle BCX\).
   - Let \(O'\) be the circumcenter of \(\triangle ADY\).

4. **Analyze the perpendiculars:**
   - Since \(YA \perp BX\) and \(YD \perp CX\), we have \(\angle BXC = \angle AYD\).

5. **Identify the ex-similicenter:**
   - Point \(K\) is the ex-similicenter of the circumcircles of \(\triangle ADY\) and \(\triangle BCX\).

6. **Symmetry and homothety:**
   - Let \(Y'\) be the symmetric point of \(Y\) with respect to \(O'\).
   - By angle chasing, \(\triangle XBC \sim \triangle Y'AD\).
   - Therefore, \(\triangle Y'AD\) is the image of \(\triangle XBC\) under the homothety centered at \(K\).

7. **Determine the locus of \(Y'\):**
   - The locus of \(Y'\) is a line, say \(\ell'\).

8. **Symmetry with respect to the bisector:**
   - Since \(O'\) is on the bisector of \(AD\), the locus of \(Y\) is the symmetric of \(\ell'\) with respect to the bisector of \(AD\).

The final answer is the symmetric of \( \boxed{ \ell' } \) with respect to the bisector of \(AD\).