Let $$l$$ be a fixed line. $$ABCD$$ be a square having a fixed side length, moving such that its vertex $$D$$ is at a fixed point on the line $$l$$ and such that the square is on one side of the line $$l$$. Let $$A'$$, $$C'$$ be the feet of the perpendiculars of $$A$$ and $$C$$ on $$l$$ respectively. Then $$\triangle AA'D$$ and $$\triangle DC'C$$ are congruent if angle $$BDA' =$$

Let $$\angle ADA' = \theta$$; since $$\angle ADC = 90^\circ$$, $$\angle CDC' = 90^\circ - \theta = \angle DAA'$$. Both triangles are right angled with hypotenuse equal to the side of the square, so $$\triangle AA'D \cong \triangle DC'C$$ for every non-degenerate position, $$0^\circ < \theta < 90^\circ$$. As $$\angle BDA' = \theta + 45^\circ$$, it lies strictly between $$45^\circ$$ and $$135^\circ$$; so $$90^\circ$$ and $$60^\circ$$ are possible, while $$30^\circ$$ is impossible and $$45^\circ$$ makes the triangles degenerate. Hence options c and d are correct.