Solution:

a. Suppose that $CD$ is the minor base of the trapezoid, that $X$ is on $AD$ and $Y$ on $BC$, as in the figure. Since the lines $AB$, $XY$, $DC$ are parallel, by Thales' theorem we have the proportion $DX : XA = CY : YB$, and hence $DX : (DX + XA) = CY : (CY + YB)$, that is $DX : DA = CY : CB$.

Note now that the triangles $ABD$ and $XPD$ are similar, since $XP$ is parallel to $AB$ (hence $\angle DXP = \angle DAB$ and $\angle DPX = \angle DBA$); from this follows the proportion between corresponding sides $XP : AB = DX : DA$.

In exactly the same way, the triangle $ABC$ is similar to the triangle $PYC$ ($PY$ is parallel to $AB$, the corresponding angles that are formed are congruent), and the proportion $PY : AB = CY : CB$ holds.

Combining the proportions written so far, $XP : AB = DX : DA = CY : CB = PY : AB$, hence $XP = PY$, as was to be shown.

b. Let $M$ be the point of intersection between $QP$ and the minor base. By the parallelism between $DC$ and $XY$ we have, in the manner of the previous proof, the similarity between the triangle $QDM$ and the triangle $QXP$, as well as between the triangle $QMC$ and the triangle $QPY$. From this we derive the proportions $DM : MQ = XP : PQ$ and $CM : MQ = YP : PQ$; since, by part (a), $XP = YP$, we obtain $DM : MQ = CM : MQ$, and hence finally $DM = CM$ ($M$ is the midpoint of $DC$).

Note that the claims of the problem (both that of part (a) and that of part (b)) are invariant under affine transformations of the plane: indeed, affinities preserve the ratios between the lengths of segments on the same line, and therefore it suffices to show the claims on an affine image of the initial construction.

Every trapezoid $ABCD$ can be transformed by an affinity into an isosceles trapezoid; take for example the affinity that fixes $A$ and $B$ and sends $Q$ to a point (different from the midpoint of $AB$) on the perpendicular bisector of $AB$. The triangle $ABQ$ is sent to an isosceles triangle, the trapezoid to an isosceles trapezoid; since affinities send lines to lines and preserve parallelism, the construction of the problem remains the same. We have thus reduced ourselves to showing that $XP = PY$ and that $QP$ meets the minor base at its midpoint in the case where $ABCD$ is isosceles. In this case, however, the claims are evident by symmetry: $P$ lies on the perpendicular bisector of $AB$, $CD$ and $XY$, which passes through $Q$.