Let $X,Y$ be the midpoints of $AB,CD$, and let $AB$ intersects $DC$ at $E$. Then $E,M,N$ are collinear. Since $XY\parallel BC$, $EN$ also passes through the midpoint of $XY$. Suppose the intersection of the perpendicular bisector is at $O$ on $MN$. Note that $EO$ is the diameter of the circumcircle of $\triangle EXY$, so it passes through the circumcenter. But it is also the median, so it must be isosceles*. It follows that $BA=CD$. (b) should be true by above? Edit: *it is either isosceles OR right at $E$. But when it is right at $E$, $O$ is outside $MN$. So (b) is false.

Nah, the answer to (b) is false. The trapezoid's lateral sides should be perpendicular to each other in order to make the perpendicular bisector of each lateral side meet on the line $MN$, but outside the segment $MN$.

a) + b): Let the midpoins of $AB$ and $CD$ be $E,F$ respectively. Denote the intersection of the perpendicular bisectors with $MN$ by $K$. Note that $S(AKB)=S(CKD)$, as $S(APN)=S(DPN)$, $S(BPM)=S(CPM)$ and $S(BKM)=S(CKM)$. So, if $AB=2x$ and $CD=2y$, $\frac{EK}{FK}=\frac{y}{x}$, but on the other hand, by pythagorean theorem, $EK^{2}-FK^{2}=PF^{2}-PE^{2}=(y^{2}-x^{2})(k+1)$, where $k=\frac{BP}{BE}=\frac{CP}{PF}$. So, $EK=PF, FK=PE$ or $x=y$. The last one easily leads to the desired result. For the first case, $PEKF$ is parallelogram, so it's rectangle as it contains right angle. But than angle $APD$ is right and the perpendicular bisector of $AP$ intersects $MN$ outside the trapezoid and so, this case is not avaliable and we get the desired result for part a). For part b), it's easy from above to see the counterexample. Just take right angled non-isosceles triangle $APD$ and take some points $B,C$ on the segments $AP$ and $DP$ such that $ABCD$ is trapezoid. It's quite easy to see that the intersection of perpendicular bisectors is on the line $MN$, not on the segment. Hence, the answer for part b is negative. P.S. However, the condition seems too easy for being problem on such kind of olympiad.

oneplusone wrote: so it passes through the circumcenter. But it is also the median, so it must be isosceles. Or, it could be right angled.