Let $I$ be the midpoint of $AB$ and $M$, $Q$ be points on the base $CD$, such that $DM = QC = AI$.

Construct squares $IMNP$ and $IQRS$, externally to the triangle $IMQ$.

Since $\angle PIS + \angle PIM + \angle MIQ + \angle QIS = 360^\circ$ and $IPUS$ is a parallelogram, we infer that $\angle PIS + \angle MIQ = 180^\circ = \angle IPU + \angle PIS$, so $\angle IPU = \angle MIQ$. As $IP = IM$ and $PU = IS = IQ$, triangles $IPU$ and $MIQ$ are congruent (S.A.S.), so $\angle PIU = \angle IMQ$.
Triangle $IMV$ is right-angled at $V$, so $\angle MIV = 90^\circ - \angle IMV = 90^\circ - \angle PIU$. It follows that $\angle MIV + \angle PIU + \angle MIP = 180^\circ$, therefore $U$, $I$, $V$ are collinear.
Hence, $UI$ is the perpendicular bisector of segment $AB$. Since $IPUS$ is a parallelogram, the line $UI$ passes through the midpoint $J$ of the diagonal $PS$.
As $\angle DAF = \angle MIP = 90^\circ$ and $AD \parallel IM$, it follows that $AF \parallel IP$. But $AF = IP$, so $AIPF$ is a parallelogram, which means that $FP \parallel AI$ and $FP = AI$. Analogously, we prove that $SH \parallel IB$ and $SH = IB$. Since $I$ is the midpoint of $AB$, we infer that $FP \parallel SH$ and $FP = SH$, so $FPHS$ is parallelogram. Therefore, the diagonals $FH$ and $PS$ have the same midpoint $J$, so $UI$ (the perpendicular bisector of $AB$) passes through the midpoint of $FH$.