$(1)$证明：$\because$正方形$ABCD$中，对角线$AC$，$BD$交于点$O$，
$\therefore \angle BOC=90^{\circ}$，$\angle OBE=45^{\circ}$，
$\because \angle COE=22.5^{\circ}$
$\therefore \angle BOH=67.5^{\circ}$，
$\because \angle BHO=90^{\circ}$，
$\therefore \angle OBH=22.5^{\circ}$，$\angle EBH=22.5^{\circ}$，
$\therefore \angle OBH=\angle EBH$，
$\because \left\{\begin{array}{l}∠BHO=∠BHE=90°\\ BH=BH\\∠OBH=∠EBH\end{array}\right.$，
$\therefore \triangle OBH$≌$\triangle EBH\left(ASA\right).$
$(2)$证明：过点$C$作$CG\bot OE$，交$OE$的延长线于点$G$，
$\because$正方形$ABCD$中，对角线$AC$，$BD$交于点$O$，
$\therefore \angle BOC=90^{\circ}$，$OB=CO$，
$\therefore \angle BOH+\angle COE=90^{\circ}$，
$\because \angle BHO=90^{\circ}$，
$\therefore \angle BOH+\angle OBH=90^{\circ}$，
$\therefore \angle COG=\angle OBH$，
$\because \left\{\begin{array}{l}∠BHO=∠OGC=90°\\∠HBO=∠GOC\\ OB=CO\end{array}\right.$，
$\therefore \triangle OBH$≌$\triangle COG\left(AAS\right)$，
$\therefore OH=CG$，
$\because OH=FM$，
$\therefore FM=CG$，
$\because \left\{\begin{array}{l}∠FEM=∠CEG\\∠EMF=∠EGC\\ FM=CG\end{array}\right.$，
$\therefore \triangle FEM$≌$\triangle CEG\left(AAS\right)$，
$\therefore FE=CE$.

$(3)$$BH-OH=2FM$.理由如下：
线段$BH$，$OH$，$FM$之间的数量关系为$BH-OH=2FM$.理由如下：过点$C$作$CG\bot OE$，交$OE$的延长线于点$G$，
$\because$正方形$ABCD$中，对角线$AC$，$BD$交于点$O$，
$\therefore \angle BOC=90^{\circ}$，$OB=CO$，
$\therefore \angle BOH+\angle COE=90^{\circ}$，
$\because \angle BHO=90^{\circ}$，
$\therefore \angle BOH+\angle OBH=90^{\circ}$，
$\therefore \angle COG=\angle OBH$，
$\because \left\{\begin{array}{l}∠BHO=∠OGC=90°\\∠HBO=∠GOC\\ OB=CO\end{array}\right.$，
$\therefore \triangle OBH$≌$\triangle COG\left(AAS\right)$，
$\therefore OH=CG$，
连接$EG$，并延长$EG$，交$BH$于点$N$，
$\because BH\bot HG$，$CG\bot HG$，$BF=CF$，
$\therefore BH$∥$CG$，
$\therefore \angle NBF=\angle GCF$，
$\because \left\{\begin{array}{l}∠NBF=∠GCF\\ BF=CF\\∠BFN=∠CEG\end{array}\right.$，
$\therefore \triangle BFN$≌$\triangle CFG\left(ASA\right)$，
$\therefore FN=FG$，$BN=CG=OH$，
连接$FH$，
则$FH=FG$，
$\because FM\bot HG$，
$\therefore MH=MG$
$\therefore FM=\frac{1}{2}NH$，
$\because NH=BH-BN=BH-CG=BH-OH$，
$\therefore FM=\frac{1}{2}(BH-OH)$，
即$BH-OH=2FM$.

$(1)$证明：$\because$正方形$ABCD$中，对角线$AC$，$BD$交于点$O$，
$\therefore \angle BOC=90^{\circ}$，$\angle OBE=45^{\circ}$，
$\because \angle COE=22.5^{\circ}$
$\therefore \angle BOH=67.5^{\circ}$，
$\because \angle BHO=90^{\circ}$，
$\therefore \angle OBH=22.5^{\circ}$，$\angle EBH=22.5^{\circ}$，
$\therefore \angle OBH=\angle EBH$，
$\because \left\{\begin{array}{l}∠BHO=∠BHE=90°\\ BH=BH\\∠OBH=∠EBH\end{array}\right.$，
$\therefore \triangle OBH$≌$\triangle EBH\left(ASA\right).$
$(2)$证明：过点$C$作$CG\bot OE$，交$OE$的延长线于点$G$，
$\because$正方形$ABCD$中，对角线$AC$，$BD$交于点$O$，
$\therefore \angle BOC=90^{\circ}$，$OB=CO$，
$\therefore \angle BOH+\angle COE=90^{\circ}$，
$\because \angle BHO=90^{\circ}$，
$\therefore \angle BOH+\angle OBH=90^{\circ}$，
$\therefore \angle COG=\angle OBH$，
$\because \left\{\begin{array}{l}∠BHO=∠OGC=90°\\∠HBO=∠GOC\\ OB=CO\end{array}\right.$，
$\therefore \triangle OBH$≌$\triangle COG\left(AAS\right)$，
$\therefore OH=CG$，
$\because OH=FM$，
$\therefore FM=CG$，
$\because \left\{\begin{array}{l}∠FEM=∠CEG\\∠EMF=∠EGC\\ FM=CG\end{array}\right.$，
$\therefore \triangle FEM$≌$\triangle CEG\left(AAS\right)$，
$\therefore FE=CE$.

$(3)$$BH-OH=2FM$.理由如下：
线段$BH$，$OH$，$FM$之间的数量关系为$BH-OH=2FM$.理由如下：过点$C$作$CG\bot OE$，交$OE$的延长线于点$G$，
$\because$正方形$ABCD$中，对角线$AC$，$BD$交于点$O$，
$\therefore \angle BOC=90^{\circ}$，$OB=CO$，
$\therefore \angle BOH+\angle COE=90^{\circ}$，
$\because \angle BHO=90^{\circ}$，
$\therefore \angle BOH+\angle OBH=90^{\circ}$，
$\therefore \angle COG=\angle OBH$，
$\because \left\{\begin{array}{l}∠BHO=∠OGC=90°\\∠HBO=∠GOC\\ OB=CO\end{array}\right.$，
$\therefore \triangle OBH$≌$\triangle COG\left(AAS\right)$，
$\therefore OH=CG$，
连接$EG$，并延长$EG$，交$BH$于点$N$，
$\because BH\bot HG$，$CG\bot HG$，$BF=CF$，
$\therefore BH$∥$CG$，
$\therefore \angle NBF=\angle GCF$，
$\because \left\{\begin{array}{l}∠NBF=∠GCF\\ BF=CF\\∠BFN=∠CEG\end{array}\right.$，
$\therefore \triangle BFN$≌$\triangle CFG\left(ASA\right)$，
$\therefore FN=FG$，$BN=CG=OH$，
连接$FH$，
则$FH=FG$，
$\because FM\bot HG$，
$\therefore MH=MG$
$\therefore FM=\frac{1}{2}NH$，
$\because NH=BH-BN=BH-CG=BH-OH$，
$\therefore FM=\frac{1}{2}(BH-OH)$，
即$BH-OH=2FM$.