Chinese Mathematical Olympiad

(1) Denote by $Q$, $R$ the midpoints of $OB$, $OC$, respectively. It is easy to see that $$EQ=\frac{1}{2}OB=RM,MQ=\frac{1}{2}OC=RF,$$ and $$\angle EQM=\angle EQO+\angle OQM=2\angle EBO+\angle OQM,$$ $$\angle MRF=\angle FRO+\angle ORM=2\angle FCO+\angle ORM.$$ Because $A$, $B$, $C$, $D$ are concyclic, and $Q$, $R$ are the midpoints of $OB$, $OC$, we have $$\angle EBO=\angle FCO,\angle OQM=\angle ORM.$$ So $\angle EQM=\angle MRF$, which implies that $△EQM\cong △MRF$, and $EN=FN$. Similarly, we have $EN=FN$, so $EM\times FN=EN\times FM$ holds.

(2) Suppose that $OA=2a$, $OB=2b$, $OC=2c$, $OD=2d$ and $$\angle OAB=\alpha ,\angle OBA=\beta ,\angle ODC=\gamma ,\angle OCD=\theta .$$ Then $$\begin{array}{rl}\cos \angle EQM& =\cos (\angle EQO+\angle OQM)\\ & =\cos (2\beta +\angle AOB)\\ & =-\cos (\alpha -\beta ).\end{array}$$ So $$\begin{gathered} E{M}^2=E{Q}^2+Q{M}^2-2EQ\times QM\times \cos \angle EQM \\ =b^2+c^2+2bc\cos (\alpha -\beta ). \end{gathered}$$ Making similar equations for $EN$, $FN$, $FM$, we have $$\begin{array}{rl}& EN\times FM=EM\times FN\\ \iff & E{N}^2\times F{M}^2=E{M}^2\times F{N}^2\\ \iff & (a^2+d^2+2ad\cos (\alpha -\beta ))\times (b^2+c^2+2bc\cos (\gamma -\theta ))\\ & =(a^2+d^2+2ad\cos (\gamma -\theta ))\times (b^2+c^2+2bc\cos (\alpha -\beta ))\\ \iff & (\cos (\gamma -\theta )-\cos (\alpha -\beta ))(ab-cd)(ac-bd)=0.\end{array}$$ Because $\alpha +\beta =\gamma +\theta$, $\cos (\gamma -\theta )-\cos (\alpha -\beta )=0$ holds if and only if $\alpha =\gamma$, $\beta =\theta$ (i.e. $A$, $B$, $C$, $D$ are concyclic) or $\alpha =\theta$, $\beta =\gamma$ (which follows $AB\parallel CD$, a contradiction); $ab-cd=0$ holds if and only if $AD\parallel BC$; $ac-bd=0$ holds if and only if $A$, $B$, $C$, $D$ are concyclic. So, when $AD\parallel BC$ holds, we also have $$EM\times FN=EN\times FM.$$ We know that $A$, $B$, $C$, $D$ are not concyclic in this case because $PB\ne PC$, so the answer is "false".