Prove a equality geometry $EN\cdot IM=IN\cdot EM $ Let $O$ be the intersection of two diagonals of rectangle $ABCD$. $M$ and $N$ are respectively the midpoints of $CD$ and $A$B. $E$ is chosen arbitrarily in $AO$. The intersection of $EN$ and $BC$ is $I$. Prove that $$EN\cdot IM=IN\cdot EM $$

We need to prove $$EN\cdot IM=IN\cdot EM \Leftrightarrow \frac{EM}{MI}=\frac{EN}{NI}$$
Or $MN$ is bisector of $\angle EMI$. Or $\angle EMN=\angle NMI$.So I drew $PF//MI$ and the intersection of $EM$ and $AB$ is $Q$ and we have $\angle FPM=\angle PMI$ so it's enough to prove $\Delta PQM$ is isosceles triangle or $P;Q;F$ are colinear.
I am stuck here. I tried some method such as: Ménelaus's theorem, $Q$ is middle point $PF$, $\angle NPQ=\angle NPF$,... but I failed.

 A: I have a different approach.  Ultimately, this proof also shows that $MN$ internally bisects $\angle EMI$.  As a consequence, your other claims (such as $\triangle PQM$ is isosceles and $P,Q,F$ are collinear) follow.  
Let $AB=CD=a$ and $BC=DA=b$.  Project $E$ onto $AB$ and $CD$ at $X$ and $Y$, respectively.  Suppose that $x=NX$.  We have $\triangle ENX\sim \triangle INB$, so
$$\frac{EN}{NI}=\frac{NX}{NB}=\frac{x}{(a/2)}=\frac{2x}{a}.$$
Now, observe that $\triangle AXE\sim \triangle ABC$, so
$$\frac{EX}{\frac{a}{2}-x}=\frac{EX}{AX}=\frac{BC}{AB}=\frac{b}{a}.$$
Consequently, $EX=\frac{b}{a}\left(\frac{a}{2}-x\right)$, and
$$YE=XY-EX=b-EX=\frac{b}{a}\left(\frac{a}{2}+x\right).$$
Also, from $\angle ENX\sim \triangle INB$, we get
$$\frac{IB}{(a/2)}=\frac{IB}{NB}=\frac{EX}{NX}=\frac{\frac{b}{a}\left(\frac{a}{2}-x\right)}{x}.$$
This shows that
$$IB=\frac{b}{2x}\left(\frac{a}{2}-x\right),$$
making
$$IC=BC+IB=b+IB=\frac{b}{2x}\left(\frac{a}{2}+x\right).$$
Hence,
$$\frac{IC}{MC}=\frac{IC}{(a/2)}=\frac{b}{ax}\left(\frac{a}{2}+x\right)=\frac{EY}{x}=\frac{EY}{NX}=\frac{EY}{MY}.$$
Therefore, $\triangle EYM$ and $\triangle ICM$ are right triangles (with $\angle EYM=90^\circ=\angle ICM$) such that the sides adjacent to the right angles are in proportion.  This means $\triangle EYM\sim \triangle ICM$, and so
$$\frac{EM}{MI}=\frac{MY}{MC}=\frac{x}{(a/2)}=\frac{2x}{a}=\frac{EN}{NI}.$$
A: This is very easy if you set a coordinate system so that $O=(0,0)$, $M= (0,1)$ and $N=(0,-1)$. Further $A= (-a,-1)$, $B=(a,-1)$, $C=(a,1)$ and $D=(-a,1)$ for some positive $a$. 

If we say $I = (a,c)$ then line $NI:\;\;y= {c+1\over a}x-1$ cuts diagonal $AC:\;\;y={x\over a}$ at $E = ({a\over c},{1\over c})$. Now, it should not be difficult to check that equality.
In fact, since the slope of $MI$ is ${c-1\over a}=k$ and the slope of $ME$ is $${{1\over c}-1\over {a\over c}} = {1-c\over a} = {1-c\over a}=-k$$ we see that $MN$ is angle bisector of angle $\angle EMI$ so we are done if we use angle bisector theorem. 
A: As you noted, we need to prove that $MN$ is the bisector of angle $\angle EMI$. Since $MD \perp MN$, this is equivalent to the statement that the pencil of lines $ME, MI, MN, MD$ is harmonic. Denoting $X=EN \cap DC$ we have
\begin{align*}
(ME,MI;MN,MD) & = (E,I;N,X) \\
& = (CE, CI; CN, CX) \\
& = (A, B; N, \infty_{AB}) \\
& = -1,
\end{align*}
so indeed, the pencil $ME, MI, MN, MD$ is harmonic.
A: Now that projective solution appear, I have another one with projective transfomation aproach. 
Transformation $\pi: E\mapsto I$ from line $AC$ to line $BC$ is perspective with respect to $N$. Now this transformation induces projective map of pencil of lines through $M$ to it self $\rho : ME\mapsto MI$. Now this one coincident wit reflection across $MN$ in three different situation, namely when: 
$\bullet$ $E=A$, then $ME=MA$ and $MI = MC$;
$\bullet$ $E=O$, then $ME=MN$ and $I = \infty$ so $MI = MN$;
$\bullet$ $E=C$, then $ME=BC$ and $I = C$ so $MI = BC$;
So $\rho $ maches with reflection across $MN$ and thus it is a reflection across $MN$ (if two projective transfomation match in three different points then they mathc in all points i.e. they are the same transformation).
Now we finish with angle bisector theorem.
