Line from incenter bisects side I got a problem recently, and have been unable to solve it.

Let $\Delta ABC$ with incenter $I$ and the incircle tangent to $BC$ at $D$. Let $M$ be the midpoint of $AD$. Prove that $MI$ bisects $BC$.

I tried to prove that the midpoint of $BC$, $M$, and $I$ are collinear, used Menelaus's theorem, used this particular lemma I'd read recently, and was unable to get anywhere.
Please help.

The lemma:
The incircle of $\Delta ABC$ is tangent to $BC, CA, AB$ at $D, E, F$ respectively. Let $M$ and $N$ be the midpoints of $BC$ and $AC$. If $K$ is the intersection of lines $BI$ and $EF$, then $BK \perp KC$.
 A: Let $E$ be the reflection of $D$ with respect to $I$. Note that the tangent line $\ell$ at $E$ to the incircle is parallel to $BC$. Consider a homothety centered at $A$ which transforms $\ell$ to $BC$. Then it maps the incircle to the $A$-excircle, and point $E$ is mapped to the point of tangency of the excircle with $BC$, say $F$. 
Since $A,E,F$ are collinear, $M$ is the midpoint of $AD$ and $I$ is the midpoint of $DE$, it follows that the midpoint of $DF$ lies on $MI$. Therefore you need to show that the midpoints of $DE$ and $BC$ coincide. In other words, the task is to prove that $BD=CE$. But this is well-known (and easy to show) that $BD = \frac{AB+BC-AC}{2} = CE$.
Let me know whether you need more details.
A: Can you find the proof with the help of the figure below？

A: In unnormalized barycentric coordinates,
$$I = (a, b, c), \\
D = (0, a + b - c, a - b + c), \\
M = (2a, a + b - c, a - b + c), \\
E = (0, 1, 1).$$
Then the problem reduces to verifying the identity
$$\begin{vmatrix}
 0 & 1 & 1 \\
 a & b & c \\
 2 a & a + b - c & a - b + c
\end{vmatrix} = 0.$$
A: 
Given $|BC|=a$, $|AC|=b$, $|AB|=c$
we can use known expressions to find that
\begin{align} 
I&=\frac{aA+bB+cC}{a+b+c}
,\\
D&=\tfrac12(B+C)+\frac{b-c}{2a}\cdot(B-C)
,\\
M&=\tfrac12(A+D)=\tfrac12 A+\tfrac14(B+C)+\frac{b-c}{4a}\cdot(B-C)
,\\
E&=\tfrac12(B+C)
.
\end{align}  
If $MI$ bisects $BC$, we must always have 
\begin{align} 
I&=E(1-t)+Mt
\tag{1}\label{1}
\end{align}  
for some real $t\in(0,1)$.
And indeed, \eqref{1} has one real solution
\begin{align} 
t&=\frac{2a}{a+b+c}
.
\end{align}
Since $a,b,c$ are the sides of $\triangle ABC$,
\begin{align} 
a&<b+c
,\\
2a&<a+b+c
,\\
\text{hence, }\quad
t\in(0,1)
.
\end{align}
