Line from incenter bisects side

Question

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.

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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 $BK \perp KC$.

Answer 1

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.

Answer 2

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 $$\begin{vmatrix} 0 & 1 & 1 \\ a & b & c \\ 2 a & a + b - c & a - b + c \end{vmatrix} = 0.$$

Answer 3

Can you find the proof with the help of the figure below？

Answer 4

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}