just a few computations show that: $|MI|=|MT|\leftrightarrow |MI|^{2}=|MT|^{2}\leftrightarrow r^{2}+\left( \frac{b-c}{2}\right)^{2}=R^{2}(1-\cos{A})^{2}\leftrightarrow (\cos{A}+\cos{B}+\cos{C}-1)^{2}=(1-\cos{A})^{2}+(\sin{B}-\sin{C})^{2}\leftrightarrow$ $(1-\cos{A})(\cos{B}+\cos{C}-1)=0$

Lemma (well-known or prove easily). Let $ABC$ be a triangle for which denote the incircle $C(I,r)$, the orthocenter $H$, the middlepoint $M$ of the side $[BC]$ and the intersection $S\in AH\cap MI$. Then there is the relation $AS=r\ .$ Proof of the proposed problem (in the case of an acute triangle). Are well-known the relations $OM=R\cos A$ and $\cos A+\cos B+\cos C=1+\frac{r}{R}\ \ \ (*)\ .$ Denote the circumcircle $C(O,R)$. Therefore, $MI=MT$ $\Longleftrightarrow$ $\widehat{SAO}\equiv\widehat{SMO}$, i.e. the quadrilateral $AOMS$ is parallelogram $\Longleftrightarrow$ $\boxed{\ OM=AS\ }$ $\Longleftrightarrow$ (using the first relation $*$ and the above lemma) $R\cos A=r$ $\Longleftrightarrow$ $\boxed{\ \cos A=\frac{r}{R}\ }$ $\Longleftrightarrow$ (using the second relation $*$) $\boxed{\ \cos B+\cos C=1\ }\ .$ Remark. $MI=MT\Longleftrightarrow OA\parallel IM\Longleftrightarrow \cos A=\frac{r}{R}$ $\Longleftrightarrow$ $\cos B+\cos C=1$ $\Longleftrightarrow$ ${\frac{a}{b+c}+\frac{r}{R}}=1\ .$

Omid Hatami wrote: $M$ is midpoint of side $BC$ of triangle $ABC$, and $I$ is incenter of triangle $ABC$, and $T$ is midpoint of arc $BC$, that does not contain $A$. Prove that \[\cos B+\cos C=1\Longleftrightarrow MI=MT\] Sorry I have a solution at http://www.artofproblemsolving.com/Forum/viewtopic.php?p=640832#640832 But it's in Spanish And as $IO\parallel BC$ the conclusion follows

This is the solution of a my student: We have $ \cos B + \cos C = 1$ $ \Longleftrightarrow \cos\angle BTY + \cos\angle CTY = 1$ $ \Longleftrightarrow \frac {TB^2 + TI^2 - IB^2}{2TB.TI} + \frac {TC^2 + TY^2 - IC^2}{2TI.TC}=1$ $ \Longleftrightarrow IB^2 + IC^2 = TB^2 + TC^2$ $ \Longleftrightarrow IM^2 = TM^2.\Box$ Note that $ TB = TI = TC$.

Here's a more intuitive approach: Let the projections of $B,C,M$ onto $IT$ be $B',C',M'$. So obviously $M'$ is the midpoint of $B'C'$. By well-known relation $TI=TB=TC$, $\angle BTI = C$, $\angle CTI = B$, we have $\cos B+\cos C=1 \Leftrightarrow B'T=C'I \Leftrightarrow M'$ midpoint of $B'C'$ and hence midpoint of $IT \Leftrightarrow MT=MI$