Let $M'$ be projection of $M$ on $AB$ $\implies$ $MM'$ $=$ $BM$ $\sin B$ $=$ $2BI \sin \frac{B}{2}$ $ =2r$

Let $E$ be the projection of $M$ on $AB$. Pick a point $F$ on $AB$ such that $IF \perp AB$ let $G$ be the intersection point of line $IF$ and the incircle, we have $\Delta BIM \sim \Delta BFI$, so \[\frac{IB}{IM}=\frac{FB}{FI}\implies \frac{IB}{FB}=\frac{IM}{FI}=\frac{IM}{IG}\]and also since $FG || EM$ and quadrilateral $BEIM$ is cyclic, we have $\angle MIG=\angle IME=\angle EBI=\angle FBI$. Thus, $\Delta BFI \sim \Delta MIG \implies \angle MGI = 90^{\circ}$ and quadrilateral $FGME$ is a rectangle so $EM=FG$ which is the diameter of the incircle.

Let IE be perpendicular to BC. Let MD be perpendicular to AB. MD = BM . sin ∠B and IE = IB.IM/BM so we have to prove 2IB.IM/BM = BM . sin ∠B. 2IB.IM/BM^2 = 2sin ∠B/2 . cos ∠B/2 = sin ∠B. we're Done.