We compute the radius of $(IBI_{A}C)$ by the Incenter-Excenter Lemma by using the Law of Sines. $$\frac{DB}{sin\frac{A}{2}} = 2R \Leftrightarrow DB = 2R sin\frac{A}{2}.$$Here, $A \neq D = AI \cap (ABC)$. So, if $H$ is the orthocenter of $\Delta BCI_{A}$, we have $$\frac{I_{A}H}{I_{A}M} = \frac{4Rsin^{2}\frac{A}{2}}{8Rsin\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2}} = \frac{sin\frac{A}{2}}{2cos\frac{B}{2}cos\frac{C}{2}}.$$Also, $$\frac{I_{A}D}{I_{A}A} = \frac{2Rsin\frac{A}{2}}{4Rsin\frac{A}{2} + rcosec\frac{A}{2}} = \frac{sin\frac{A}{2}}{2(sin\frac{A}{2} + sin\frac{B}{2}sin\frac{C}{2})}.$$It can be shown that these two ratios are equal and the conclusion follows.

parmenides51 wrote: Let $ABC$ be a triangle, $I_a$ the center of the excircle at side $BC$, and $M$ its reflection across $BC$. Prove that $AM$ is parallel to the Euler line of the triangle $BCI_a$. This is much older than the source mentioned, as far as I can remember, but I don't remember a reference. For a solution, consider the inversion of power $I_aB \cdot I_aC$ and a reflection over the angle bisector of $\angle BI_aC$. Then the circumcenter $O$ of $I_aBC$ goes to $M$ and its orthocenter $H$ is mapped to $A$. Since $I_aO, I_aH$ are isogonal, $A$ lies on $OI_a$ and $M$ lies on $HI_a$. By the inversion, note that $I_aO \cdot I_aM = I_aH \cdot I_aA$, so by Thales' theorem, we are done.

Let $I_c$ and $I_b$ denote $C$ and $B$ excenters of $\Delta ABC$ Let $I, O$ and $H$ denote incenter of $\Delta ABC$, circumcenter and orthocenter of $\Delta BI_aC$ Let $D$ be foot of altitude from $I_a$ to $BC$ By duality of orthocenters and excenters $\Delta ABC$ is orthic triangle of $\Delta I_aI_bI_c$ and $I$ is orthocenter of $\Delta I_aI_bI_c$ $I$ and $H$ are corresponding points in similar triangles $\Delta BCI_a$ and $\Delta I_aI_bI_c$, and so are $A$ and $D$ $\implies \frac{I_aI}{I_aA}=\frac{I_aH}{I_aD}$ $BICI_a$ is cyclic with diameter $II_a$ $\implies I_aO=\frac{I_aI}{2}$ $\frac{I_aO}{I_aA}=\frac{I_aI}{2\cdot I_aA}=\frac{I_aH}{2\cdot I_aD}=\frac{I_aH}{I_aM}$ $\implies AM\parallel OH$ $OH$ is Euler line of $\Delta BCI_a$ so we are done

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