Given triangle $ABC$, $O$ and $I$ are circumcenter and incenter, respectively, and $M$ is the midpoint of segment $BC$. Prove that $OI$ is perpendicular to $AM$ if and only if $$\dfrac{2}{BC}=\dfrac{1}{AB}+\dfrac{1}{AC}$$
Please give me the strategy with this geometry problem, as I quite don't know what to do with the circumcenter and incenter.
$\require{begingroup} \begingroup$ $\def\i{\mathbf{i}}\def\Re{\operatorname{Re}}$ Let $|BC|=a$, $|AC|=b$, $|AB|=c$, $|BM|=|CM|=\tfrac12\,a$.
Considering the vertices $A,B,C$ of triangle as complex numbers, let
\begin{align} B&=(0,0)=0 ,\quad C=(a,0)=a ,\quad A=(u,v)=u+v\cdot\i ,\quad M=\tfrac12\,(B+C)=\tfrac12\,a \tag{1}\label{1} . \end{align}
Then the coordinates of the incenter $I$ and the circumcenter $O$ can be found as
\begin{align} I&=\frac{a\,A+b\,B+c\,C}{a+b+c} = \frac{a(u+c)}{a+b+c}+\frac{av}{a+b+c}\,\i \tag{2}\label{2} ,\\ O&= \frac{a^2\,(b^2+c^2-a^2)\,A+b^2\,(a^2+c^2-b^2)\,B+c^2\,(b^2+a^2-c^2)\,C} {a^2\,(b^2+c^2-a^2)+b^2\,(a^2+c^2-b^2)+c^2\,(b^2+a^2-c^2)} \\ &= \frac{a(au(b^2+c^2-a^2)+c^2(a^2+b^2-c^2))}{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} +\frac{a^2v(b^2+c^2-a^2)}{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)} \,\i \tag{3}\label{3} . \end{align}
In case when $OI\perp AM$, we must have
\begin{align} \Re\left( \frac{O-I}{A-M} \right)&=0 \tag{4}\label{4} . \end{align}
Using \begin{align} v&=\sqrt{c^2-u^2} \tag{5}\label{5} ,\\ u &=c\cos\angle ABC =\frac{a^2-b^2+c^2}{2a} \tag{6}\label{6} , \end{align}
\begin{align} \Re\left( \frac{O-I}{A-M} \right)&= \frac{2bc-a(b+c)}{2b^2-a^2+2c^2} \tag{7}\label{7} , \end{align}
so the condition \eqref{4} holds (that is, $OI\perp AM$) when
\begin{align} a&=\frac{2bc}{b+c} \tag{8}\label{8} , \end{align}
as expected.
$\endgroup$