Find $k$ in $BC : CD = 2 : k$
$AB:AC=3:2$
$\vec a = OA, \vec b = OB, \vec c = OC$
What i know $OA$ perpendicular to $AD; BD = (2+k)/2 BC; OA=OB=OC$. But geometry is just my weak point, how to find $k$?
Edit : Radii = 1 As shown in the figure, let $D$ be a point on the extension of side $BC$ and $k$ be the number where
$BC:CD=2:k$.
Moreover, set
$\vec{OA}=\vec{a},$ $\vec{OB}=\vec{b},$ $\vec{OC}=\vec{c},$
Since the equality
$|\vec{b}-\vec{a}|=\frac{3}{2}|\vec{c}-\vec{a}|$
Let $|AB|=c=3y$, $|AC|=b=2y$, $|BC|=a=2x$, $|CD|=d=kx$, $|AD|=e$.
By the power of the point $D$, \begin{align} |AD|^2=e^2&=|CD|\cdot|BD| ,\\ e^2&=k\,(k+2)\,x^2 . \end{align}
By the cosine rule
\begin{align} \triangle ABC:\quad \phantom{-}\cos\gamma&=\frac{a^2+b^2-c^2}{2ab} =\frac{4x^2-5y^2}{8xy} ,\\ \triangle ACD:\quad -\cos\gamma&=\frac{d^2+b^2-e^2}{2db} =\frac{2\,y^2-k\,x^2}{2kxy} ,\\ -\frac{y(5k-8)}{8kx}&=0 ,\\ k&=\frac85 . \end{align}
And in general case,
\begin{align} \frac da&=\frac{b^2}{c^2-b^2} . \end{align}