<s>[INST] What is the relationship between the length of the circumradius and the inradius in $ \triangle ABC $? For reference: In a right angle triangle ABC, an interior bisector BD is traced, where I is or incenter, $\measuredangle B = 90 ^ o$ and $3BI = 4ID$. Find the relationship between the circumraio and inraio lenght of $\triangle ABC$. (Answer:3)
My progress:
I made the drawing
Inradius = r
Circumradius = R
$r=\frac{a+c-b}{2}=\frac{ab}{a+b+c}\\
R = \frac{b}{2}\\
\frac{R}{r} = \frac{b}{a+c-b}$

 [/INST] Using the bisector’s property:
$$
\begin{align}
\frac{AB}{BC}&=\frac{AD}{DC}\\
\\
\frac{BC}{DC}&=\frac{BI}{ID}
\end{align}
$$
And $AD+DC=AC$ & $\frac{BI}{ID}=\frac{3}{4}$ can you get $\frac{3}{4}\left(AB+BC\right)=AC$?
</s>