<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] T.Poncelet:$BA+BC = AC+2r\\
R = \frac{AC}{2}\\
\frac{BA}{BI}=\frac{DA}{DI}\\
\frac{BC}{BI}=\frac{DC}{DI}\\
\therefore BA+BC = AC+2r\\
\frac{DA \cdot BI}{DI}+\frac{DC \cdot BI}{DI} = 2R+2r\\
\frac{4}{3}\cdot (DA+DC) = 2R+2r\\
\frac{4}{3}\cdot (AC) = 2R+2r\\
\frac{4}{3}\cdot (2R) = 2R+2r\\
\boxed{R =3r}$
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