<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 standard notations for side lengths,
Please note that in $\triangle ADB$, as $AI$ is the angle bisector,
$AD:c = 3:4 \implies AD = \frac{3c}{4}$
Similarly in $\triangle CBD, CD = \frac{3a}{4}$
So, $b = 2R = AD + CD = \frac{3}{4} (a + c)$
$a + c = \frac{8R}{3} \tag1$
Now use the fact that $\triangle ABC$ is a right triangle and hence,
$r = \frac{1}{2} (a + c - 2R)$
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