What is the relationship between the length of the circumradius and the inradius in $ \triangle ABC $?

Question

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

$r=\frac{a+c-b}{2}=\frac{ab}{a+b+c}\\ R = \frac{b}{2}\\ \frac{R}{r} = \frac{b}{a+c-b}$

Answer 1

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,
$r = \frac{1}{2} (a + c - 2R)$

Answer 2

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 $\frac{3}{4}\left(AB+BC\right)=AC$?

Answer 3

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