Find the measure of the relationship $\frac{1}{r_1} - \frac{1}{r} $in the figure below In In a right triangle $ABC$ ($A=90°$) with inradio $r$, cevian $AD$ is drawn in such a way that the inradium of $ABD$ and $ADC$ are equal to $r1$.If $AD=2$, calculate $\frac{1}{r1}-\frac{1}{r}$ (Answer:0,5).
My progress:

$\triangle CED \sim \triangle CAB  \\
\frac{CE}{AC}=\frac{DE}{AB}=\frac{CD}{BC}\\
\triangle BDL \sim \triangle BCA\\
\frac{DL}{AC}=\frac{BD}{BC}=\frac{LB}{AB}\\
CE = CI\\
BK = BL$
but I still haven't found the necessary relationship to finalize

In a right triangle, $~b + c = 2r + a$, where $a$ is hypotenuse. If $s$ represents sub-perimeter, then by law of cotangent,
$ \displaystyle \frac{s_{\triangle ABC} - b}{r} = \frac{s_{\triangle ABD} - 2}{r_1}$
$ \displaystyle \frac{s_{\triangle ABC} - c}{r} = \frac{s_{\triangle ACD} - 2}{r_1}$
Adding both,
$~\displaystyle \frac{a}{r} = \frac{s_{\triangle ABC} - 2}{r_1} = \frac{r + a - 2}{r_1}$
$ \implies a (r - r_1) = 2r - r^2 \tag1$
By equating area of $\triangle ABC$ to sum of areas of $\triangle ABD$ and $\triangle ACD$,
$r_1 (a + b + c + 4) = r( a + b + c)$
i.e. $~r_1 (r + a + 2) = r (r + a)$
$\implies a (r - r_1) = r_1 (r + 2) - r^2\tag2$
Equating $(1)$ and $(2)$ and simplifying we get $ \displaystyle \frac{1}{r_1} - \frac{1}{r} = \frac12$
