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
 A: Here's what seems to be an unnecessarily-complicated solution.

Define $b:=|AC|$, $c:=|AB|$, $d:=|AD|$, $p:=|BD|$, $q:=|CD|$. Let $r$ be the inradius of $\triangle ABC$, and let $s$ be the common inradius of $\triangle ABD$ and $\triangle ACD$.
We know $$\text{inradius}\cdot \text{perimeter} = 2\,\text{area}$$ so we can write
$$\begin{align}
s(c+d+p) &= 2|\triangle ABD|=\frac{p}{p+q}\cdot 2|\triangle ABC| = \frac{p}{p+q}\, r (b+c+p+q) \tag1\\[8pt]
s(b+d+q) &= 2|\triangle ACD|=\frac{q}{p+q}\cdot 2|\triangle ABC| = \frac{q}{p+q}\, r (b+c+p+q) \tag2
\end{align}$$
Solving this linear system for $b$ and $c$ gives
$$
b = -q-d + \frac{2 d q r}{(p + q)(r - s)} \qquad\qquad
c = -p-d + \frac{2 d p r}{(p + q)(r - s)} \tag3
$$
Since $\triangle ABC$ is right, we also know
$$\begin{align}
2r = |AC|+|AB|-|BC| &= b+c-(p+q) \\
&= 2\,\frac{ ds- (p+q)(r-s)}{r - s} \\
\to \qquad (p+q)(r-s) &= ds -r(r-s)\tag4
\end{align}$$
By Stewart's Theorem, we have
$$b^2p+c^2q=(p+q)(d^2+pq) \quad\underset{(3)}{\to}\quad 
 (p + q) s (r - s)= d r (2s-r) \tag5$$
Combining $(4)$ and $(5)$ to eliminate $p+q$ gives

$$r s (r - s) = d (r - s)^2 \quad\to\quad r s = d (r - s) \quad\to\quad
\frac1d=\frac1s-\frac1r \tag{$\star$}$$

Substituting $d=2$ gives the specific result for the question as stated. $\square$

There's almost-certainly a quicker path to the target relation. Note that
$$s(b+c+2d+p+q)=2|\triangle ABC| = r(b+c+p+q) \qquad\to\qquad \frac1s-\frac1r=\frac{d}{|\triangle ABC|}$$
So, really, "all we have to do" is show $d^2=|\triangle ABC|$. I'm not seeing a particularly good way to do that. Even so, looking at this as $2d^2=bc$ gives an easy way to construct an accurate figure in, say, GeoGebra, for further investigation.
A: 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$
A: 
Particular case, where triangle is isosceles:
As can be seen in figure a in this case $BC=4$ ; now you can find measures of AB and AC, then use this formula:
$r=\frac Sp$
(where S is area and p is half perimeter) to find $r$ and $r_1$.
$S_{ABC}=\frac{AD\times BC}2=\frac{2\times4}2=4$
$S_{ADB}=\frac 42=2$
$2p_{ABC}=4+4\sqrt 2$
$2p_{ABD}=4+2\sqrt 2$
So we have:
$$x=\frac1{\frac 2{2+\sqrt2}}-\frac 1{\frac 4{2+2\sqrt 2}}=\frac{2+\sqrt2}2-\frac{1+\sqrt 2}2=\frac 12$$
You may also use the fact that the centers or circles $O$ and $O_1$ are collinear and use another method.

Update: Figure C shows no- isosceles triangle and we have:
$\frac 1{0.569}-\frac 1{0,796}=0.5$
