Is the $(3,4,5)$ triangle the only rectangular triangle with this property? While solving a loosely related exercise, by luck I found out that the $(3,4,5)$ triangle has the following property:
The product of the lengths ($\sqrt{2}$ and $\sqrt{5}$) of the two shorter line segments from a corner to the center of the inscribed circle equals the length ($\sqrt{10}$) of the longest one.
Somewhat satisfied with this, for me own new found, result I now wonder if any other rectangular $(a,b,c)$ triangles have this particular property. $(a,b,c)$ does not need to be a Pythagorean triple (but it would be extra nice).
It tried some straightforward algebraic equations but failed to find answer ... Maybe finding non rectangular such triangles is easier, but ideally I ask for rectangular ones.

update
Can this property of certain pythagorean triples in relation to their inner circle be generalized for other values of $n$?
is already linked to this one but I take the freedom to explicitly mention it here in post for following reasons

*

*the question asked there is about generalizing answer given here

*the answers to both questions always left some exercises for reader

*myself I am not able (I continue to try) to do these exercises

Maybe someone can fully write out the missing gaps.

In a Pythagorean right triangle $\triangle ABC$, we know that $a^2 + b^2 = c^2$ where $a, b, c$ are positive integers.  We also know that $|\triangle ABC| = rs$, where $r$ is the inradius and $s = (a+b+c)/2$ is the semiperimeter.  Thus we have $$\begin{align}
r &= \frac{ab}{a+b+c} \\
&= \frac{ab}{a+b+\sqrt{a^2+b^2}} \\
&= \frac{ab(a+b-\sqrt{a^2+b^2})}{(a+b+\sqrt{a^2+b^2})(a+b-\sqrt{a^2+b^2})} \\
&= \frac{ab(a+b-\sqrt{a^2+b^2})}{2ab} \\
&= \frac{1}{2}(a+b-c) \\
&= s-c.
\end{align}$$  Denoting $I$ as the incenter, the respective distances from the incenter to the vertices are $$IA = \sqrt{r^2 + (s-a)^2}, \\ IB = \sqrt{r^2 + (s-b)^2}, \\ IC = \sqrt{r^2 + (s-c)^2} = r \sqrt{2}.$$
Then assuming $a < b < c$, we require $IB \cdot IC = IA$, or $$\begin{align}
0 = IB^2 \cdot IC^2 - IA^2 = \left(r^2 + (s-b)^2\right)(2r^2) - \left(r^2 + (s-a)^2\right).
\end{align}$$  I leave it as an exercise to show that this condition is nontrivially satisfied if and only if $b = (a^2-1)/2$, hence $a$ must be an odd positive integer for $b$ to be an integer.  Then $c$ will automatically be an integer since $$c^2 = a^2 + b^2 = \left(\frac{a^2+1}{2}\right)^2.$$  Therefore, the solution set is parametrized by the triple
$$(a,b,c) = \bigl(2r+1, 2r(r+1), 2r(r+1)+1\bigr), \quad r \in \mathbb Z^+,$$  where $r$ is the inradius of such a triangle.  In particular, this leads to the triples
$$(3,4,5), \\ (5,12,13), \\ (7,24,25), \\ (9,40,41), \\ \ldots.$$
