It is important to state the axiomatic framework the OP is coming from. It appears that they are working within the confines of 'Euclidean Plane Geometry' (a.k.a. high school plane geometry), and not the $\mathbb R \times \mathbb R$ Cartesian Coordinate Space.
The OP's proof is completely valid in that setting, and if carefully argued there is no circular reasoning.
The next section is just for fun.
Another title for the OP's question:
New Proof of Pythagorean Theorem (using the incenter of a triangle)?
(they can erase the picture of the circle).
The OP's proof doesn't rely on the concept of a circle or tangential distances.
A Theory of (tick-marked) Ray Lines could be postulated that describes the plane, and using the OP's logic, the simultaneous truth of the two equations
$$\tag 1 \frac{ar}{2}+\frac{br}{2}+\frac{cr}{2}=\frac{ab}{2}$$
$$\tag 2 (a−r)+(b−r)=c$$
can be argued.
You will find the two-dimensional Pythagorean Theorem to be true if you believe in the following:
$\quad \text{The Bisection of Two Rays Emanating from the Same Point}$
$\quad \text{The Perpendicular Distance From a Point to a Line}$
$\quad \text{The Area of a Rectangle}$
$\quad \text{Similar Triangles}$
$\quad \text{The Area of a Triangle}$
With that under your belt, you prove the following:
Theorem 1: Concurrency of Angle Bisectors of a Triangle.
In a triangle, the angle bisectors intersect at a point that is equidistant from the sides of the triangle; this point is called the incenter of the triangle.
If want to feel more comfortable before throwing out your Compass-and-Straightedge, please read
How to Bisect an Angle Using Only a Ruler
From the Cosine Law,
$$AB^2+AC^2-BC^2=2\cdot AB\cdot AC\cdot \cos A \implies AB = 5 \text{ or } 16$$
The area of $\triangle ABC$ can be written as the following two ways: $$S_{ABC}=\frac{1}{2}\cdot AB\cdot AC\cdot \sin A = \frac{1}{2}\cdot(AB+AC+BC)\cdot r$$
where $r$ is the radius of the incircle of $\triangle ABC$. This will give us $$r=\frac{AB\cdot AC\cdot \sin60^\circ}{AB+AC+BC}$$ By plugging in the values of $AB, AC, BC$ and $\sin A$, we can get $r$. Further, because $BP=BQ,\ CQ=CR,\ AR=AP$, and $$BP+AP=AB\\BQ+CQ=BC\\AR+RC=AC$$ we know that $$BQ = \frac{BA+BC-AC}{2}$$Therefore, the following two cases are:
Case 1: $AB=5$, then $r=\frac{7\sqrt{3}}{6}$ and $BQ=\frac{3}{2}$.
Case 2: $AB = 16$, then $r = 3\sqrt{3}$ and $BQ = 7$.
Best Answer
Your example fully answers the question as put.
For a primitive triple, suppose it has the standard representation $s^2-t^2,2st,s^2+t^2$ where $s$ and $t$ obey the usual conditions. Then the perimeter $p$ is $2st+2s^2$, and the area $A$ is $st(s^2-t^2)$.
If $r$ is the radius of the incircle, then $rp/2=A$, so $rp=2A$. For an incircle of radius $n$, we want $n(2st+2s^2)=2st(s^2-t^2)$. Cancellation simplifies this to $n=t(s-t)$.
The simplest way to achieve this is to let $s=n+1$ and $t=n$. That ensures $s$ and $t$ are relatively prime and of opposite parity. For $n$ odd we can let $t=1$ and $s=n+1$.
These are the only possibilities for $n=1$ or $n$ an odd prime. Other alternatives are available for $n$ with a more complicated multiplicative structure.