Fact. If $x$, $y$, $z$ are points on the complex plane, then the circumcenter of $\triangle xyz$ is the point
$$ i \frac{x(z\overline{z} - y\overline{y})+ y(x\overline{x}-z\overline{z})+z(y\overline{y}-x\overline{x})}{x(\overline{z} - \overline{y})+ y(\overline{x}-\overline{z})+z(\overline{y}-\overline{x})}$$
where "$\overline{x}$" indicates the complex conjugate of $x$.
Let us consider $\triangle abc$, defining $d$, $e$, $f$ on sides $bc$, $ca$, $ab$, respectively:
$$d := b + \alpha(c-b) \qquad e := c+\beta(a-c) \qquad f := a+\gamma(b-a)$$
for real scalars $\alpha, \beta, \gamma$.
Let $p$, $q$, $r$ be the respective circumcenters of $\triangle aef$, $\triangle dbf$, $\triangle dec$; for simplicity, we take $a$, $b$, $c$ on the unit circle so that $\overline{a}=1/a$, etc. Then,
$$\begin{align}
p &= i\;\frac{c(b-a)+ \beta b ( a-c) + \gamma c( a - b )}{b-c} \\[6pt]
q &= i\;\frac{a(c-b)+ \gamma c ( b-a) + \alpha a( b - c )}{c-a} \\[6pt]
r &= i\;\frac{b(a-c)+ \alpha a ( c-b) + \beta b( c - a )}{a-b}
\end{align}$$
We deduce
$$\frac{p-q}{a-b} = \frac{q-r}{b-c} = \frac{r-p}{c-a} =: u$$
which, by taking the modulus, proves $\triangle abc \sim \triangle pqr$. A little algebra reveals that
$$u + \overline{u} = 1$$
Now ... A point, $m$, (which may or may not be the Miquel point) is the center of spiral similarity from $\triangle abc$ and $\triangle pqr$, if and only if
$$\frac{a-m}{p-m} = \frac{b-m}{q-m} = \frac{c-m}{r-m}$$
(This characterization is what inspired me to approach the problem in the complex plane.) Solving for $m$ via the first equality gives
$$m = \frac{aq-bp}{(a-b)-(p-q)} = \frac{r - c u}{1-u}$$
whence
$$m-r = \frac{u(r-c)}{1-u} = \frac{u}{\overline{u}}(r-c) \qquad\qquad
\overline{m}-\overline{r} = \frac{\overline{u}}{u}(\overline{r}-\overline{c})$$
so that
$$|m-r|^2 = (m-r)(\overline{m}-\overline{r}) = (r-c)(\overline{r}-\overline{c}) = |c-r|^2$$
indicating that $c$ and $m$ are equidistant from $r$: thus, $m$ is on the circumcircle of $\triangle dec$. By symmetry, it is also on the circumcircles of $\triangle aef$ and $\triangle dbf$; the center of spiral similarity is in fact the Miquel point.
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
Best Answer
OK here's a bit of criticism, meant in the spirit of improvement. Reworking the steps of your attempted proof with the same numbering (plus some commentary):
(Your diagram is hard to read; the letters are very small relative to the diagram. I suspect you were sometimes misreading it yourself. The square angle mark is a useful device for avoiding too many numbers:)
$\triangle BDF$ is a right triangle with right angle at $D \quad$ - (introduce the main triangle)
$DE\;$ is the altitude from $D$ to $E$ on $BF$, perpendicular to $BF \quad$ - (key construction line)
$ $ - (I didn't really understand what you were trying to set up in point 2)
$\triangle BDE$ and $\triangle FDE $ both contain $90^\circ$ by construction; therefore, $\triangle BDE$ and $\triangle FDE$ are both right triangles.
$ $- (basically OK, could go into next point)
$\triangle BDE$ shares the angle at $B$ with $\triangle BDF$ and both are right triangles, therefore $\triangle BDE \sim \triangle BDF$. In the same way $\triangle DEF$ shares the angle at $F$ with $\triangle BDF$ so $\triangle DEF\sim \triangle BDF$. Thus also $\triangle BDE \sim \triangle DEF$.
$ $ - (You did not establish this similarity - it works via the main triangle, and the fact that a right triangle can be cut into two similar triangles is key)
Thus the ratio between sides of $\triangle DEF$ and the corresponding sides of both $\triangle BDE$ and $\triangle BDF$ are equal, giving:
$ $ - (No need to say "if $P$" for something you've established; "since $P$" works but it's implied anyway. You neglected to note that one of your ratios here was with the main triangle. Having said you're equating ratios, you should probably do that explicitly. You were using vector-style notation for scalar length.)
Note that $|BE|+|EF| = |BF| \quad$
$ $- (I might have said this back in point 2 when we split $BF$ )
Then: \begin{align}|DF|^2 &= |EF|\cdot \left(|BE|+|EF|\right) \\[1ex] |DF|^2 &= |EF|\cdot \left(\frac{|DE|^2}{|EF|}+|EF|\right) \\[1ex] |DF|^2 &= |DE|^2+|EF|^2 \end{align} as required since $\triangle DEF \sim \triangle BDF$.
$ $- (You can usually just let the equations speak for themselves but linking words here or there are OK. For more involved arguments, equation numbering is very helpful. Probably the link back to the main triangle through similarity should be stated here.)
Clean version with a variation from your point 5 to solve direct for the main triangle:
$\triangle BDF$ is a right triangle with right angle at $D$ - construct altitude $DE\;$ from $D$ to $E$ on $BF$; this is perpendicular to $BF$ and note that $|BE|+|EF| = |BF|$.
$\triangle BDE$ and $\triangle BDF$ share the angle at $B$ and both are right triangles, therefore $\triangle BDE \sim \triangle BDF$. In the same way $\triangle DEF$ shares the angle at $F$ with $\triangle BDF$ so $\triangle DEF\sim \triangle BDF$.
Thus the ratio between sides of $\triangle BDF$ and the corresponding sides of both $\triangle BDE$ and $\triangle DEF$ are equal, giving:
Then: \begin{align} |DB|^2 + |DF|^2 &= |BF|\cdot |BE| + |BF|\cdot |EF| \\ &= |BF|\cdot \left(|BE|+|EF|\right) \\ &= |BF|^2 \end{align} as required.