Is this geometric proof logically circular?

Question

Here is the problem:

Prove that in a triangle $ABC$ with $C$ as the right angle, where $a$ denote the side in front of angle $A$, $b$ denote the side in front of angle $B$, $c$ denote the side in front of angle $C$, the diameter of the incircle of $ABC$ equals to $a + b - c$.

Let $r$ be the radius of the incircle. Then, we know that $[ABC] = \frac{1}{2}r(a+b+c) = \frac{ab}{2}$, therefore $$r(a+b+c)=ab \Longleftrightarrow 2r = \frac{2ab}{a+b+c}.$$

The problem asks for a proof that $2r = a + b - c$. If this is true, then by what we found above, we have $$a + b - c = \frac{2ab}{a+b+c}$$ $$\Longleftrightarrow 2ab=(a+b-c)(a+b+c)$$ $$=a^2+ab+ac+ab+b^2+bc-ac-bc-c^2=2ab,$$

where in the last part, we know that $a^2+b^2=c^2$, so they cancel by their signs. Our equation thus becomes $2ab=2ab$, which is true.

The problem I'm having is that the statement $2ab=2ab$ can be derived from anything, not necessarily my solution. I am also assuming the problem statement, and even though it seems to be fine, I am unsure if it is circular or not.

Answer 1

It's not so much circular as written in a way that could confuse you into thinking so. With many proofs, if you present the reasoning in the exploratory way you'd get the result, it can be hard to tell which inferences are bidirectional. The way to fix this is to alter the presentation. In this case, we write$$2r=\frac{(a+b)^2-a^2-b^2}{a+b+c}=\frac{(a+b)^2-c^2}{a+b+c}=a+b-c.$$

Answer 2

Your solution is correct. You just have to rephrase the logic of your solution: Instead of:

The problem asks for a proof that 2r=a+b−c. If this is true, then by what we found above, we have

Say:

The problem asks for a proof that $2r=a+b−c$. By what we have shown above, the problem is equivelent to:...