# Is this geometric proof logically circular?

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.

• When you reach an equality like this, the question is not whether the argument is circular, but whether the steps are reversible so you can build a bridge from one statement to the other. Circularity would occur if you could, for example, only prove Pythagoras using the result you set out to establish - then you couldn't use Pythagoras in the proof. Commented May 18, 2020 at 21:10

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$$2r=\frac{(a+b)^2-a^2-b^2}{a+b+c}=\frac{(a+b)^2-c^2}{a+b+c}=a+b-c.$$

• Oh, I see, that's quite clever. I'm not sure if this is really my solution though since I didn't quite use this manipulation? Commented May 18, 2020 at 20:57
• @DavidDong I suppose we could more closely mirror your reasoning with $a+b-c=\frac{(a+b)^2-c^2}{a+b+c}=\frac{2ab}{a+b+c}=r$, or you could write that backwards, which is what I did.
– J.G.
Commented May 18, 2020 at 20:59

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