# How do you derive that the inradius in a right triangle is $r=\frac{a+b-c}2$?

If we have a right triangle then the inradius is equal to $$r=\frac{a+b-c}2,$$ where $$c$$ is the hypothenuse and $$a$$ and $$b$$ are the legs.

This formula is mentioned in various places and it can be useful both in geometric problems and in problems on Pythagorean triples.1

Question: How can one derive this formula?

1It is stated on Wikipedia (in the current revision without a reference). Some posts on this site where this equation (or something closely related) is mentioned: If the radius of inscribed circle in a right triangle is $3 cm$ and the non-hypotenuse side is $14cm$, calculate triangle's area., Prove the inequality $R+r > \sqrt{2S}$, In a Right Angled Triangle., How do I find the radius of the circle which touches three sides of a right angled triangle?, Is there a way to see this geometrically?, Range of inradius of a right Triangle.

I will mention that I would be able to derive this myself in some way. (And some of the posts linked above in fact include something which basically leads to a proof.) Still, I think that it is to have somewhere this nice fact as a reference. And I wasn't able to find on this site a question specifically about this problem. I can post an answer myself - but I wanted to give others an opportunity to make a post first.

• One possible way: let $I$ be the incenter and suppose that incircle touches legs $AC$ and $BC$ at points $P$ and $Q$, respectively. Then, $PIQC$ is a square, so $r=PI=CP$. Commented Oct 19, 2021 at 10:48
• A sketch is outlined here artofproblemsolving.com/wiki/index.php/Inradius Commented Oct 19, 2021 at 10:59
• In the last example among your links, it is neatly proved (by Intelligenti pauca) (see the edit). Can you see that he says $c=a+b-2r$?
– ACB
Commented Oct 19, 2021 at 11:01
• @ACB Yes, I have explicitly mentioned in the question that some of the links give a proof. Still, I considered posting question like this useful. 1. It is about this specific formula. 2. The answers collect various proofs. 3. If somebody is searching for a proof of this formula on MSE, the title "Is there a way to see this geometrically?" won't be the first guess for a post where to find something like that. Commented Oct 19, 2021 at 11:40
• Another way: we know area of right triangle is $ab/2$ and we know $A = r \cdot s$ where is $s$ is sub-perimeter. So $r (a+b+c) = ab$. Multiplying both sides by $(a+b-c)$, we get $r \cdot (2ab) = ab (a+b-c)$ Commented Oct 19, 2021 at 11:42

Let the tangent points be $$A'$$, $$B'$$ and $$C'$$ labelled in the usual way. Then, since tangents from a point to a circle have equal length, $$CB'=r=CA'$$.

Therefore, for the same reason, $$AB'=AC'=b-r$$ and $$BA'=BC'=a-r$$ And since $$AC'+BC'=c$$, we get $$a-r+b-r=c$$ and hence the result.

• +1. One can readily show that, for any triangle, the tangent segments from $A$, $B$, $C$ always have lengths $a'=(-a+b+c)/2$, $b'=(a-b+c)/2$, $c'=(a+b-c)/2$, respectively. (Just solve the system $a=b'+c'$, $b=c'+a'$, $c=a'+b'$.) When there's a right angle, the inradius is necessarily equal to the corresponding tangent segment.
– Blue
Commented Oct 21, 2021 at 14:37
• After noticing that it is the same picture, I wondered whether the images was taken from Wolfram MathWorld article on Right Triangle. (Basically by accident I have noticed the same picture also in another post on this site.) Commented Oct 27, 2021 at 8:59
• @Martin Sleziak Probably - I just grabbed the first one I could find Commented Oct 27, 2021 at 9:16
• As mentioned in my comment below the question, the image in this post is very comprehensive.
– ACB
Commented Oct 27, 2021 at 14:04

$$\large \Delta=\color{green}\blacktriangle+\color{red}\blacktriangle+\color{blue}\blacktriangle$$ $$\frac{ab}2=\frac{ar}2+\frac{br}2+\frac{cr}2$$ $$ab=(a+b+c)r$$ $$ab(a+b-c)=(a+b+c)(a+b-c)r$$ $$ab(a+b-c)=[(a+b)^2-c^2]r$$ $$ab(a+b-c)=(\underbrace{a^2+b^2-c^2}_{0}+2ab)r$$ $$ab(a+b-c)=2ab\cdot r$$ $$r=\frac{a+b-c}2$$

Here's an approach that I don't think I've seen before (but don't doubt exists in the literature):

By the squares method, $$r^2+r(a-r)+r(b-r)=r\,\dfrac{a+b+c}2,$$ $$2(a+b-r)=a+b+c,$$ $$\color{green}{\mathbf{r=\dfrac{a+b-c}2.}}$$

Then $$R+r=\dfrac c2+r=\dfrac{a+b}2\le\sqrt{ab\mathstrut}\,=\sqrt{2S}\,.$$

\begin{align*}r&=\frac{\Delta}s\\&=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}} \\&=\sqrt{\frac{({a+b-c})({a-b+c})({-a+b+c})}{8s}} \\&=\sqrt{\frac{({a+b-c})[c^2-({a-b})^2]}{8s}} \\&=\sqrt{\frac{({a+b-c})(2ab)}{8s}} \\&=\sqrt{\frac{({a+b-c})}{2}.\frac{\Delta}{s}} \\&=\sqrt{\frac{({a+b-c})}{2}.r}\end{align*} $$\implies r=\frac{a+b-c}2,$$

Splitting the right-angled triangle into 3 triangles of bases $$a$$, $$b$$ and $$c$$ and the same height $$r$$（the in-radius）yields \begin{aligned} \frac{r a}{2}+\frac{r b}{2}+\frac{r c}{2} &=\frac{a b}{2} \\ r(a+b+c) &=a b \\ r &=\frac{a b}{a+b+c} \cdot \frac{a+b-c}{a+b-c} \\ &=\frac{a b(a+b-c)}{a^{2}+b^{2}+2 a b-c^{2}} \\ &=\frac{a+b-c}{2}\quad \left(\because c^{2}=a^{2}+b^{2}\right) \end{aligned}

We can consider the triangle formed by $$(a, 0), (0, b), (0, 0)$$ in the Cartesian Plane and proceed to find the point of intersection of the angular bisectors.

One angular bisector is clearly the line $$y = x$$.

For the other one, assume that the angle at (a, 0) to be $$\theta$$. The angular bisector is $$y = -\tan \frac{\theta}{2}x+ a\tan \frac{\theta}{2}$$.

The point of intersection these lines is $$\left(\frac{a\tan\frac{\theta}{2}}{1 + \tan\frac{\theta}{2}}, \frac{a\tan\frac{\theta}{2}}{1 + \tan\frac{\theta}{2}}\right)$$ which means that the inradius is just $$\frac{a\tan\frac{\theta}{2}}{1 + \tan\frac{\theta}{2}}$$. Starting from $$\tan\theta = \frac{b}{a}$$, it is easy to show that $$\tan\frac{\theta}{2} = \frac{c - a}{b}$$.

\begin{align} \frac{a\tan\frac{\theta}{2}}{1 + \tan\frac{\theta}{2}} &= \frac{a}{2}\tan\theta \left(1 - \tan \frac{\theta}{2}\right) \\ &= \frac{b}{2} \left(1 - \frac{c - a}{b}\right) \\ & = \frac{a+b-c}{2} \end{align}

\begin{align} r &=\frac{\Delta}{s} \\ &=\frac{\frac{ab}2}{\frac{a+b+c}2} \\ &=\frac{ab}{a+b+c} \\ &=\frac{\frac{(a+b)^2-c^2}{2}}{a+b+c}\\ &=\frac{(a+b-c)(a+b+c)}{2(a+b+c)}\\ \Rightarrow r=\frac{a+b-c}2 \end{align}

Labeling in Linked Wiki was incorrect. Vertex $$C$$ should have been more suitably labeled/placed at the vertex containing the right angle. Vertices A,B,C ordering should correctly be:

We have in-radius $$r$$

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

on simplification using the cut & pasted image above

also depicts a correct relation for in-radius $$r$$ as can be verified by algebraic cross-multiplication leading to Pythagoras thm of the right triangle:

$$(a+b)^2-c^2= 2ab \to c^2= a^2+b^2.$$