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

Question

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?

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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.

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^{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.}

Answer 1

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.

Answer 2

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 $\tan\frac{\theta}{2} = \frac{c - a}{b}$.

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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:

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

Answer 7

\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}

Answer 8

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

Answer 9

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