Why are the inradii of similar triangles in the ratio of the corresponding sides of the triangles?
I understand why the same applies to circumradii, medians, altitudes and angle bisectors. But I have no idea how to prove this for inradii of two similar triangles.
Edit: I used SAS similarity of triangles to prove that the medians and circumradii of two similar triangles would be in the ratio of corresponding sides of the triangles. Similarly, I used AA similarity for altitudes and AA similarity for angle bisectors. I would like to know if there's a similar way to prove for the inradii of similar triangles.
Best Answer
Let $s$ be the semiperimeter of a triangle with the sides $a, b$ and $c$:
$$ s=\frac{a+b+c}{2} $$
Then, by Heron's formula, the area $A$ of the triangle is
$$ A=\sqrt{s(s-a)(s-b)(s-c)} $$
The length of the inradius $r$ is
$$ r=\frac{A}{s} $$
Not lets create a similar triangle with the stretching factor $f$, so that its sides are $a'=af;\,b'=bf;\,c'=cf$. The semiperimeter $s'$ of the similar triangle is
$$ s'=\frac{af+bf+cf}{2}\\ s'=f\frac{a+b+c}{2}\\ s'=fs $$
And here is the area $A'$:
$$ A'=\sqrt{sf(sf-af)(sf-bf)(sf-cf)}\\ A'=\sqrt{sf^4(s-a)(s-b)(s-c)}\\ A'=f^2\sqrt{s(s-a)(s-b)(s-c)}\\ A'=f^2A $$
The length of the inradius $r'$ of the similar triangel is
$$ r'=\frac{A'}{s'}=\frac{f^2A}{fs}=f\frac{A}{s}\\ r'=fr $$
So, when you stretch all sides of the triangle by the same factor, then also the inradius is stretched by the same factor.