How is it proved that 3 inradii multiplied by 3 inradii of similar triangles are equal?

Question

I have an SAT Prep question that has really stumped me. You are given triangle ABC, altitudes AD, BE, and CF. You are also given H as the orthocenter and that a, b, c, d, e, and f are the in radii of triangles AFH, BDH, CEH, CDH, AEH, BFH respectively. The task is to prove that abc = def. I have so far proven that there are three sets of similar triangles proven by AA (right angles from the altitudes and vertical angles at the orthocenter, H). I know that since they're similar, in each f the sets the inradii are similar in the same ratio that the sides of the triangles are, but, since I don't know any side lengths, I don't have any idea how to prove that abc = def.

P.S. I'm brand new to Stack Exchange so please be gentle. I don't really know how this works

Answer 1

Notice that triangle $AEH$ is similar to triangle $BDH$ with a side ratio of $\frac{AE}{BD}$, therefore $\frac{e}{b}=\frac{AE}{BD}$.

Doing the same argument yields:

$\frac{edf}{abc}=\frac{AE\cdot DC \cdot BF}{BD\cdot AF\cdot EC}$. This equal to $1$ by Ceva's theorem.

Answer 2

I think you have, from similar triangles:

$\dfrac{e}{b} = \dfrac{AH}{BH}$

$\dfrac{d}{a} = \dfrac{CH}{AH}$

$\dfrac{f}{c} = \dfrac{BH}{CH}$

$\dfrac{def}{abc} = \dfrac{AH\cdot BH\cdot CH}{AH\cdot BH\cdot CH} = 1$