Let $ABC$ and $AB'C'$ be similar right angled triangles with right angles at $C$ and $C'$, respectively. Let $l$ be the line between $C$ and $C'$, and let $D$ and $D'$ be the points on $l$ such that $BD$ and $B'D'$ are perpendicular to $l$.
Prove that $CD=C'D'$.
You can solve this in a pretty easy way just through a quick construction and similar triangle ratios. But it made me ask myself if you could argue the problem like this.
Supposing $C,A,C'$ are col-linear, then there exists a homothety centered at $A$ that maps one triangle to the other. And $CD=C'D'=0$ for any ratio between the two triangles. So can you say that upon rotation this relationship will always be the same thus finishing the problem. (Obviously I'd have to quote the properties of something or other to do this in a proof - if it works at all).
Thanks for any help