# Help on whether a geometry solution is valid.

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

• That's the whole point. You have to clearly show that for any rotation the mentioned line segments have equal lengths. I don't think this simplifies the problem and you should give a proof for an arbitrary configuration of the problem anyways. – S.B. Aug 8 '13 at 4:25

Through a series of transformations, you can reasonably assume that C, A, C’ are collinear.

But then, since $<BCA = < B’C’A’ = 90$ degrees, together with BD perpendicular to $l$ and B’D’ perpendicular to $l$ (= CAC’A’), the result will be $CD = C'D' = 0$ (as pointed out).

This is only true for a particular case when $<BCA = < B’C’A’ = 90$ degrees.

Besides making C, A, C’ collinear, one can even go further by letting A = A’ with BC // B'C'.

What if the case becomes more general? That is, $<BCA$ and $<B’C’A’$ are not 90 degrees.

If we draw triangles of the form A= A’, BC // B’C’, one can immediate tell CD is not equal to C’D’.