Proof that Isometry f Preserves Midpoints | Geometric Reflection Counterexample

[pivoxa15]

Homework Statement

Suppose f is an isometry that fixes O (origin). Prove f preserves midpoints of line segments.

The Attempt at a Solution

Geometricallly, f could be a reflection in which case it would not preserve the mid point of any line segment that does not intersect the origin anywhere.

So I don't see a proof at all and infact sees a mistake.

 

[psi^]

But in the case of a reflection the transformation of a midpoint is still a midpoint, no?

 

[pivoxa15]

That is true. I was thinking along the wrong lines (no pun intended) in that I was thinking that f maps midpoint to the exact same mid point.

Everything makes geometric sense. The only problem is to prove it algebraically. Can't see how to do it.

 

[Mathdope]

Are we in R^n?

 

[psi^]

If f is a isometry, u.v=f(u).f(v) holds. So the vector norm (u.u)^1/2 and distance stays the same.

 

[pivoxa15]

I have worked out the quesion in the OP. I now need to show that f(ru)=rf(u) with the same conditions given in the OP.