Do reflections about a triangle fix a point?

Consider a triangle \(ABC\). We perform these transformations consecutively:
(i) Reflect the plane about line \(BC\).
(ii) Reflect the plane about line \(CA\).
(iii) Reflect the plane about line \(AB\).
The question is: Is there a point that returns to itself after these three reflections?

Let \(A'\) be the reflection of \(A\) about line \(BC\), and \(C'\) the reflection of \(C\) about line \(AB\). Let \(P\) be a point that maps to itself after all transformations. After these reflections, \(A' \curvearrowright A\) and \(C \curvearrowright C'\), where \(\curvearrowright\) means "maps to".

Reflection preserves angles and lengths as an isometry. Thus, \(\triangle PA'C \cong \triangle PAC'\). This implies \(P\) is the center of spiral similarity mapping \(A'C\) to \(AC'\). Such a point is unique, though proving this requires further exploration—try it yourself!

Now, observe that \(\triangle BA'C \cong \triangle BAC'\) because:
\(BA' = BA\), \(A'C = AC = AC'\), and \(BC = BC'\) as shown. Thus, \(B\) is the unique center of spiral similarity taking \(\overline{A'C}\) to \(\overline{AC'}\). So, \(B\) seems a candidate for \(P\). However, \(B\) moves after each reflection and doesn’t return to itself, regardless of the triangle’s shape.

Hence, no point \(P\) on the plane remains fixed. Share your thoughts on this!