Problem
GEO-B3-M01-P002 Similarity of Inverse Triangles
Under an inversion centered at \(O\), points \(A\) and \(B\) map to \(A^*\) and \(B^*\). Prove that \(\angle OAB=\angle OB^*A^*\).
Hint 1. Compare the ratios \(OA:OB\) and \(OB^*:OA^*\).
Hint 2. Prove that \(\triangle OAB\) and \(\triangle OB^*A^*\) are similar.
E. Full solution. Let the radius of inversion be \(R\). Then \(OA\cdot OA^*=OB\cdot OB^*=R^2\). Hence \(\frac{OA}{OB}=\frac{OB^*}{OA^*}\). Also rays \(OA\) and \(OA^*\) coincide, and rays \(OB\) and \(OB^*\) coincide, so \(\angle AOB=\angle B^*OA^*\). By SAS similarity, \(\triangle OAB\sim\triangle OB^*A^*\). Therefore the corresponding angles are equal: \(\angle OAB=\angle OB^*A^*\).
This is a small but important lemma: it explains why inversion preserves angles and why the reversed order of vertices keeps appearing.