Let $I$ be the center of the circle inscribed in triangle $A B C$. Suppose that: $C A+A I=$ $B C$. Determine the value of the ratio $\frac{\widehat{B A C}}{\widehat{C B A}}$
When two segments are not end-to-end on the same line, we cannot say much about the sum of their lengths. Therefore, to use the hypothesis \( CA + AI = BC \), we need to construct a segment \( AJ \) on the line \((CA)\) such that: \( AJ = AI \) and \( CJ = CA + AJ \) (so \( C \) and \( J \) are on opposite sides of \( A \)). The relation \( AJ = AI \) implies that triangle \( AJI \) is isosceles, so \(\widehat{AJI} = \widehat{AIJ}\): let's call this angle \(\alpha\). \(\widehat{IAC} = \widehat{AJI} + \widehat{AIJ} = 2\alpha\). Since \((AI)\) is the bisector of \(\widehat{BAC}\), \(\widehat{IAC} = \frac{\widehat{BAC}}{2}\). Furthermore, triangles \(IJC\) and \(IBC\) have one equal angle: \(\widehat{ICJ} = \widehat{ICB}\) located between two equal sides, \(CI\) and \(CJ = CB\), so they are congruent (they have all their sides and all their angles equal), hence in particular: \(\alpha = \widehat{IJC} = \widehat{IBC} = \frac{\widehat{CBA}}{2}\). Therefore, ultimately: \(\widehat{CBA} = 2\alpha\) while \(\widehat{BAC} = 4\alpha\), so \(\widehat{CAC} = 2\).