Let $ABC$ be a triangle, $I$ the center of the inscribed circle. Suppose that: $CA + AI = BC$. Determine the ratio of the angles: $\frac{\widehat{BAC}}{\widehat{CAA}}$
![](https://cdn.mathpix.com/cropped/2024_05_10_732985225e470d5c7305g-019.jpg?height=455&width=564&top_left_y=1667&top_left_x=756)

To take advantage of the relationship of the hypothesis with a sum of lengths, the three points must be collinear. We will therefore consider a point \( J \) on \( (CA) \) such that \( AI = AJ \), with \( A \) between \( C \) and \( J \), so that \( CJ = CB \). Let's call \( \alpha \) the angle \( \widehat{AJI} \). Since triangle \( AJI \) is isosceles, \( \widehat{AIJ} = \alpha \), so \( \widehat{IAC} = 2\alpha \), which implies \( \widehat{BAC} = 4\alpha \), since \( (AI) \) is the bisector of \( \widehat{BAC} \). Moreover, \( CJ = CA + AJ = CA + AI = BC \), so triangles \( JCI \) and \( BCI \) are congruent: two equal sides and the angles between these two sides are equal, since \( (CI) \) is the bisector of \( \widehat{BCA} \). Therefore, \( \widehat{IBC} = \widehat{IJC} = \alpha \), and since \( (BI) \) is the bisector of \( \widehat{CBA} \), \( \widehat{CBA} = 2\alpha \), which implies \( \frac{\widehat{BAC}}{\widehat{CBA}} = 2 \).

## - Inscribed Angle -

The concept of an inscribed angle is a particularly important case of using angles. Let \( A, B, C \) be three points on a circle with center \( O \). The inscribed angle \( \widehat{BAC} \), which intercepts the arc \( BC \), is half the measure of the central angle \( \widehat{BOC} \). Indeed, extend \( (AO) \), which intersects the circle at \( D \). In triangle \( AOC \), the exterior angle \( \widehat{DOC} \) is equal to the sum of the base angles, \( \widehat{OAC} + \widehat{ACO} \), but these angles are equal because triangle \( AOC \) is isosceles. Therefore, \( \widehat{DOC} = 2 \times \widehat{OAC} \), and similarly, \( \widehat{BOD} = 2 \times \widehat{BAD} \). Depending on the configuration, the central angle \( \widehat{BOC} \) is the sum or the difference of the central angles \( \widehat{DOC} \) and \( \widehat{BOD} \), but in both cases, it is twice the measure of the inscribed angle \( \widehat{BAC} \).
![](https://cdn.mathpix.com/cropped/2024_05_10_732985225e470d5c7305g-020.jpg?height=676&width=1590&top_left_y=1022&top_left_x=232)

It follows essentially that if points \( B \) and \( C \) are fixed, the inscribed angle \( \widehat{BAC} \) will always have the same measure, regardless of the position of point \( A \). Provided that when \( A \) crosses the chord \( BC \), the inscribed angle is transformed into its supplementary angle, and the central angle becomes the reflex angle equal to twice the supplementary angle. This equality of inscribed angles is used in both directions: on the one hand, if \( A, B, C, D \) are on the same circle, with \( A \) and \( D \) on the same side of \( (BC) \), the angles \( \widehat{BAC} \) and \( \widehat{BDC} \) are equal; on the other hand, conversely, if \( A, B, C, D \) are four points (with \( A \) and \( D \) on the same side of \( (BC) \)) such that \( \widehat{BAC} = \widehat{BDC} \), then the four points \( A, B, C, D \) are concyclic, i.e., on the same circle. It is common to use the theorem in both directions, on the one hand to prove that four points are concyclic, and on the other hand to conclude that other angles, not initially known, are equal.