Let $I$ be the center of the circle inscribed in triangle $ABC$. Suppose that: $AB = AC + CI$. Determine the value of the ratio $\frac{\widehat{ACB}}{\widehat{ABC}}$.

![](https://cdn.mathpix.com/cropped/2024_05_10_8c44cd8b5c585c15e58bg-234.jpg?height=836&width=674&top_left_y=336&top_left_x=337)

Recall that the bisector of an angle divides it into two equal angles. Points on the bisector are equidistant from the two sides of the angle. It follows that the three bisectors of a triangle $ABC$ intersect at a point generally called $I$, which is equidistant from the three sides of the triangle. This is therefore the center of a circle tangent to the three sides of the triangle, called the "circle inscribed in triangle $ABC$".
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 $A B = A C + C I$, we need to construct a segment $C J$ on the line $(A C)$ such that: $C J = C I$ and $A J = A C + C I$ (so $A$ and $J$ are on opposite sides of $C$). The relation $C J = C I$ implies that the triangle $C J I$ is isosceles, so $\widehat{C J I} = \widehat{C I J}$: let's call this angle $\alpha$. $\widehat{I C A} = \widehat{C J I} + \widehat{C I J} = 2 \alpha$. Since $\widehat{I C A} = \frac{\widehat{B C A}}{2}$ because $(C I)$ is the bisector of $\widehat{B C A}$. Furthermore, the triangles $I J A$ and $I B A$ have an equal angle: $\widehat{I A J} = \widehat{I A B}$ situated between two equal sides, $A I$ and $A J = A B$, so they are congruent (they have all their sides and all their angles equal), hence in particular: $\alpha = \widehat{I J A} = \widehat{I B A} = \widehat{\frac{A B C}{2}}$. Therefore, ultimately: $\widehat{A B C} = 2 \alpha$ while $\widehat{B C A} = 4 \alpha$, hence $\widehat{\widehat{A C A}} = 2$.

![](https://cdn.mathpix.com/cropped/2024_05_10_8c44cd8b5c585c15e58bg-235.jpg?height=845&width=786&top_left_y=383&top_left_x=241)

## Inscribed Angles

If $B$ and $C$ are two fixed points on a circle, and we vary a third point $A$ on the same circle, the inscribed angle $\widehat{B A C}$ does not depend on the position of point $A$.

There are several ways to state this theorem: if four points $A, B, C, D$ are on the same circle, the angles $\widehat{B A C}$ and $\widehat{B D C}$ are equal if $A$ and $D$ are on the same side of the line $(B C)$, and supplementary if they are on opposite sides of $(B C)$. Conversely, any four points in the plane, $A, B, C, D$, satisfying: $\widehat{B A C} = \widehat{B D C}$ if $A$ and $D$ are on the same side of $(B C)$ or $\widehat{B A C} + \widehat{B D C} = 180^{\circ}$ if $A$ and $D$ are on opposite sides of $(B C)$ are "cyclic," meaning they lie on the same circle. The proof must consider all cases, but the key idea is that if $A$ and $B$ are on a circle with center $O$, the triangle $A O B$ is isosceles. If the line $(A O)$ intersects the circle again at $A'$, since the sum of the three angles of the triangle $A O B$ is $180^{\circ}$, $\widehat{B O A'} = \widehat{B A O} + \widehat{A B O} = 2 \cdot \widehat{B A O}$, from which we deduce that the central angle $\widehat{B O C}$, which does not depend on $A$, is twice the inscribed angle $\widehat{B A C}$.

![](https://cdn.mathpix.com/cropped/2024_05_10_8c44cd8b5c585c15e58bg-236.jpg?height=1002&width=994&top_left_y=361&top_left_x=354)

If we want a theorem that does not depend on the cases, we need to introduce angles of lines: the angle $(A B, A C)$ is the oriented angle through which the line $(A B)$ must be rotated to coincide with $(A C)$. Therefore, $(A B, A C) = -(A C, A B)$ and more generally: $(A B, A C) + (A C, A D) = (A B, A D)$ (Chasles' relation) for any points $A, B, C$, and $D$. Using these angles of lines, the inscribed angle theorem can be written: four points $A, B, C, D$ are on the same circle if and only if $(A B, A C) = (D B, D C)$.

An important special case of this theorem: the angle $\widehat{B A C}$ is a right angle if and only if $A$ is located on the circle with diameter $[B C]$ (so the central angle is a straight angle).