Exercise 3. Let $A B C$ be an isosceles triangle such that $\widehat{A B C}=60^{\circ}$; it is not known at which vertex $A B C$ is isosceles. Find all possible values of $\widehat{A C B}$.
Solution to Exercise 3 We will demonstrate that $\widehat{\mathrm{ACB}}$ can be $60^{\circ}$, and no other value.

First, if $A B C$ is equilateral, then it is indeed isosceles, and we have $\widehat{A B C}=60^{\circ}$, so $A B C$ meets the conditions of the problem, and we observe that $\widehat{\mathrm{ACB}}=60^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_29_73269b600131539d27eeg-2.jpg?height=385&width=350&top_left_y=1580&top_left_x=882)

Conversely, we will show that $A B C$ is equilateral, meaning that all three angles, including $\widehat{A C B}$ itself, are $60^{\circ}$. Indeed, let $x, y$, and $z$ be the three angles of $A B C$. The problem statement indicates that two of the angles are equal, and one of the angles is $60^{\circ}$. Without loss of generality, we assume that $x=y$, and that $x=60^{\circ}$ or $z=60^{\circ}$:

$\triangleright$ if $x=60^{\circ}$, then $y=60^{\circ}$ and $z=180^{\circ}-x-y=60^{\circ}$;

$\triangleright$ if $z=60^{\circ}$, then $2 x=x+y=180^{\circ}-z=120^{\circ}$, so $x=y=60^{\circ}$,

which concludes the proof.