(JBMO 2002 - Shorlist, Problem G3), treated in class

Let $A B C$ be an isosceles triangle at $A$, with $\widehat{C A B}=20^{\circ}$. Let $D$ be a point on the segment $[A C]$, such that $A D=B C$. Calculate the angle $\widehat{B D C}$.
Let $E$ be the point such that triangle $E D A$ is directly isometric to triangle $A B C$. Then, in oriented angles, $\widehat{E A B}=\widehat{E A D}-\widehat{B A D}=\widehat{A C B}-\widehat{B A C}=80^{\circ}-$ $20^{\circ}=60^{\circ}$. Since $A B=A C=A E$, triangle $A B C$ is therefore equilateral. We also have $\widehat{B E D}=\widehat{B E A}-\widehat{D E A}=60^{\circ}-\widehat{B A C}=60^{\circ}-20^{\circ}=40^{\circ}$. Furthermore, since $A B C$ is equilateral, we have $B E=A E=D E$, from which we conclude that $\widehat{E D B}=\widehat{D B E}=\frac{1}{2}\left(180^{\circ}-\widehat{B E D}\right)=70^{\circ}$.

It follows that $\widehat{B C D}=180^{\circ}-\widehat{A D E}-\widehat{E D B}=180^{\circ}-\widehat{C B A}-70^{\circ}=180^{\circ}-80^{\circ}-$ $70^{\circ}=30^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_10_831e903be1b8a1180f82g-61.jpg?height=526&width=572&top_left_y=2128&top_left_x=663)