Sharygin I.F.

The angle at the vertex $A$ of the isosceles triangle $ABC (AB = AC)$ is $20^{\circ}$. On the side $AB$, we lay off the segment $AD$, equal to $BC$. Find the angle $BCD$.
We will sequentially construct points $E, F$, and $G$ on sides $AB$ and $AC$ such that $BC = CE = EF = FG$ (see the left figure).

This will result in isosceles triangles $BCE$ (with an angle of $20^{\circ}$ at vertex $E$), $CEF$ (with an angle of $60^{\circ}$ at the base $CF$), $EFG$ (with an angle of $100^{\circ}$ at vertex $F$), and $FGA$ (with angles of $20^{\circ}$ at the base $FA$).

Since $AG = AD$, point $G$ coincides with point $D$. Because triangle $CEF$ is equilateral, triangle $CFG$ is isosceles with an angle of $10^{\circ}$ at the base, and $\angle BCD = \angle BCG = 80^{\circ} - 10^{\circ} = 70^{\circ}$.
![](https://cdn.mathpix.com/cropped/2024_05_06_2f46c6bf14fbb7f478d4g-49.jpg?height=836&width=1372&top_left_y=572&top_left_x=389)

Answer

$70^{\circ}$.