3. In triangle $ABC$, side $AB$ is the longest. On side $AB$, points $D$ and $E$ are chosen such that $\overline{AC}=\overline{AD}$ and $\overline{BE}=\overline{BC}$. Determine the angle $\measuredangle ACB$, if $\measuredangle ECD=20^{\circ}$.
Solution. Let $x=\measuredangle B C D$ and $y=\measuredangle A C E$. Since triangle $B C E$ is isosceles, $2\left(20^{\circ}+x\right)+\measuredangle E B C=180^{\circ}$, hence

![](https://cdn.mathpix.com/cropped/2024_06_05_6d6f67207af7682f8d1dg-4.jpg?height=409&width=1086&top_left_y=1783&top_left_x=375)

$$
2\left(20^{\circ}+x\right)=180^{\circ}-\measuredangle E B C
$$

Similarly, since triangle $A C D$ is isosceles,

$$
2\left(20^{\circ}+y\right)=180^{\circ}-\measuredangle C A D
$$

By adding these two equations, we get

$$
2\left(20^{\circ}+x+20^{\circ}+y\right)=180^{\circ}+\left(180^{\circ}-\measuredangle E B C-\measuredangle C A D\right)
$$

which simplifies to $2\left(\measuredangle A C B+20^{\circ}\right)=180^{\circ}+\measuredangle A C B$, from which it follows that $\measuredangle A C B=140^{\circ}$.