3. In triangle $ABC$, side $AB$ is the longest. Points $D$ and $E$ are taken on side $AB$ such that $AD = AC$ and $BE = BC$. Determine $\measuredangle ACB$ if $\measuredangle ECD = 20^{\circ}$.
Solution. Let $\measuredangle B C D=x$ and $\measuredangle A C E=y$ (make a diagram). From the isosceles triangle $B C E$ we get $2\left(20^{\circ}+x\right)=180^{\circ}-\measuredangle E B C$, and from the isosceles triangle $A C D$ we have $2\left(20^{\circ}+y\right)=180^{\circ}-\measuredangle C D A$. By adding the last two equations, we find

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

which simplifies to

$$
2\left(\measuredangle A C B+20^{\circ}\right)=180^{\circ}+\measuredangle A C B \text {, i.e., } \angle A C B=140^{\circ} \text {. }
$$