3. Given a convex quadrilateral $A B C D$ such that

$$
\measuredangle A B D=50^{\circ}, \measuredangle A D B=80^{\circ}, \measuredangle A C B=40^{\circ}, \measuredangle D B C=\measuredangle B D C+30^{\circ}
$$

Calculate $\Varangle D B C$.
Solution. Since in triangle $ABD$ we have $\measuredangle ABD=50^{\circ}$ and $\measuredangle ADB=80^{\circ}$, we get $\measuredangle DAB=50^{\circ}$. Therefore, $DA=DB$ (see the diagram on the right). Now, since

$$
\measuredangle ADB=80^{\circ}=2 \measuredangle BCA
$$

and points $C$ and $D$ are on the same side of line $AB$, we conclude that point $C$ lies on the circle with center $D$ and radius $DA=DB$. Therefore, triangle $BCD$ is isosceles, and from

$$
\measuredangle BDC + 2 \measuredangle DBC = 180^{\circ} \text{ and } \Varangle DBC = \Varangle BDC + 30^{\circ}
$$

it follows that $\measuredangle DBC=70^{\circ}$.