In the quadrilateral $\mathrm{ABCD}$, the following angles are known:

$B A C \angle=50^{\circ}, A B D \angle=60^{\circ}, D B C \angle=20^{\circ}, B D C \angle=30^{\circ}$. Is it true that the angle $C A D$ is $30^{\circ}$?
Let's construct an equilateral triangle over the segment $AB$, and denote the vertex on the $BD$ segment as $G$. Draw a line through $D$ parallel to $AB$, intersecting $AG$ at $E$. $\angle EDG = \angle ABD = 60^\circ$ (so $E$ is above $C$), and $\angle DGE = \angle BGA = 60^\circ$, thus $\triangle DEG$ is equilateral, so $DE = EG$. According to the condition, $\angle GDC = \angle BDC = 30^\circ$, so $DC$ is the perpendicular bisector of segment $GE$.

![](https://cdn.mathpix.com/cropped/2024_05_02_84753f4460507e91dd1cg-1.jpg?height=729&width=588&top_left_y=346&top_left_x=758)

In $\triangle ABC$, $\angle BCA = 180^\circ - (50^\circ + 60^\circ + 20^\circ) = 50^\circ$, so $\triangle ABC$ is isosceles with $BC = BA$, which is equal to $BG$ since $\triangle ABG$ is equilateral. Therefore, in $\triangle GBC$, $BG = BC$ and $\angle BCG = \angle BGC = \frac{180^\circ - 20^\circ}{2} = 80^\circ$, so $\angle ACG = \angle BCG - \angle BCA = 80^\circ - 50^\circ = 30^\circ$.

In $\triangle GCE$, the perpendicular bisector of side $GE$ passes through the opposite vertex, so $CG = CE$. $\angle EGC = 180^\circ - (60^\circ + 80^\circ) = 40^\circ$, so $\angle ECG = 180^\circ - 2 \cdot 40^\circ = 100^\circ$, which implies $\angle ECG + \angle BCG = 100^\circ + 80^\circ = 180^\circ$, so points $E$, $C$, and $B$ are collinear.

Therefore, quadrilateral $ABED$ is an isosceles trapezoid, with equal base angles, so $\angle DBA = \angle ABC = 80^\circ$, which implies that $\angle DAC$ is indeed $30^\circ$.