Measure of the angle

Question

Let $ABCD$ an convex quadrilateral such that $m(\hat{ABD})=10^{\circ}$, $m(\hat{DBC})=20^{\circ}$, $m(\hat{BAC})=100^{\circ}$ and $m(\hat{CAD})=40^{\circ}$. Find the measure of the angle $\hat{CDB}$. I solved the problem using just the sine theorem but I want to solve using syntethic method. What is necessary to construct? Thank you!

Answer 1

Rotate $B$ around $A$ for $60^{\circ}$. We get new point $S$. Since $SA = SB =: r$ and $\angle ASB = 2\angle ADB$ we see that $D$ is on circle with center at $S$ and radius $r$, so we have also $DS =r$. Since $S$ is reflection of $A$ across $BC$ we have $CS = CA$ and $\angle SAC = \angle ASC = 40^{\circ}$.

Now triangle $ASD$ is isosceles ($SD=SA$) so $\angle ASD = 20^{\circ}$ and thus $BSCD$ is cyclic. This means that $$\angle BDC = 180^{\circ}-\angle BSC = 80 ^{\circ}$$