TITLE: Finding angle in a given triangle.
QUESTION [5 upvotes]: In the picture above:


$\overset{\Delta}{ACD}$ is a triangle.
$B$ is a point on $[CD]$.
$m(\widehat{ABC})=140^\circ$
$|AB|=|BC|$.
$|AC|=|BD|$.
What is $\color{red}{m(\widehat{ADB})}$?


There is probably a short answer, but i can't find it.
[Answer is $\color{red}{30^\circ}$.]

REPLY [3 votes]: Obviously, $\angle BAC = \angle BCA = 20^{\circ}$, and $\angle DBA = 40^{\circ}$.
Choose point $E$ such that $\triangle EAB$ forms an equilateral triangle.
Since $\angle DBA = 40^{\circ}$, and $\angle ABE = 60^{\circ}$, then $\angle DBE = 20^{\circ}$.
Now notice: $\overline {AC} = \overline {DB}$; $\angle ACB = \angle DBE = 20^{\circ}$; and $\overline {CB} = \overline {BE}$.  We have side-angle-side equivalence, so, if we draw in segment $DE$, we can say that $\triangle ACB \cong \triangle DBE$.
Hence, $\angle DEB = 140^{\circ}$.  Since $\angle AEB = 60^{\circ}$, then $\angle DEA = 80^{\circ}$.  Since side $\overline {DE} = \overline {EA}$ in $\triangle DEA$, then $\angle EDA = \angle EAD = 50^{\circ}$.
Now, $\angle EDA$ ($50^{\circ}$) - $\angle EDB$ ($20^{\circ}$) = $\angle CDA$ ($30^{\circ}$).