How to find measurements of angle CBD? 
I got $x + CBD = 110^{o}$. I think I should find measurements of angle CBD, so how to find it? 
Do you have any ideas? Please help me! 




*

*$\angle ACB = 180^\circ-(10^\circ+70^\circ)-(60^\circ+20^\circ) = 20^\circ$ and $\angle AEB = 180^\circ-60^\circ-(50^\circ+30^\circ) = 40^\circ$.

*Draw a line from point $E$ parallel to $AB$, labeling the intersection with $AC$ as a new point $F$ and conclude $\Delta CEF\sim\Delta  ABC$,
\begin{align}
\angle CEF &= \angle CBA = 50^\circ+30^\circ = 80^\circ\\
\angle FEB &= 180^\circ-80^\circ = 100^\circ\\
\angle AEF &= 100^\circ-40^\circ = 60^\circ\\
\angle CFE &= CAB = 60^\circ+20^\circ = 80^\circ\\
\angle EFA &= 180^\circ-80^\circ = 100^\circ
\end{align}

*Draw a line $FB$ labeling the intersection with $AE$ as a new point $G$ and conclude $\Delta AFE\cong\Delta  BEF$, 
\begin{align}  
\angle AFB &=\angle BEA = 40^\circ\\
\angle BFE &= \angle AEF = 60^\circ\\
\angle FGE &= 180^\circ-60^\circ-60^\circ = 60^\circ = \angle AGB\\
\angle ABG &= 180^\circ-60^\circ-60^\circ = 60^\circ\\
\end{align}

*Draw a line $DG$. Since $AD=AB$ (leg of isosceles) and $AG=AB$ (leg of equilateral), conclude $AD = AG$, $\Delta DAG$ is isosceles and
$$
\angle ADG =\angle AGD = \frac{180^\circ-20^\circ}2 = 80^\circ.
$$

*Since $\angle DGF = 180^\circ-80^\circ-60^\circ = 40^\circ$, conclude $\Delta FDG$ (with two $40^\circ$ angles) is isosceles, so $DF = DG$.

*With $EF = EG$ (legs of equilateral) and $DE = DE$ (same line segment) conclude $\Delta DEF\cong\Delta  DEG$ by side-side-side rule, $\angle DEF =\angle DEG = x$, and
$$
\angle FEG = 60^\circ = x+x\quad\Rightarrow\quad \large\color{blue}{x=30^\circ}.
$$

