TITLE: In the following diagram of a triangle, AB = BC = CD and AD = BD. Find the measure of angle D.
QUESTION [5 upvotes]: In the following diagram of a triangle, $\overline{AB} = \overline{BC} = \overline{CD}$ and $\overline{AD} = \overline{BD}$. Find the measure of angle $D$.
I know this should be easy but I am stuck.  I started by saying angle $\widehat{ACB} = \theta$ and that the supplement angle $\widehat{BCD} = 180^\circ-\theta$.  I know that angle $\widehat{CAB}=\theta$ as well and that angle $\widehat{ABC} = 180^\circ-2\theta$.  In addition, angles $\widehat{CDB}$ and $\widehat{CBD}$ are equal.  I am not sure how to solve for angle $\widehat{CDB}$ ... is it possible to find an exact numerical measure?  I hate overlooking something obvious.  Thank you for your help.

REPLY [2 votes]: \begin{align} 
5\,\delta&=180^\circ
\end{align}  
Edir
\begin{align} 
|AB|&=|BC|=|CD|
,\\
|AD|&=|BD|
.
\end{align}  
Let $\angle BDA=\delta$.
Then, from isosceles $\triangle BDC$, 
$\angle CBD=\delta$, $\angle DCB=180^\circ-2\,\delta$.
In  $\triangle ABC$,
$\angle BCA=180^\circ-\angle DCB=2\,\delta$,
$\angle BAC=\angle BCA=2\,\delta$.
Also, 
$\angle BAC=\angle BAD=\angle ABD$,
hence 
$\angle ABD=2\,\delta$.