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.
Best Answer
\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 $\angle ABD=2\,\delta$.