+833 In pentagon ABCDE, \overline{AC} bisects \angle BCE and $\overline{AD}$ bisects $\angle BDE.$ If $\angle CBD = 30^\circ$ and $\angle CED = 60^\circ,$ then find $\angle CAD,$ in degrees.
+15090 \(\color{blac}Find\ \angle CAD\ in\ degrees.\\\)
Thanks webbinsight for the useful suggestions to solve this tricky task!
\(\angle ADB =\angle ADE=\alpha\\ \angle ACB=\angle ACE=\beta\)
The angles intersection AD/EC are twice:
\((120°-\alpha )\ and\ (60°+\alpha )\)
The angles intersection AC/BD are twice:
\((30°+\beta)\ and\ (150°-\beta)\\ \angle CAD=60°+\alpha -\beta\)
In the triangle (intersection AC/BD) - A-D the following applies:
\(\alpha =180°-(150°-\beta )-(60°+\alpha -\beta)\\ \color{blue}\beta = 15°\)
The angles intersection BD/CE are twice:
\((180°-30°-2\beta=120°)\ and\ (60°)\)
In the triangle (intersection AD/CE) - (intersection CE/BD) - D the following applies:
\((120°)+(60°+\alpha )+(\alpha )=180°\\ \color{blue} \alpha =0\ ?\ ?\ ?\)
I ask everyone for help.
!
