For reference: In the figure shown P and Q are points of tangency and $ \overset{\LARGE{\frown}}{BC}+\overset{\LARGE{\frown}}{FE}=130^o. $
Calculate "x".
My progress:
I think the starting point would be the interior angle theorem
$ \measuredangle BIC = \frac{\overset{\LARGE{\frown}}{BC}+\overset{\LARGE{\frown}}{FE}}{2}=\frac{130}{2}=65^\circ \implies \measuredangle BIF = 115^\circ \\
\triangle BDF(isosceles) \\
\triangle ACE(isoscels)\\
\measuredangle IBF=\measuredangle IFB = \measuredangle ICE = \measuredangle IEC=\frac{65^o}{2}$
Best Answer
$\angle BAC + \angle FAE = 65^0$
$\angle PAQ = \frac{\angle POQ}{2} = 90^\circ - \frac{x}{2}$
$\angle BAF + \angle BDF = 180^\circ$
$\implies 65^\circ + 90^\circ - \frac{x}{2} + x = 180^\circ$
$\implies x = 50^\circ$