What is the measure of $\measuredangle x$ in the figure below?

Question

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}$

Answer 1

$\angle BAC + \angle FAE = 65^0$

$\angle PAQ = \frac{\angle POQ}{2} = 90^\circ - \frac{x}{2}$

$\angle BAF + \angle BDF = 180^\circ$

$\implies 65^\circ + 90^\circ - \frac{x}{2} + x = 180^\circ$

$\implies x = 50^\circ$

Answer 2

Let $\measuredangle PQD=y$.

Thus, $$2y+x=180^{\circ}.$$

In another hand, since $ACDE$ is cyclic, we obtain: $$y+x+\measuredangle CDB+\measuredangle FDE=180^{\circ}$$ or
$$y+x+65^{\circ}=180^{\circ}.$$ Can you end it now?

I got $x=50^{\circ}.$