Tangential and cyclic quadrilaterals

$\triangle BCD$ and $\triangle BAD$ are right triangles according to the Thales' Theorem .

$BD$ bisects $\angle ABC$ since the center of the inscribed circle lies on it, hence, $\triangle ABD \cong \triangle BCD$ $(A.S.A)$ , so $AB=BC$ and $\angle CBD = \angle ABD$ .

Since $BCEF$ is a tangential quadrilateral and applying Pitot's Theorem , we have

$BC+EF=CE+BF$

$BC-BF=CE-EF=FA=x=10-7=$ $\color{plum}\large \boxed{3}$

This problem is an interesting one