Angle $x$

We note that in $\triangle ACD$ , the two sides about $\angle A$ has a ratio of $AD:AC = 4:6=2:3$ . In $\triangle ABC$ , the two sides about $\angle A$ also has ratio of $AC:AB = 6:9 = 2:3$ . Therefore, the two triangles are similar, and $\angle ABC = \angle ACD = 35^\circ$ . Since $\triangle BDE$ is isosceles, $\angle DEB = DBE = 35^\circ$ and $\angle EDB = 180^\circ - 2\times 35^\circ = \boxed{110}^\circ$ .

$\angle {BED}=\angle {EBD}=35\degree\implies \angle {EDB}=180\degree-2\times 35\degree=\boxed {110\degree}$ .