Find $x^\circ$

You can easily get $|\overline {AC}|=2|\overline {AD}|\cos 35\degree$ . Apply sine rule to $\triangle {BAD}$ to obtain $|\overline {AB}|=|\overline {AD}|\times \dfrac{\sin \angle {ADB}}{\sin \angle {ABD}}$ , with $\angle {ADB}=60\degree+x, \angle {ABD}=55\degree-x$ . Applying sine rule to $\triangle {ABC}$ you get $|\overline {AB}|=|\overline {AC}|\times \dfrac{\sin 95\degree}{\sin 55\degree}$ . Comparing the two expressions for $|\overline {AB}|$ you can get $55\degree-x=30\degree$ .

Because $\angle {DBC}=25\degree\implies \angle {BDC}=50\degree-25\degree=25\degree$ .