In the given figure, $AB\parallel DE, \angle ABC=110°, \angle CDE=100°$then $\angle BCD$ is

From the figure

$AF\parallel DE,DF$ is transversal

$\angle EDF+\angle DFG=180°$ ($\therefore$cointerior angles)

$$\begin{gathered} 100°+\angle DFG=180° \\ \angle DFG=180°-100°=80° \end{gathered}$$

$\angle DFG=\angle BFC=80°$ ($\therefore$vertically opposite angles)

$\angle BFC+\angle BCF=\angle ABC$

($\therefore$Sum of interior angles of the triangle is equal to its exterior angle)

$$\begin{gathered} 80°+x°=110° \\ x°=110°-80°=30° \end{gathered}$$