In the given figure, AB || CD || EF. Find the value of x.

Since $AB \parallel CD$ and $BC$ is a transversal.

So, $ \angle \text{ABC} = \angle \text{BCD} $ [alternate interior angles]

$ \Rightarrow 70^\circ = x + \angle \text{ECD} \, ....(\text{i}) $

Now, $CD \parallel EF$ and $CE$ is transversal.

So, $ \angle \text{ECD} + \angle \text{CEF} = 180^\circ $ [sum of consecutive interior angles is $180^\circ$]

$ \therefore \angle \text{ECD} + 130^\circ = 180^\circ $

$ \Rightarrow \angle \text{ECD} = 180^\circ - 130^\circ = 50^\circ $

Putting $ \angle \text{ECD} = 50^\circ $ in (i) we get,

$ 70^\circ = x^\circ + 50^\circ $

$ \Rightarrow x = 70 - 50 = 20 $