Question

In figure show that AB || EF.

Answer 1

We have,
$\angle BCD=\angle BCE+\angle ECD$
$={36}^{\circ }+{30}^{\circ }={66}^{\circ }$
$\therefore \angle ABC=\angle BCD$
Thus, lines AB and CD are intersected by the line BC such that $\angle ABC=\angle BCD$ i.e. the alternate angles are equal. Therefore,
AB || CD ....(i)
Now, $\angle ECD+\angle CEF={30}^{\circ }+{150}^{\circ }={180}^{\circ }$
This shows that the sum of the interior angles on the same side of the transversal CE is ${180}^{\circ }$ i.e. they are supplementary.
$\therefore$ EF || CD ...(ii)
From (i) and (ii), we have
AB || CD and CD || EF $\Rightarrow$ AB || EF.
Hence, AB || EF.