Question

In the figure, show that AB ||EF.

Answer 1

Given in figure, AB ||EF
$\angle BAC={57}^0,\angle ACE={22}^0$
$\angle ECD={35}^{0}and\angle ACE={22}^0$
To prove : AB||EF,
Proof : $\angle ECD+\angle CEF={35}^0+{145}^0={180}^0$
But these are co-interior angles
$\therefore EF||CD$
But AB||CD
$\therefore AB||EF$