Question

In the given figure, $\angle ABC={65}^{\circ },\angle BCE={30}^{\circ },\angle DCE={35}^{\circ }$ and $\angle CEF={145}^{\circ }$. Prove that $AB\parallel EF$.

Answer 1

$\angle ABC={65}^{\circ }$

$\angle BCD=\angle BCE+\angle ECD={30}^{\circ }+{35}^{\circ }={65}^{\circ }$
$\angle ABC=\angle BCD$ (Altemate Angles)

$\to AB\parallel CD$ ...(1)

$\angle FEC+\angle ECD=l{45}^{\circ }+{35}^{\circ }={180}^{\circ }$

But these angles are consecutive interior angles formed on the same side of the transversal.

$\to CD\parallel EF$ ...(2)

From (1) and (2),

$AB\parallel EF$

Hence, proved.