Question

In the given figure, $AB\parallel CD$ and a transversal $t$ cuts them at $E$ and $F$ respectively. If $EP$ and $FQ$ are the bisectors of $\angle AEF$ and $\angle EFD$ respectively, prove that $EP\parallel FQ$

Answer 1

We know that $AB\parallel CD$ and $t$ is a transversal

from the figure we know that $\angle AEF$ and $\angle EFD$ are alternate angles

So we get

$\angle AEF=\angle EFD$

Dividing both side by 2 we get

$\frac{1}{2}\angle AEF=\frac{1}{2}\angle EFD$

So we get

$\angle PEF=\angle EFQ$

The alternate interior angles are formed only when the transversal $EF$ cuts both $FQ$ and $EP$

Therefore it is proved that $EP\parallel FQ$