Question

In Figure, $BA||ED$ and $BC||EF$. Show that $\angle ABC+\angle DEF={180}^o$

Answer 1

Given$BA\parallel EDandBC\parallel EF$

We have to show $\angle ABC+\angle DEF={180}^0$

Construct a ray $EA$ opposite $EF$

$\therefore$ $\angle EFD=\angle EPB$ (corresponding angle)

$\angle ABC=\angle EPB$

$BC\parallel EF$ or $PB\parallel ED$

$\therefore$ $\angle DEF=\angle EPB$ (corresponding angle)

$\therefore$ $\angle ABC+\angle DEF={180}^0$.