Question

AB and CD are two parallel lines. PQ cuts AB and CD at E and F respectively. EL is the bisector of $\angle FEB$. If $\angle LEB={35}^0$, then $\angle CFQ$ will be

Answer 1

AB || CD and PQ is the transversal EL is the bisector of $\angle FEB$ and $\angle LEB={35}^0$

$\therefore \angle FEB=2\times {35}^0={70}^0$
$\because AB||CD$
$\therefore \angle FEB+\angle EFD={180}^0$
(Co-interior angles)
${70}^0+\angle EFD={180}^0$
$\therefore \angle EFD={180}^0-{70}^0={110}^0$
But $\angle CFQ=\angle EFD$
(Vertically opposite angles)
$\therefore \angle CFQ={110}^0$