Question

A cyclic quadrilateral ABCD whose opposite sides when produced intersect at the [points P and Q]. The bisectors PF and QF of $\angle P$ and $\angle Q$ respectively at F
produce QF to meet AB at E. Prove that $\angle PFG={90}^{\circ }$.

Answer 1

A cyclic quadrilateral ABCD whose opposite sides when produced intersect at the [points P and Q]. The bisectors PF and QF of $\angle P$ and $\angle Q$ respectively at F
produce QF to meet AB at E
In $\Delta QDGand\Delta QBE$
$\angle 1=\angle 2$ (QF is bisector of \angle DQC)
$\angle 3=\angle 4$
Exterior angles of cyclic quadrilateral = its interior opposite angles
$\therefore \angle 5=\angle 6$ .........(i)
But $\angle 5=\angle 7$ (Vertically opposite angles) .........(ii)
$\therefore \angle 6=\angle 7$ (using (i) and (ii))
Now in $\Delta PGFand\Delta PEF$
$\angle 9=\angle 8$
$and\angle 7=\angle 6$
$\therefore \angle PFG=\angle PFE$
But $\angle PFG+\angle PFE={180}^{\circ }$
$\therefore \angle PFG=\angle PFE={90}^{\circ }$
So $\angle PFG={90}^{\circ }$