Question

prove that the bisectors of the angle form by producing the opposite sides of cyclic quadrilateral intersect at right angle

Answer 1

A cyclic quadrilateral ABCD whose opposite sides when produced intersect at the [points P and Q]. The bisectors PF and QF of ∠P and ∠Q respectively at F
produce QF to meet AB at E
In ΔQDG and ΔQBE
∠1 = ∠2 (QF is bisector of ∠DQC)
∠3 = ∠4
Exterior angles of cyclic quadrilateral = its interior opposite angles
∴ ∠5 = ∠6 .........(i)
But ∠5 = ∠7 (Vertically opposite angles) .........(ii)
∴ ∠6 = ∠7 (using (i) and (ii))
Now in ΔPGF and ΔPEF
   ∠9 = ∠8
and ∠7 = ∠6
∴ ∠PFG = ∠PFE
But ∠PFG + ∠PFE = 180°
∴ ∠PFG = ∠PFE = 90°
So ∠PFG = 90°    

Answer 2

n ΔQDG and ΔQBE
∠1 = ∠2 (QF is bisector of ∠DQC)
∠3 = ∠4
Exterior angles of cyclic quadrilateral = its interior opposite angles
∴ ∠5 = ∠6 .........(i)
But ∠5 = ∠7 (Vertically opposite angles) .........(ii)
∴ ∠6 = ∠7 (using (i) and (ii))
Now in ΔPGF and ΔPEF
∠9 = ∠8
and ∠7 = ∠6
∴ ∠PFG = ∠PFE
But ∠PFG + ∠PFE = 180°
∴ ∠PFG = ∠PFE = 90°
So ∠PFG = 90°