Given : ABCD is a cyclic quadrilateral. Its opposite sides DA and CB are produced to meet at P and opposite sides AB and DC are produced to meet at Q. The bisectors of ∠P and ∠Q meet is F.
To Prove : ∠PFQ = 900.
Construction : Produce PF to meet DC is G.
Proof : In ΔPEB, ∠5 = ∠2 + ∠6 .....(i)
[∵ Exterior angle of a triangle is equal to the sum of interior opposite angles]
But ∠2 = ∠1 And,
∠6 = ∠D [∵ In a cyclic quadrilateral, exterior angle = interior opposite angle]
∴ ∠5 = ∠1 + ∠D ....(ii) [From (i)] Now in ΔPDG, ∠7 = ∠1 + ∠D ...(iii) [∵ Exterior angle of a triangle is equal to the sum of interior opposite angles]
Frim (ii) and (iii), we have
∠5 = ∠7
Now, in ΔQEF and ΔQGF, [Proved above]
∠5 = ∠7 [Common side]
QF = QF [Given]
∠3 = ∠4 [AAS criterion]
∴ ΔQEF ≡ ΔQGE [By cpctc]
∴ ∠8 = ∠9 But ∠8 + ∠9 = 1800
∴ ∠8 = ∠9 = 900[Linear Pair Axiom]
∴∠PFQ = 900
