In the following figure, ABCD is a cyclic quadrilateral in which AD is parallel to BC. If the bisector of angle A meets BC at point E and the given circle at point F, prove that :

(i) EF = FC

(ii) BF = DF

AF is the angle bisector of ∠A,

∴ ∠BAF = ∠DAF

Also,

∠DAE = ∠BAE …….(1)

and

∠DAE = ∠AEB ……..(2) [Alternate angles are equal]

From (1) and (2) we get,

∠BAE = ∠AEB

(i) In △ABE,

⇒ ∠BAE + ∠AEB + ∠ABE = 180°

⇒ ∠AEB + ∠AEB + ∠ABE = 180°

⇒ 2∠AEB + ∠ABE = 180°

⇒ ∠ABE = 180° - 2∠AEB

Since, ABCD is a cyclic quadrilateral and sum of opposite angles in a cyclic quadrilateral = 180°.

∴ ∠ABC + ∠ADC = 180°

⇒ ∠ABE + ∠ADC = 180°

⇒ 180° - 2∠AEB + ∠ADC = 180°

⇒ ∠ADC = 180° - 180° + 2∠AEB

⇒ ∠ADC = 2∠AEB ………(3)

Since, ADCF is also a cyclic quadrilateral,

∴ ∠AFC + ∠ADC = 180°

⇒ ∠AFC = 180° - ∠ADC

⇒ ∠AFC = 180° - 2∠AEB [From (3)]

In △ECF,

⇒ ∠EFC + ∠ECF + ∠FEC = 180°

⇒ ∠AFC + ∠ECF + ∠FEC = 180° [From figure, ∠EFC = ∠AFC]

⇒ ∠AFC + ∠ECF + ∠AEB = 180° [∠FEC = ∠AEB (Vertically opposite angles are equal)]

⇒ ∠ECF = 180° - (∠AFC + ∠AEB)

⇒ ∠ECF = 180° - (180° - 2∠AEB + ∠AEB)

⇒ ∠ECF = ∠AEB

⇒ ∠ECF = ∠FEC

∴ EF = FC [As sides opposite to equal angles are equal.]

Hence, proved that EF = FC.

(ii) AF is the angle bisector of ∠A,

∴ ∠BAF = ∠DAF

∴ arc BF = arc DF [As equal arcs subtends equal angles]

∴ BF = DF [Equal arcs have equal chords]

Hence, proved that BF = DF.