#### Question

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

#### Solution

Given – ABCD is a cyclic quadrilateral in which AD || BC

Bisector of ∠A meets BC at E and the given circle at F. DF and BF are joined.

To prove –

(i) EF = FC

(ii) BF = DF

Proof – ABCD is a cyclic quadrilateral and AD ∥ BC

∵ AF is the bisector of ∠A, ∠BAF = ∠DAF

Also, ∠DAE = ∠BAE

∠DAE = ∠AEB [Alternate angles]

(i) In ΔABE, ∠ABE = 180° - 2∠AEB

∠CEF = ∠AEB [vertically opposite angles]

∠ADC = 180° - ABC = 180° - (180° - 2∠AEB)

∠ADC = 2∠AEB

∠AFC = 180° - ∠ADC

= 180° - 2∠AEB [since ADFC is a cyclic quadrilateral]

∠ECF = 180° - (∠AFC + ∠CEF)

= 180 - (180 - 2∠AEB + ∠AEB)

= ∠AEB

∴ EC = EF

(ii) ∴ Arc BF = Arc DF [Equal arcs subtends equal angles]

⇒ BF = DF [Equal arcs have equal chords]