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:

1. EF = FC
2. BF = DF

Answer 1

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 –

1. EF = FC
2. 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

∴ EF = FC

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

`=>` BF = DF    ...[Equal arcs have equal chords]

Hence proved.