Question

Angle Properties of a Circle 26. The adjoining figure shows a quadrilateral ABCD in which AD || BC inscribed in a circle. The bisector of ZBAD cuts BC at E and the circle at F. Prove that (a) EF = FC, and (b) BF =DF. E B​

Answer 1

step by step

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)

1) 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

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

⇒ BF = DF [Equal arcs have equal chords]

Answer

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Answer 2

Answer:

2)Therefore Arc BF = arc DF (equal arc subtends equal angles)

= BF = DF (equal arc subtends equal angles)