In the following figure,

(i) if ∠BAD = 96°, find ∠BCD and ∠BFE.

(ii) Prove that AD is parallel to FE.

(i) ABCD is a cyclic quadrilateral.

So, ∠BAD + ∠BCD = 180° [As sum of opposite angles in a cyclic quadrilateral = 180°]

⇒ ∠BCD = 180° - ∠BAD

= 180° - 96°

= 84°.

As DCE is a straight line.

∴ ∠BCE = 180° - ∠BCD

= 180° - 84° = 96°.

BCEF is a cyclic quadrilateral,

So, ∠BCE + ∠BFE = 180°

⇒ ∠BFE = 180° - ∠BCE

= 180° - 96°

= 84°.

Hence, ∠BCD = 84° and ∠BFE = 84°.

(ii) Now, ∠BAD + ∠BFE = 96° + 84° = 180°.

But these two are interior angles on the same side of a pair of lines AD and FE.

Hence, proved that AD || FE.