In the following figure, AD is the diameter of the circle with centre O. Chords AB, BC and CD are equal. If ∠DEF = 110°, calculate :

(i) ∠AEF, (ii) ∠FAB.

Join AE, OB and OC.

(i) As AOD is the diameter

∠AED = 90° [Angle in a semi-circle is a right angle]

But, given ∠DEF = 110°

So,

∠AEF = ∠DEF - ∠AED = 110° - 90° = 20°.

Hence, ∠AEF = 20°.

(ii) Also given, Chord AB = Chord BC = Chord CD

So,

∠AOB = ∠BOC = ∠COD [Equal chords subtends equal angles at the centre]

From figure,

⇒ ∠AOB + ∠BOC + ∠COD = 180° [AOD is a straight line]

⇒ ∠AOB = ∠BOC = ∠COD = $\dfrac{180°}{3}$ = 60°

Now, in ∆OAB we have

OA = OB [Radii of same circle]

So, ∠OAB = ∠OBA [Angles opposite to equal sides are equal]

In ∆OAB,

⇒ ∠OAB + ∠OBA + ∠AOB = 180° [By angle sum property of triangle]

⇒ ∠OAB + ∠OBA + 60° = 180°

⇒ ∠OAB + ∠OBA = 180° - 60° = 120°.

Since, ∠OAB = ∠OBA

∴ ∠OAB = ∠OBA = $\dfrac{120°}{2}$ = 60°.

Now, in cyclic quadrilateral ADEF

⇒ ∠DEF + ∠DAF = 180° [As sum of opposite angles in cyclic quadrilateral = 180°]

⇒ ∠DAF = 180° - ∠DEF

⇒ ∠DAF = 180° - 110° = 70°.

From figure,

∠FAB = ∠DAF + ∠OAB = 70° + 60° = 130°.

Hence, ∠FAB = 130°.