ABCD is a cyclic quadrilateral in which BC = CD and EF is a tangent at A. ∠CBD = 43° and ∠ADB = 62°. Find :

(a) In △ BDC,

⇒ BC = CD (Equal sides)

⇒ ∠BDC = ∠DBC = 43° (Angles opposite to equal sides are equal)

From figure,

⇒ ∠ADC = ∠ADB + ∠BDC = 62° + 43° = 105°.

Hence, ∠ADC = 105°.

(b) We know that,

Opposite angles of a cyclic quadrilateral are supplementary.

⇒ ∠ABC + ∠ADC = 180°

⇒ ∠ABC + 105° = 180°

⇒ ∠ABC = 180° - 105° = 75°.

From figure,

⇒ ∠ABD = ∠ABC - ∠DBC = 75° - 43° = 32°.

Hence, ∠ABD = 32°.

(c) From figure,

⇒ ∠FAD = ∠ABD = 32°. (Angles in alternate segment are equal)

Hence, ∠FAD = 32°.