In the figure (ii) given below, sides AB and DC of a cyclic quadrilateral are produced to meet at a point P and the sides AD and BC produced to meet at a point Q. If ∠ADC = 75° and ∠BPC = 50°, find ∠BAD and ∠CQD.

Since sum of angles in a triangle = 180°.

In △ADP,

⇒ ∠ADP + ∠DAP + ∠DPA = 180°
⇒ 75° + ∠DAP + 50° = 180°
⇒ ∠DAP + 125° = 180°
⇒ ∠DAP = 180° - 125° = 55°.

From figure,

∠BAD = ∠DAP = 55°.

In cyclic quadrilateral sum of opposite angles = 180°

In ABCD,

∴ ∠ADC + ∠CBA = 180°
⇒ 75° + ∠CBA = 180°
⇒ ∠CBA = 180° - 75° = 105°.

In △ABQ,

⇒ ∠ABQ + ∠BAQ + ∠AQB = 180°

From figure, ∠BAQ = ∠BAD and ∠ABQ = ∠CBA or,

⇒ 105° + 55° + ∠AQB = 180°
⇒ ∠AQB + 160° = 180°
⇒ ∠AQB = 180° - 160° = 20°.

From figure,

∠CQD = ∠AQB = 20°.

Hence, the value of ∠BAD = 55° and ∠CQD = 20°.