In the figure (i) given below, ∠CBP = 40°, ∠CPB = q° and ∠DAB = p°. Obtain an equation connecting p and q. If AC and BD meet at Q so that ∠AQD = 2q° and the points C, P, B and Q are concyclic, find the values of p and q.

From figure,

∠ADC = ∠CBP = 40°. (∵ angles in alternate segments are equal.)

Since sum of angles in a triangle = 180°.

In △ADP,

∠DAP + ∠APD + ∠ADP = 180°

From figure, ∠ADP = ∠ADC = 40°.

⇒ p° + q° + 40° = 180°
⇒ p° + q° = 180° - 40°
⇒ p° + q° = 140° …..(i)

Join AC and BD as shown in the figure below:

∠CQB = ∠AQD = 2q° (∵ vertically opposite angles are equal.)

Given C, P, B, Q are concyclic.

∴ ∠CPB + ∠CQB = 180°
⇒ q° + 2q° = 180°
⇒ 3q° = 180°
⇒ q° = 60°.

Putting value of q in equation (i) we get,

⇒ p° + 60° = 140°
⇒ p° = 140° - 60° = 80°.

Hence, the value of p = 80 and q = 60 and the relation between p and q is p + q = 140.