Solve this:

​Q27. Two tangents PQ and PR are drawn from an external point to a circle with centre O. Prove that QORP is a cyclic quadrilateral.

                                                   OR

   In the given figure, ABCD is a cyclic quadrilateral and PQ is tangent to the circle at C. If BD is a diameter, $\angle$DCO and $\angle$ABD = 60°, find $\angle$BCE.

                     

Dear Student,
 

∠DAB = 90°  (Angle in a semi-circle)

In ΔABD,

∠ABD  + ∠ADB + ∠DAB = 180°

⇒ ∠ADB = 180° – (60° + 90°)

⇒∠ADB = 180° – 150°

⇒∠ADB = 30°

∠DCQ = ∠CBD = 40°

(Alternate segment theorem)

∠BCD = 90°

In ΔBCD,

∠BCD + ∠BDC + ∠DBC = 180°

⇒ ∠BDC = 180° – (90° + 40°) = 50°

∠BDC = ∠BCP = 50° (Alternate segment theorem)

∴ (i) ∠ADB = 30°

(ii) ∠BCP = 50°

Hope this will help you and clear your doubt.
Regards,