I'm having some difficulty to solve this sum....i'm nt able to form the diagram nor solve d question... please help.....

Prove that the sum of the angls formed in the four segments exterior to a cyclic quadrilateral by the sides is equal to six right angles.

Let ABCD be the given cyclic quadrilateral.

∠P, ∠Q, ∠R and ∠S are in the four external segments.

To Prove: ∠P + ∠Q + ∠R + ∠S = 6 right angles.

Construction: Join SB and SC.

It can be clearly seen that, APBS is a cyclic quadrilateral.

∴ ∠ASB + ∠P = 180°  ............ (1)

Similarly, BQCS and CRDS are cyclic quadrilaterals.

∴ ∠CSB + ∠Q = 180°  ............ (2)

and, ∠CSD + ∠R = 180°  ............ (3)

Adding (1), (2) and (3),

∠ASB  + ∠CSB + ∠P + ∠Q + ∠CSD + ∠R = 180° + 180° + 180°

⇒ ∠P + ∠Q + ∠R + (∠ASB + ∠CSB + ∠CSD) = 3 × 180°

⇒ ∠P + ∠Q + ∠R + ∠S = 3 × 2 × 90°

⇒ ∠P + ∠Q + ∠R + ∠S = 6 × 90°

⇒ ∠P + ∠Q + ∠R + ∠S = 6 right angles.