Let's label the diagram, AB and CD are parallel chords, and tangents from points A, B, C and D make a quadrilateral PQRS

Construction: Join OA, OB, OC and OD where O is radius

To Prove: PQRS is cyclic i.e. ∠P + ∠R = 180°

Proof:

We know, By alternate segment theorem

angle between chord and tangent is equal to the angle in the other segment.

Therefore,

∠1 = ∠3 and ∠2 = ∠3 [As AC is a chord]

⇒ ∠1 = ∠2 = ∠3

Also, in ΔACP, By angle sum property

∠1 + ∠2 + ∠P = 180°

⇒ ∠3 + ∠3 + ∠P = 180°

⇒ ∠P = 180° - 2 ∠3 …[1]

In Δ BIC, By angle sum property

∠3 + ∠5 + ∠CIB = 180°

⇒ ∠3 + ∠5 = 90° [∠CIB = 90°, Since AB ⊥ CD]

⇒ ∠3 = 90° - ∠5 …[2]

From [1] and [2]

⇒ ∠P = 180 - 2(90 - ∠5)

⇒ ∠P = 2 ∠5 …[A]

Also, Considering arc BD

⇒ ∠4 = 2 ∠5 …[3]

[The angle subtended by an arc at the center of the circle is double the angle subtended by arc at any point on circumference of the circle]

Also, In Quadrilateral OBRD, By angle sum property

∠4 + ∠OBD + ∠R + ∠ODR = 360°

[∠OBD = ∠ODR = 90, as tangent at any point on circle is perpendicular to the radius through point of contact]

⇒ ∠4 + 90° + 90° + ∠R = 360°

⇒ ∠R = 180° - ∠4

⇒ ∠R = 180° - 2 ∠5 …[B] [From 3]

Adding [A] and [B], we get

⇒ ∠P + ∠R = 180° - 2 ∠5 + 2 ∠5

⇒ ∠P + ∠R = 180°

Hence Proved.