In the given figure, PAT is tangent to the circle with center O, at point A on its circumference and is parallel to chord BC. If CDQ is a line segment, show that :

(i) ∠BAP = ∠ADQ

(ii) ∠AOB = 2∠ADQ

(iii) ∠ADQ = ∠ADB.

(i) Since, PAT || BC

∠BAP = ∠ABC [Alternate angles are equal] ………(1)

In cyclic quadrilateral ABCD,

∠ABC + ∠ADC = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°] …………….(2)

From figure,

∠ADQ + ∠ADC = 180° [Linear pairs] ………(3)

From (2) and (3), we get :

∠ADQ = ∠ABC …………(4)

From (1) and (4), we get :

∠BAP = ∠ADQ.

Hence, proved that ∠BAP = ∠ADQ.

(ii) We know that,

Angle subtended by an arc at the center is twice the angle subtended at any other point of circumference.

Arc AB subtends ∠AOB at the center and ∠ADB at the remaining part of the circle.

⇒ ∠AOB = 2∠ADB …………(3)

From figure,

∠ADB = ∠PAB [Angles in alternate segment are equal.]

Substituting above value in (3), we get :

⇒ ∠AOB = 2∠PAB

∠PAB = ∠ADQ [Proved above]

⇒ ∠AOB = 2∠ADQ.

Hence, proved that ∠AOB = 2∠ADQ.

(iii) From figure,

∠BAP = ∠ADB [Angles in alternate segment are equal.] ………(4)

∠BAP = ∠ADQ [Proved in part (i)] ………..(5)

From (4) and (5), we get :

∠ADQ = ∠ADB.

Hence, proved that ∠ADQ = ∠ADB.