#### Question

In the given figure, PAT is tangent to the circle with centre 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

#### Solution

i) Since PAT ∥ BC

∴ `∠`PAB = `∠`ABC (alternate angles) .........(i)

In cyclic quadrilateral ABCD,

Ext `∠`ADQ = `∠`ABC ………..(ii)

From (i) and (ii)

`∠`PAB = `∠`ADQ

ii) Arc AB subtends AOB at the centre and ADB at the remaining part of the circle.

∴ `∠`AOB = 2`∠`ADB

⇒ `∠`AOB = 2`∠`PAB (angles in alternate segments)

⇒ `∠`AOB = 2`∠`ADQ (proved in (i) part)

iii)

∴ `∠`BAP = `∠`ADB (angles in alternate segments)

But

`∠`BAP = `∠`ADQ (proved in (i) part)

∴ `∠`ADQ = `∠`ADB

Is there an error in this question or solution?

Solution In the Given Figure, Pat is Tangent to Circle with Centre O at Point a on Its Circumference and is Parallel to Chord Bc. If Cdq is a Line Segment (I) ∠Bap = ∠Adq (Ii) ∠Aob = 2∠Adq (Iii) ∠Adq = ∠Adb Concept: Cyclic Properties.