#### Question

The given figure shows a semi-circle with centre O and diameter PQ. If PA = AB and ∠BCQ =140°; Find measures of angles PAB and AQB. Also, show that AO is parallel to BQ.

#### Solution

Join PB.

i) In cyclic quadrilateral PBCQ,

`∠`BPQ + `∠`BCQ = 180°

⇒ `∠`BPQ + 140° = 180°

⇒ `∠`BPQ = 40° ……. (1)

Now in ΔPBQ,

`∠`PBQ + BPQ + `∠`BQP = 180°

⇒ 90° + 40° + `∠`BQP = 180°

⇒ `∠`BQP = 50°

In cyclic quadrilateral PQBA,

`∠`PQB + `∠`PAB = 180°

⇒ 50° + `∠`PAB = 180°

⇒ `∠`PAB = 130°

ii) Now in ΔPAB,

`∠`PAB + `∠`APB + ABP = 180°

⇒ 130° + `∠`APB + `∠`ABP = 180°

⇒ `∠`APB + `∠`ABP = 50°

But

`∠`APB = `∠`ABP ( ∵ PA = PB)

∴ `∠`APB = `∠`ABP = 25°

`∠`BAQ = `∠`BPQ = 40°

`∠`APB = 25° = `∠`AQB (angles in the same segment)

∴ `∠`AQB = 25° ………. (2)

iii) Arc AQ subtends `∠`AOQat the centre and `∠`APQ at the remaining part of the circle.

We have,

`∠`APQ = `∠`APB + `∠`BPQ ……. (3)

From (1), (2) and (3), we have

`∠`APQ = 25° + 40° = 65°

∴ `∠`AOQ = 2`∠`APQ = 2 × 65° = 130°

Now in ΔAOQ,

`∠`OAQ = `∠`OQA = ( ∵ OA = OQ)

But

`∠` OAQ + `∠`OQA + `∠`AOQ = 180°

⇒ `∠`OAQ + `∠`OAQ +130° = 180°

⇒ 2 `∠`OAQ = 50°

⇒ `∠`OAQ = 25°

∴ `∠`OAQ = `∠`AQB

But these are alternate angles.

Hence, AO is parallel to BQ.