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.

Answer 1

Join PB.

i. In cyclic quadrilateral PBCQ,

∠BPQ + ∠BCQ = 180°

${\displaystyle \Rightarrow }$ ∠BPQ + 140° = 180°

${\displaystyle \Rightarrow }$ ∠BPQ = 40°  ...(1)

Now in ΔPBQ,

∠PBQ  + ∠BPQ + ∠BQP = 180°

${\displaystyle \Rightarrow }$ 90° + 40° + ∠BQP = 180°

${\displaystyle \Rightarrow }$ ∠BQP = 50°

In cyclic quadrilateral PQBA,

∠PQB + ∠PAB = 180°

${\displaystyle \Rightarrow }$ 50° + ∠PAB = 180°

${\displaystyle \Rightarrow }$ ∠PAB = 130°

ii. Now in ΔPAB,

∠PAB + ∠APB + ∠ABP = 180°

${\displaystyle \Rightarrow }$ 130° + ∠APB  + ∠ABP = 180°

${\displaystyle \Rightarrow }$ ∠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 ∠AOQ at 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°

${\displaystyle \Rightarrow }$ ∠OAQ + ∠OAQ + 130° = 180°

${\displaystyle \Rightarrow }$ 2∠OAQ  = 50°

${\displaystyle \Rightarrow }$ ∠OAQ = 25°

∴ ∠OAQ = ∠AQB

But these are alternate angles.

Hence, AO is parallel to BQ.