Q) In the given figure, PQ is tangent to a circle centred at O and ∠BAQ = 30°; show that BP = BQ.

Ans: Here, We need to prove that BP = BQ,
it means we need to get ∠ BQP = ∠ BPQ

Step 1: Let’s start with given diagram.
Since AB is a straight line and passing thru the circle’s center O, hence AB is a diameter

∴ ∠ AQB = 90 …….. (i)
since it is given that ∠ QAB = 30,
∴ ∠ ABQ = 60 (sum of all angles in Δ ABQ is 180)

Since OP is a straight line,

∴ ∠ ABQ + ∠ QBP = 180
∴ ∠ QBP = 180 – 60 = 120

Step 2: Next, we look at Δ QBP,

∴ ∠ QBP + ∠ BPQ + ∠ PQB = 180
∴ 120 + ∠ BPQ + ∠ PQB = 180
∴ ∠ BPQ + ∠ PQB = 180 – 120
∴ ∠ BPQ + ∠ PQB = 60 ………. (ii)

Step 3: Next, let’s connect OQ: 

Now we look in Δ OQP,
Since OQ is a radius and PQ is tangent to the circle,
∴ ∠ OQP = 90 …………… (iii)

Step 4: Now we look in Δ AOQ,
Since OA and OQ are radii of same circle,
∴ OA = OQ
∴ ∠ AQO = ∠ QAO = 30 (given) …… (iv)

Step 5: From equation (i) we calculated:
∠ AQB = 90
∴ ∠ AQO + ∠ OQB = 90
∴ 30 + ∠ OQB = 90 [from equation (iv)]
∴ ∠ OQB = 60

Step 6: From equation (iii) we calculated:
∠ OQP = 90
∴ ∠ OQB + ∠ PQB = 90
∴ 60 + ∠ PQB = 90 (from above)
∴ ∠ PQB = 30

Step 7: From equation (ii) we calculated:
∴ ∠ BPQ + ∠ PQB = 60
∴ ∠ BPQ + 30 = 60
∴ ∠ BPQ = 60 – 30 = 30

Step 8: Now we get, ∠ BPQ = ∠ PQB
Therefore in Δ BPQ, opposite sides to equal angles will also be equal
∴ BP = BQ

Hence Proved !

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