**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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