Q) Two tangents PA and PB are drawn to a circle with centre O from an external point P. Prove that ∠APB= 2(∠OAB).

Q26 B – Sample Question Paper – Set 1 – Maths Standard – CBSE 2026

Ans:

Step 1: Let’s make a diagram for better understanding of the question:

Step 2:

Next, we compare Δ OAQ and Δ OBQ:

Since OA and OB are radii of same circle,

∴ OA = OB

AQ = BQ        (since OP is perpendicular bisector to AB)

OQ = OQ       (being common arm)

∴ Δ OAQ ≅ Δ OBQ

∴ ∠ OAQ = ∠ OBQ              (by CPCT) …………(i)

Step 3: Next, we take Δ OAP:

In Δ OAP,  ∠ OAP + ∠ APO + ∠ POA = 180 ⁰

∴ 90 + ∠ APO + ∠ POA = 180 ⁰     (since tangent drawn on a circle is perpendicular to radius)

∴ ∠ APO + ∠ POA = 90 ⁰   ………………. (ii)

Step 4: In Δ OAQ,  ∠ OAQ + ∠ AQO + ∠ QOA = 180 ⁰

∴ ∠ OAQ + 90 + ∠ QOA = 180 ⁰     (since cord AB is perpendicular to OP)

∴ ∠ OAQ + ∠ QOA = 90 ⁰ ………………. (iii)

Step 5: By comparing equation (ii) and equation (iii), we get:

∠ APO + ∠ POA = ∠ OAQ + ∠ QOA

∵ ∠ POA and ∠ QOA are name of same angles

∴ ∠ APO = ∠ OAQ   …………. (iv)

Similarly, we can prove that ∠ BPO = ∠ OBQ  ………….. (v)

Step 6: By adding equations (iv) and equation (v), we get:

∠ APO + ∠ BPO = ∠ OAQ + ∠ OBQ

∴ ∠ APB = ∠ OAQ + ∠ OBQ

From equation (i), we have ∠ OAQ = ∠ OBQ

∴ ∠ APB = 2 ∠ OAQ

∵ ∠ OAQ and ∠ OAB are name of same angles

∴ ∠ APB = 2 ∠ OAB  

Hence Proved!

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