Mathematics
In the figure (ii) given below, PQ is a tangent to the circle with centre O and AB is a diameter of the circle. If QA is parallel to PO, prove that PB is tangent to the circle.
Circles
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Answer
Join OQ as shown in the figure below:
In △OAQ,
OA = OQ (Radius of the same circle.)
∠OAQ = ∠OQA.
Given QA || PO
∴ ∠OAQ = ∠POB (∵ corresponding angles are equal.)
and ∠OQA = ∠QOP (∵ alternate angles are equal.)
But ∠OAQ = ∠OQA,
∴ ∠POB = ∠QOP
Now in △OPQ and △OBP
OP = OP (Common sides)
OQ = OB (Radius of the same circle.)
∠QOP = ∠POB
∴ △OPQ ≅ △OBP (S.A.S. axiom of congruency)
As corresponding parts of congruent triangles are congruent,
∴ ∠OQP = ∠OBP
But ∠OQP = 90°
∴ ∠OBP = 90°
∴ PB is the tangent of the circle.
Hence, proved that PB is the tangent of the circle.
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