Mathematics
The figure given below shows two circles with centres A, B and a transverse common tangent to these circles meet the straight line AB in C. Prove that :
AP : BQ = PC : CQ.
Circles
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Answer
In △APC and △BQC
∠PCA = ∠QCB (∵ vertically opposite angles are equal)
∠APC = ∠BQC (∵ both are equal to 90 as radius and tangent to a circle at the point of contact are perpendicular to each other.)
∴ △APC ~ △BQC (By AA axiom of similarity)
Since triangles are similar hence the ratio of their corresponding sides are equal.
Hence, proved that AP : BQ = PC : CQ
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From a point outside a circle, with centre O, tangents PA and PB are drawn. Prove that
(i) ∠AOP = ∠BOP
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