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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.

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, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

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.

APBQ=PCCQ\therefore \dfrac{AP}{BQ} = \dfrac{PC}{CQ}

Hence, proved that AP : BQ = PC : CQ

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