# Two parallel tangents of a circle meet a third tangent at points P and Q.Prove that PQ subtends right angle at the centre..

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by Pruthvi11

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by Pruthvi11

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=> CO = CO (common)

=> angle OPC = angle OBC = 90 (both are angles formed by radius-tangent intersection)

=> OP = OB (radii of same circle)

Therefore, triangle OPC is congruent to triangle OBC, by RightAngle-Hypotenuse-Side test.

=> angle PCO = angle BCO (CPCT) ...(1)

Similarly prove triangles AQO and ABO congurent.

=> angle OAQ = angle OAB (CPCT) ...(2)

In quadrilateral PCAQ,

angle OPC = angle OQA = 90

angles PCA+CAQ+AQP+QPC = 360 (sum of interior angles)

=> (PCO+BCO)+(OAB+OAQ)+90+90 = 360

=> 2BCO+2OAB+180 = 360 (From 1 and 2)

=> 2(BCO+OAB) = 360-180

=> BCO+OAB = 180/2 = 90 ...(3)

Now, in triangle AOC,

angles AOC+BCO+OAB = 180

=> AOC+90 = 180 (From 3)

=> AOC = 180-90 = 90