# Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

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by TPS

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by TPS

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therefore OQ is longer than OP so it is true that OP is shortest of all points.so our assumption is wrong .

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CONSTRUCTION : Take a point B , other than A ,on the tangent L .Join OB .

Suppose OB meets the circle in C.

PROOF : We know that , among all line segment joining the point O to a point on L , the perpendicular is shortest to L .

OA = OC ( Radius of same circle)

Now OB = OC + BC

Therefore OB greater than OC

⇒ OB greater than OA

⇒ OA is shorter than OB

B is an arbitrary point on the tangent L. Thus OA is shorter than any other line segment joining O to any point on L . Hence here OA is perpendicular to L.