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

**Given :** A circle C (0,* r*) and a tangent *l* at point A.

**To prove :** OA ⊥ *l*

**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 the same circle)

Now, OB = OC + BC.

∴ OB > OC

⇒ OB > OA

⇒ OA < 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*.

Here, OA ⊥ *l*

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