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Prove that the tangent at any point of a circle is perpendicular to the radius through the point of the contact.
Given: A tangent to a circle.
To do: To prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Solution:
![](/assets/questions/media/148618-32647-1606140540.png)
Take a point B, other than point A on the tangent P.
Join OB.
Suppose OB meets the circle on the point C.
Proof: We know that, among all line segment joining the point 0 to a point on P,
The perpendicular is shortest to P
$OA = OC\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( Radius\ of\ the\ same\ circle)$
Now, $OB\ =\ OC\ +\ BC$
$\Rightarrow \ OB\ >\ OC$
$\Rightarrow OB\ >\ OA$
$\Rightarrow OA\
B is an arbitrary point on the tangent P.
Thus, OA is shorter than any other line segment joining O to any point on P Here,
$OA\ \perp P$
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