Question

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

Answer 1

Given : A circle $C(0,r)$ and a tangent at point $A$.
To prove : $OA\perp l$
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, $OB=OC+BC$
Therefore, $OB>OC$
$OB>OA$
$OA<OB$
$B$ is an arbitrary point on the tangent $l$, thus, $OA$ is shorter than any other line segment. Thus, we can say that $OA\perp l$.
Hence, proved.