Question

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

Answer 1

Given: Line $l$ is a tangent to the circle with centre $O$ at the point of contact $A$.
To prove: line $l\perp$ radius $OA$.
Proof: Assume that, line $l$ is not perpendicular to seg $OA$.
Suppose, seg OB is drawn perpendicular to line $l$.
Of course, B is not same as A.
Now take a point $C$ on line $l$ such that $A-B-C$ and $BA=BC$.
Now in, $△OBC$ and $△OBA$
Seg $BC\cong \mathrm{seg}BA\dots$ (construction)
$\angle OBC\cong \angle OBA\dots$ (each right angle)
Seg $OB\cong \mathrm{seg}OB$
$\therefore △OBC\cong △OBA\dots$ (SAS test)
$\therefore OC=OA$
But seg $OA$ is a radius.
$\therefore$ seg OC must also be radius.
$\therefore$ C lies on the circle.
That means line $l$ intersects the circle in two distinct points
$A$ and $C$.
But line $l$ is a tangent. ... (given)
$\therefore$ it intersects the circle in only one point.
Our assumption that line $l$ is not perpendicular to radius $OA$ is wrong.
$\therefore$ line $l\perp$ radius OA.