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

Let there be a circle with centre $O$ and radius $r$. Let $P$ be any point on the circle. Let $PT$ be the tangent to the circle at point $P$.

To Prove: $OP \perp PT$ (i.e., the radius $OP$ is perpendicular to the tangent $PT$ at the point of contact $P$).

We use proof by contradiction (indirect proof). Assume $OP$ is NOT perpendicular to $PT$.

Then there exists some other line $OQ$ drawn from $O$ to the tangent $PT$ such that $OQ \perp PT$ (where $Q$ is the foot of perpendicular from $O$ to $PT$, and $Q \neq P$).

In right triangle $OQP$ (with right angle at $Q$):

(In any right triangle, the hypotenuse is the longest side, so $OP$ is the hypotenuse and $OQ$ is a leg — $OQ < OP$.)

Since $OP = r$ (radius), we have $OQ < r$.

Since $Q$ lies on the tangent line $PT$ and $OQ < r$, the point $Q$ lies inside the circle.

But $PT$ is a tangent to the circle — it touches the circle at exactly one point ($P$) and no other point of $PT$ lies inside or on the circle.

However, $Q$ is a point on $PT$ and it lies inside the circle — this is a contradiction.

Our assumption (that $OP$ is not perpendicular to $PT$) must be false.

Therefore, $OP \perp PT$. $\square$