Radius at Right Angle to Tangent

Theorem

If a straight line touch a circle, and a straight line joined from the center to the point of contact, the straight line so formed will be perpendicular to the tangent.

Proof

Let $DE$ be tangent to the circle $ABC$ at $C$.

Let $F$ be the center of $ABC$.

Let $FC$ be the radius in question.

Suppose $FC$ were not perpendicular to $DE$.

Instead, suppose $FG$ were drawn perpendicular to $DE$.

Since $\angle FGC$ is a right angle, then from Two Angles of Triangle Less than Two Right Angles it follows that $\angle FCG$ is acute.

From Greater Angle of Triangle Subtended by Greater Side it follows that $FC > FG$.

But $FC = FB$ and so $FB > FG$, which is impossible.

So $FG$ can not be perpendicular to $DE$.

Similarly it can be proved that no other straight line except $FC$ can be perpendicular to $DE$.

Therefore $FC$ is perpendicular to $DE$.

$\blacksquare$

Historical Note

This is Proposition 18 of Book III of Euclid's The Elements.