Radius at Right Angle to Tangent

Theorem

In the words of Euclid:

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

(The Elements: Book $\text{III}$: Proposition $18$)

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.

Aiming for a contradiction, 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 are 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 proof is Proposition $18$ of Book $\text{III}$ of Euclid's The Elements.
It is the converse of Proposition $19$: Right Angle to Tangent of Circle goes through Center.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{III}$. Propositions