Right Angle to Tangent of Circle goes through Center

Theorem

In the words of Euclid:

If a straight line touch a circle, and from the point of contact a straight line be drawn at right angles to the tangent, the centre will be on the straight line so drawn.

(The Elements: Book $\text{III}$: Proposition $19$)

Proof

Let $DE$ touch the circle $ABC$, and let $AC$ be drawn at right angles to $DE$.

Suppose the center is not on $AC$.

Let $F$ be the center of $ABC$, and let $FC$ be joined from there to the point of contact.

By Radius at Right Angle to Tangent, $FC$ is perpendicular to $DE$.

Therefore $\angle FCE$ is a right angle.

But by hypothesis $\angle ACE$ is also a right angle.

Thus $\angle FCE = \angle ACE$, which is impossible unless $F$ lies on $AC$.

Therefore the center of $ABC$ lies on $AC$.

$\blacksquare$

Historical Note

This proof is Proposition $19$ of Book $\text{III}$ of Euclid's The Elements.

Sources

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