Right Angle to Tangent of Circle goes through Center

Proof

 * Euclid-III-19.png

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 $AC$.